THE 8051 MICROCONTROLLERSimple comparison: Pentium vs.
8051FEATURE8051PENTIUMCOMMENT
Clock Speed12Mhz. typical
but 60MHz. ICs available1,000 MHz. (1GHz.)8051 internally
divides clock by 12 so for 12MHz. clock effective clock rate is
just 1MHz.
Address bus16 bits32 bits8051 can address 216, or 64Kbytes of
memory.
Pentium can address 232, or
4 GigaBytes of memory.
Data bus8 bits64 bitsPentiums wide bus allows very fast data
transfers.
ALU width8 bits32 bitsBut - Pentium has multiple 32 bit ALUs
along with floating-point units.
ApplicationsDomestic appliances,
Peripherals, automotive etc.Personal Computers
And other high performance areas.
Power consumptionSmall fraction of a wattTens of wattsPentium
runs hot as power consumption increases with frequency.
Cost of chipAbout 2 Euros. In volumeAbout 200 Euros
Depending on spec.
The Intel 8051 is an 8-bit microcontroller which means that most
available operations are limited to 8 bits. There are 3 basic
"sizes" of the 8051: Short, Standard, and Extended. The Short and
Standard chips are often available in DIP form, but the Extended
8051 models often have a different form factor, and are not
"drop-in compatable". All these things are called 8051 because they
can all be programmed using 8051 assembly language, and they all
share certain features (although the different models all have
their own special features).
Some of the features that have made the 8051 popular are:
8-bit data bus
16-bit address bus
34 general purpose registers each of 8 bits
16 bit timers (usually 2, but may have more, or less).
3 internal and 2 external interrupts.
Bit as well as byte addressable RAM area of 16 bytes.
4 8-bit ports, (short models have 2 8-bit ports).
16-bit program counter and data pointer
As shown on the previous picture, the 8051 microcontroller has
nothing impressive at first sight:
4 Kb program memory is not much at all.
128Kb RAM (including SFRs as well) satisfies basic needs, but it
is not imposing amount.
4 ports having in total of 32 input/output lines are mostly
enough to make connection to peripheral environment and are not
luxury at all.
As it is shown on the previous picture, the 8051 microcontroller
have nothing impressive at first sight:
The whole configuration is obviously envisaged as such to
satisfy the needs of most programmers who work on development of
automation devices. One of advantages of this microcontroller is
that nothing is missing and nothing is too much. In other words, it
is created exactly in accordance to the average users taste and
needs. The other advantage is the way RAM is organized, the way
Central Processor Unit (CPU) operates and ports which maximally use
all recourses and enable further upgrading.
Exercise:
1. Compare and contrast microprocessors and microcontrollers in
detail.
2. Describe the advantages of microcontrollers for some
applications.
8051 MICROCONTROLLER'S PINSPins 1-8: Port 1 Each of these pins
can be configured as input or output.
Pin 9: RS Logical one on this pin stops microcontrollers
operating and erases the contents of most registers. By applying
logical zero to this pin, the program starts execution from the
beginning. In other words, a positive voltage pulse on this pin
resets the microcontroller.
Pins10-17: Port 3 Similar to port 1, each of these pins can
serve as universal input or output . Besides, all of them have
alternative functions:
Pin 10: RXD Serial asynchronous communication input or Serial
synchronous communication output.
Pin 11: TXD Serial asynchronous communication output or Serial
synchronous communication clock output.
Pin 12: INT0 Interrupt 0 input
Pin 13: INT1 Interrupt 1 input
Pin 14: T0 Counter 0 clock input
Pin 15: T1 Counter 1 clock input
Pin 16: WR Signal for writing to external (additional) RAM
Pin 17: RD Signal for reading from external RAM
Pin 18, 19: X2, X1 Internal oscillator input and output. A
quartz crystal which determines operating frequency is usually
connected to these pins. Instead of quartz crystal, the miniature
ceramics resonators can be also used for frequency stabilization.
Later versions of the microcontrollers operate at a frequency of 0
Hz up to over 50 Hz.
Pin 20: GND Ground
Pin 21-28: Port 2 If there is no intention to use external
memory then these port pins are configured as universal
inputs/outputs. In case external memory is used then the higher
address byte, i.e. addresses A8-A15 will appear on this port. It is
important to know that even memory with capacity of 64Kb is not
used ( i.e. note all bits on port are used for memory addressing)
the rest of bits are not available as inputs or outputs.
Pin 29: PSEN If external ROM is used for storing program then it
has a logic-0 value every time the microcontroller reads a byte
from memory.
Pin 30: ALE Prior to each reading from external memory, the
microcontroller will set the lower address byte (A0-A7) on P0 and
immediately after that activates the output ALE. Upon receiving
signal from the ALE pin, the external register (74HCT373 or
74HCT375 circuit is usually embedded ) memorizes the state of P0
and uses it as an address for memory chip. In the second part of
the microcontrollers machine cycle, a signal on this pin stops
being emitted and P0 is used now for data transmission (Data Bus).
In this way, by means of only one additional (and cheap) integrated
circuit, data multiplexing from the port is performed. This port at
the same time used for data and address transmission.
Pin 31: EA By applying logic zero to this pin, P2 and P3 are
used for data and address transmission with no regard to whether
there is internal memory or not. That means that even there is a
program written to the microcontroller, it will not be executed,
the program written to external ROM will be used instead.
Otherwise, by applying logic one to the EA pin, the microcontroller
will use both memories, first internal and afterwards external (if
it exists), up to end of address space.
Pin 32-39: Port 0 Similar to port 2, if external memory is not
used, these pins can be used as universal inputs or outputs.
Otherwise, P0 is configured as address output (A0-A7) when the ALE
pin is at high level (1) and as data output (Data Bus), when logic
zero (0) is applied to the ALE pin.
Pin 40: VCC Power supply +5V
INPUT/OUTPUT PORTS (I/O PORTS)All 8051 microcontrollers have 4
I/O ports, each consisting of 8 bits which can be configured as
inputs or outputs. This means that the user has on disposal in
total of 32 input/output lines connecting the microcontroller to
peripheral devices.
A logic state on a pin determines whether it is configured as
input or output: 0=output, 1=input. If a pin on the microcontroller
needs to be configured as output, then a logic zero (0) should be
applied to the appropriate bit on I/O port. In this way, a voltage
level on the appropriate pin will be 0.
Similar to that, if a pin needs to be configured as input, then
a logic one (1) should be applied to the appropriate port. In this
way, as a side effect a voltage level on the appropriate pin will
be 5V (as it is case with any TTL input). This may sound a bit
confusing but everything becomes clear after studying a simplified
electronic circuit connected to one I/O pin.
Input/Output (I/O) pinThis is a simplified overview of what is
connected to a pin inside the microcontroller. It concerns all pins
except those included in P0 which do not have embedded pullup
resistor.
Output pinA logic zero (0) is applied to a bit in the P
register. By turning output FE transistor on, the appropriate pin
is directly connected to ground.
Input pinA logic one (1) is applied to a bit in the P register.
Output FE transistor is turned off. The appropriate pin remains
connected to voltage power supply through a pull-up resistor of
high resistance.
A logic state (voltage) on any pin can be changed or read at any
moment. A logic zero (0) and logic one (1) are not equal. A logic
one (0) represents almost short circuit to ground. Such a pin is
configured as output.
A logic one (1) is loosely connected to voltage power supply
through resistors of high resistance. Since this voltage can be
easily pulled down by an external signal, such a pin is configured
as input.
Port 0
It is specific to this port to have a double purpose. If
external memory is used then the lower address byte (addresses
A0-A7) is applied on it. Otherwise, all bits on this port are
configured as inputs or outputs.
Another characteristic is expressed when it is configured as
output. Namely, unlike other ports consisting of pins with embedded
pull-up resistor ( connected by its end to 5 V power supply ), this
resistor is left out here. This, apparently little change has its
consequences:
If any pin on this port is configured as input then it performs
as if it floats. Such input has unlimited input resistance and has
no voltage coming from inside.
When the pin is configured as output, it performs as open drain,
meaning that by writing 0 to some ports bit, the appropriate pin
will be connected to ground (0V). By writing 1, the external output
will keep on floating. In order to apply 1 (5V) on this output, an
external pull-up resistor must be embedded.
Only in case P0 is used for addressing external memory ( only in
that case), the microcontroller will provide internal power supply
source in order to establish logical ones on pins. There is no need
to add external pullup resistors.
Port 1
This is a true I/O port, because there are no role assigning as
it is the case with P0. Since it has embedded pull-up resistors it
is completely compatible with TTL circuits.
Port 2
Similar to P0, when using external memory, lines on this port
occupy addresses intended for external memory chip. This time it is
the higher address byte with addresses A8-A15. When there is no
additional memory, this port can be used as universal input-output
port similar by its features to the port 1.
Port 3
Even though all pins on this port can be used as universal I/O
port, they also have an alternative function. Since each of these
functions use inputs, then the appropriate pins have to be
configured like that. In other words, prior to using some of
reserve port functions, a logical one (1) must be written to the
appropriate bit in the P3 register. From hardwares perspective ,
this port is also similar to P0, with the difference that its
outputs have a pull-up resistor embedded.
Current limitations on pins
When configured as outputs ( logic zero (0) ), single port pins
can "receive" current of 10mA. If all 8 bits on a port are active,
total current must be limited to 15mA (port P0: 26mA). If all ports
(32 bits) are active, total maximal current must be limited to
71mA. When configured as inputs (logic 1), embedded pull-up
resistor provides very weak current, but strong enough to activate
up to 4 TTL inputs from LS series.
It may be seen from description of some ports, that even though
all pins have more or less similar internal structure, it is
necessary to pay attention to which of them will be used for what
and how.
For example: If they are used as outputs with high voltage level
(5V), then port 0 should be avoided because its pins do not have
added resistor for connection to +5V. Only low logic level can be
obtained therefore, if another port is used for the same purpose,
one should have in mind that pull-up resistors have a relatively
high resistance. Consequentaly it can be counted on only several
hundreds microamperes of current coming out of a pin.
Exercise:
3. List the four ports of the 8051.
4. Explain the dual role of port 0 and port 2.
5. Code 8051 instructions for I/O handling.
6. Code Assembly language to use the ports for input or
output.
8051 MICROCONTROLLER MEMORY ORGANISATIONThe microcontroller
memory is divided into Program Memory and Data Memory. Program
Memory (ROM) is used for permanent saving program being executed,
while Data Memory (RAM) is used for temporarily storing and keeping
intermediate results and variables. Depending on the model in use (
still referring to the whole 8051 microcontroller family) at most a
few Kb of ROM and 128 or 256 bytes of RAM can be used. However
All 8051 microcontrollers have 16-bit addressing bus and can
address 64 kb memory. It is neither a mistake nor a big ambition of
engineers who were working on basic core development. It is a
matter of very clever memory organization which makes these
controllers a real programmers tidbit .
Program Memory
The oldest models of the 8051 microcontroller family did not
have internal program memory . It was added from outside as a
separate chip. These models are recognizable by their label
beginning with 803 ( for ex. 8031 or 8032). All later models have a
few Kbytes ROM embedded, Even though it is enough for writing most
of the programs, there are situations when additional memory is
necessary. A typical example of it is the use of so called lookup
tables. They are used in cases when something is too complicated or
when there is no time for solving equations describing some
process. The example of it can be totally exotic (an estimate of
self-guided rockets meeting point) or totally common( measuring of
temperature using non-linear thermo element or asynchronous motor
speed control). In those cases all needed estimates and
approximates are executed in advance and the final results are put
in the tables ( similar to logarithmic tables ).
How does the microcontroller handle external memory depends on
the pin EA logic state:
EA=0 In this case, internal program memory is completely
ignored, only a program stored in external memory is to be
executed.
EA=1 In this case, a program from builtin ROM is to be executed
first ( to the last location). Afterwards, the execution is
continued by reading additional memory.
in both cases, P0 and P2 are not available to the user because
they are used for data nd address transmission. Besides, the pins
ALE and PSEN are used too.
Data Memory
As already mentioned, Data Memory is used for temporarily
storing and keeping data and intermediate results created and used
during microcontrollers operating. Besides, this microcontroller
family includes many other registers such as: hardware counters and
timers, input/output ports, serial data buffers etc. The previous
versions have the total memory size of 256 locations, while for
later models this number is incremented by additional 128 available
registers. In both cases, these first 256 memory locations
(addresses 0-FFh) are the base of the memory. Common to all types
of the 8051 microcontrollers. Locations available to the user
occupy memory space with addresses from 0 to 7Fh. First 128
registers and this part of RAM is divided in several blocks.
The first block consists of 4 banks each including 8 registers
designated as R0 to R7. Prior to access them, a bank containing
that register must be selected. Next memory block ( in the range of
20h to 2Fh) is bit- addressable, which means that each bit being
there has its own address from 0 to 7Fh. Since there are 16 such
registers, this block contains in total of 128 bits with separate
addresses (The 0th bit of the 20h byte has the bit address 0 and
the 7th bit of th 2Fh byte has the bit address 7Fh). The third
group of registers occupy addresses 2Fh-7Fh ( in total of 80
locations) and does not have any special purpose or feature.
ADDITIONAL MEMORY BLOCK OF DATA MEMORY In order to satisfy the
programmers permanent hunger for Data Memory, producers have
embedded an additional memory block of 128 locations into the
latest versions of the 8051 microcontrollers. Naturally, its not so
simpleThe problem is that electronics performing addressing has 1
byte (8 bits) on disposal and due to that it can reach only the
first 256 locations. In order to keep already existing 8-bit
architecture and compatibility with other existing models a little
trick has been used.
Using trick in this case means that additional memory block
shares the same addresses with existing locations intended for the
SFRs (80h- FFh). In order to differentiate between these two
physically separated memory spaces, different ways of addressing
are used. A direct addressing is used for all locations in the
SFRs, while the locations from additional RAM are accessible using
indirect addressing.
Addressing
While operating, processor processes data according to the
program instructions. Each instruction consists of two parts. One
part describes what should be done and another part indicates what
to use to do it. This later part can be data (binary number) or
address where the data is stored. All 8051 microcontrollers use two
ways of addressing depending on which part of memory should be
accessed:
Direct Addressing
On direct addressing, a value is obtained from a memory location
while the address of that location is specified in instruction.
Only after that, the instruction can process data (howdepends on
the type of instruction: addition, subtraction, copy). Obviously, a
number being changed during operating a variable can reside at that
specified address. For example:
Since the address is only one byte in size ( the greatest number
is 255), this is how only the first 255 locations in RAM can be
accessed in this case the first half of the basic RAM is intended
to be used freely, while another half is reserved for the SFRs.
MOV A,33h; Means: move a number from address 33 hex. to
accumulator
Indirect Addressing
On indirect addressing, a register which contains address of
another register is specified in the instruction. A value used in
operating process resides in that another register. For
example:
Only RAM locations available for use are accessed by indirect
addressing (never in the SFRs). For all latest versions of the
microcontrollers with additional memory block ( those 128 locations
in Data Memory), this is the only way of accessing them. Simply,
when during operating, the instruction including @ sign is
encountered and if the specified address is higher than 128 ( 7F
hex.), the processor knows that indirect addressing is used and
jumps over memory space reserved for the SFRs.
MOV A,@R0; Means: Store the value from the register whose
address is in the R0 register
into accumulator
On indirect addressing, the registers R0, R1 or Stack Pointer
are used for specifying 8-bit addresses. Since only 8 bits are
avilable, it is possible to access only registers of internal RAM
in this way (128 locations in former or 256 locations in latest
versions of the microcontrollers). If memory extension in form of
additional memory chip is used then the 16-bit DPTR Register
(consisting of the registers DPTRL and DPTRH) is used for
specifying addresses. In this way it is possible to access any
location in the range of 64K.
Exercise:
7. List the five addressing modes of the 8051
microcontroller.
8. How do we select Program memory and data memory
sepeartely.
SFRs (Special Function Registers)
SFRs are a kind of control table used for running and monitoring
microcontrollers operating. Each of these registers, even each bit
they include, has its name, address in the scope of RAM and clearly
defined purpose ( for example: timer control, interrupt, serial
connection etc.). Even though there are 128 free memory locations
intended for their storage, the basic core, shared by all types of
8051 controllers, has only 21 such registers. Rest of locations are
intensionally left free in order to enable the producers to further
improved models keeping at the same time compatibility with the
previous versions. It also enables the use of programs written a
long time ago for the microcontrollers which are out of production
now.
A Register (Accumulator) This is a general-purpose register
which serves for storing intermediate results during operating. A
number (an operand) should be added to the accumulator prior to
execute an instruction upon it. Once an arithmetical operation is
preformed by the ALU, the result is placed into the accumulator. If
a data should be transferred from one register to another, it must
go through accumulator. For such universal purpose, this is the
most commonly used register that none microcontroller can be
imagined without (more than a half 8051 microcontroller's
instructions used use the accumulator in some way).
B Register
B register is used during multiply and divide operations which
can be performed only upon numbers stored in the A and B registers.
All other instructions in the program can use this register as a
spare accumulator (A).
INCLUDEPICTURE
"http://www.mikroe.com/en/books/8051book/images/note.gif" \*
MERGEFORMATINET During programming, each of registers is called by
name so that their exact address is not so important for the user.
During compiling into machine code (series of hexadecimal numbers
recognized as instructions by the microcontroller), PC will
automatically, instead of registers name, write necessary addresses
into the microcontroller.
R Registers (R0-R7)
This is a common name for the total 8 generalpurpose registers
(R0, R1, R2 ...R7). Even they are not true SFRs, they deserve to be
discussed here because of their purpose. The bank is active when
the R registers it includes are in use. Similar to the accumulator,
they are used for temporary storing variables and intermediate
results. Which of the banks will be active depends on two bits
included in the PSW Register. These registers are stored in four
banks in the scope of RAM.
The following example best illustrates the useful purpose of
these registers. Suppose that mathematical operations on numbers
previously stored in the R registers should be performed: (R1+R2) -
(R3+R4). Obviously, a register for temporary storing results of
addition is needed. Everything is quite simple and the program is
as follows :
MOV A,R3; Means: move number from R3 into accumulator
ADD A,R4; Means: add number from R4 to accumulator (result
remains in accumulator)
MOV R5,A; Means: temporarily move the result from accumulator
into R5
MOV A,R1; Means: move number from R1 into accumulator
ADD A,R2; Means: add number from R2 to accumulator
SUBB A,R5; Means: subtract number from R5 ( there are R3+R4
)
PSW Register (Program Status Word) This is one of the most
important SFRs. The Program Status Word (PSW) contains several
status bits that reflect the current state of the CPU. This
register contains: Carry bit, Auxiliary Carry, two register bank
select bits, Overflow flag, parity bit, and user-definable status
flag. The ALU automatically changes some of registers bits, which
is usually used in regulation of the program performing.
P - Parity bit. If a number in accumulator is even then this bit
will be automatically set (1), otherwise it will be cleared (0). It
is mainly used during data transmission and receiving via serial
communication.
- Bit 1. This bit is intended for the future versions of the
microcontrollers, so it is not supposed to be here.
OV Overflow occurs when the result of arithmetical operation is
greater than 255 (deci mal), so that it can not be stored in one
register. In that case, this bit will be set (1). If there is no
overflow, this bit will be cleared (0).
RS0, RS1 - Register bank select bits. These two bits are used to
select one of the four register banks in RAM. By writing zeroes and
ones to these bits, a group of registers R0-R7 is stored in one of
four banks in RAM.
RS1RS2Space in RAM
00Bank0 00h-07h
01Bank1 08h-0Fh
10Bank2 10h-17h
11Bank3 18h-1Fh
F0 - Flag 0. This is a general-purpose bit available to the
user.
AC - Auxiliary Carry Flag is used for BCD operations only.
CY - Carry Flag is the (ninth) auxiliary bit used for all
arithmetical operations and shift instructions.
DPTR Register (Data Pointer)
These registers are not true ones because they do not physically
exist. They consist of two separate registers: DPH (Data Pointer
High) and (Data Pointer Low). Their 16 bits are used for external
memory addressing. They may be handled as a 16-bit register or as
two independet 8-bit registers.Besides, the DPTR Register is
usually used for storing data and intermediate results which have
nothing to do with memory locations.
SP Register (Stack Pointer) A value of the Stack Pointer ensures
that the Stack Pointer will point to valid RAM and permits Stack
availability. By starting each subprogram, the value in the Stack
Pointer is incremented by 1. In the same manner, by ending
subprogram, this value is decremented by 1. After any reset, the
value 7 is written to the Stack Pointer, which means that the space
of RAM reserved for the Stack starts from this location. If another
value is written to this register then the entire Stack is moved to
a new location in the memory.
P0, P1, P2, P3 - Input/Output Registers In case that external
memory and serial communication system are not in use then, 4 ports
with in total of 32 input-output lines are available to the user
for connection to peripheral environment. Each bit inside these
ports coresponds to the appropriate pin on the microcontroller.
This means that logic state written to these ports appears as a
voltage on the pin ( 0 or 5 V). Naturally, while reading, the
opposite occurs voltage on some input pins is reflected in the
appropriate port bit.
The state of a port bit, besides being reflected in the pin,
determines at the same time whether it will be configured as input
or output. If a bit is cleared (0), the pin will be configured as
output. In the same manner, if a bit is set to 1 the pin will be
configured as input. After reset, as well as when turning the
microcontroller on , all bits on these ports are set to one (1).
This means that the appropriate pins will be configured as
inputs.
Conditionally said, I/O ports are directly connected to the
microcontrollers pins. This means that a logic state of these
registers can be checked by voltmeter and vice versa-voltage on the
pins can be checked by testing their bits!
Exercise:
9. How do you switch between register banks?
10. Discuss how to access the SFR.
11. Discuss how to access the extra 128 bytes of RAM space in
the 8051.
Counters and Timers
The main oscillator of the microcontroller uses quartz crystal
for its operating. As the frequency of this oscillator is precisely
defined and very stable, these pulses are the most suitable for
time measuring (such oscillators are used in quartz clocks as
well). In order to measure time between two events it is only
needed to count up pulses from this oscillator. That is exactly
what the timer is doing. Namely, if the timer is properly
programmed, the value written to the timer register will be
incremented or decremented after each coming pulse, i.e. once per
each machine cycle cycle. Taking into account that one instruction
lasts 12 quartz oscillator periods (one machine cycle), by
embedding quartz with oscillator frequency of 12MHz, a number in
the timer register will be changed million times per second, i.e.
each microsecond.
The 8051 microcontrollers have 2 timer counters called T0 and
T1. As their names tell, their main purpose is to measure time and
count external events. Besides, they can be used for generating
clock pulses used in serial communication, i.e. Baud Rate.
Timer T0
As it is shown in the picture below, this timer consists of two
registers TH0 and TL0. The numbers these registers include
represent a lower and a higher byte of one 16-digit binary
number.
This means that if the content of the timer 0 is equal to 0
(T0=0) then both registers it includes will include 0. If the same
timer contains for example number 1000 (decimal) then the register
TH0 (higher byte) will contain number 3, while TL0 (lower byte)
will contain decimal number 232.
Formula used to calculate values in registers is very simple:TH0
256 + TL0 = TMatching the previous example it would be as follows
:3 256 + 232 = 1000
Since the timers are virtually 16-bit registers, the greatest
value that could be written to them is 65 535. In case of exceeding
this value, the timer will be automatically reset and afterwords
that counting starts from 0. It is called overflow. Two registers
TMOD and TCON are closely connected to this timer and control how
it operates.
TMOD Register (Timer Mode)
This register selects mode of the timers T0 and T1. As
illustrated in the following picture, the lower 4 bits (bit0 -
bit3) refer to the timer 0, while the higher 4 bits (bit4 - bit7)
refer to the timer 1. There are in total of 4 modes and each of
them is described here in this book.
Bits of this register have the following purpose:
GATE1 starts and stops Timer 1 by means of a signal provided to
the pin INT1 (P3.3):
1 - Timer 1 operates only if the bit INT1 is set
0 - Timer 1 operates regardless of the state of the bit INT
1
C/T1 selects which pulses are to be counted up by the
timer/counter 1:
1 - Timer counts pulses provided to the pin T1 (P3.5)
0 - Timer counts pulses from internal oscillator
T1M1,T1M0 These two bits selects the Timer 1 operating mode.
T1M1T1M0ModeDescription
00013-bit timer
01116-bit timer
1028-bit auto-reload
113Split mode
GATE0 starts and stops Timer 1, using a signal provided to the
pin INT0 (P3.2):
1 - Timer 0 operates only if the bit INT0 is set
0 - Timer 0 operates regardless of the state of the bit INT0
C/T0 selects which pulses are to be counted up by the
timer/counter 0:
1 - Timer counts pulses provided to the pin T0(P3.4)
0 - Timer counts pulses from internal oscillator
T0M1,T0M0 These two bits select the Timer 0 operating mode.
T0M1T0M0ModeDescription
00013-bit timer
01116-bit timer
1028-bit auto-reload
113Split mode
Timer 0 in mode 0 (13-bit timer)
This is one of the rarities being kept only for compatibility
with the previuos versions of the microcontrollers. When using this
mode, the higher byte TH0 and only the first 5 bits of the lower
byte TL0 are in use. Being configured in this way, the Timer 0 uses
only 13 of all 16 bits. How does it operate? With each new pulse
coming, the state of the lower register (that one with 5 bits) is
changed. After 32 pulses received it becomes full and automatically
is reset, while the higher byte TH0 is incremented by 1. This
action will be repeated until registers count up 8192 pulses. After
that, both registers are reset and counting starts from 0.
Timer 0 in mode 1 (16-bit timer)
All bits from the registers TH0 and TL0 are used in this mode.
That is why for this mode is being more commonly used. Counting is
performed in the same way as in mode 0, with difference that the
timer counts up to 65 536, i.e. as far as the use of 16 bits
allows.
Timer 0 in mode 2 (Auto-Reload Timer)
What does auto-reload mean? Simply, it means that such timer
uses only one 8-bit register for counting, but it never counts from
0 but from an arbitrary chosen value (0- 255) saved in another
register.
The advantages of this way of counting are described in the
following example: suppose that for any reason it is continuously
needed to count up 55 pulses at a time from the clock
generator.
When using mode 1 or mode 0, It is needed to write number 200 to
the timer registers and check constantly afterwards whether
overflow occured, i.e. whether the value 255 is reached by counting
. When it has occurred, it is needed to rewrite number 200 and
repeat the whole procedure. The microcontroller performs the same
procedure in mode 2 automatically. Namely, in this mode it is only
register TL0 operating as a timer ( normally 8-bit), while the
value from which counting should start is saved in the TH0
register. Referring to the previous example, in order to register
each 55th pulse, it is needed to write the number 200 to the
register and configure the timer to operate in mode 2.
Timer 0 in Mode 3 (Split Timer)
By configuring Timer 0 to operate in Mode 3, the 16-bit counter
consisting of two registers TH0 and TL0 is split into two
independent 8-bit timers. In addition, all control bits which
belonged to the initial Timer 1 (consisting of the registers TH1
and TL1), now control newly created Timer 1. This means that even
though the initial Timer 1 still can be configured to operate in
any mode ( mode 1, 2 or 3 ), it is no longer able to stop, simply
because there is no bit to do that. Therefore, in this mode, it
will uninterruptedly operate in the background .
The only application of this mode is in case two independent
'quick' timers are used and the initial Timer 1 whose operating is
out of control is used as baud rate generator.
TCON - Timer Control Register
This is also one of the registers whose bits directly control
timer operating.Only 4 of all 8 bits this register has are used for
timer control, while others are used for interrupt control which
will be discussed later.
TF1 This bit is automatically set with the Timer 1 overflow
TR1 This bit turns the Timer 1 on
1 - Timer 1 is turned on
0 - Timer 1 is turned off
TF0 This bit is automatically set with the Timer 0 overflow.
TR0 This bit turns the timer 0 on
1 - Timer 0 is turned on
0 - Timer 0 is turned off
How to start Timer 0 ?
Normally, first this timer and afterwards its mode should be
selected. Bits which control that are resided in the register
TMOD:
This means that timer 0 operates in mode 1 and counts pulses
from internal source whose frequency is equal to 1/12 the quartz
frequency.In order to enable the timer, turn it on:
Immediately upon the bit TR0 is set, the timer starts operating.
Assuming that a quartz crystal with frequency of 12MHz is embedded,
a number it contains will be incremented every microsecond. By
counting up to 65.536 microseconds, the both registers that timer
consists of will be set. The microcontroller automatically reset
them and the timer keeps on repeating counting from the beginning
as far as the bits value is logic one (1).
How to 'read' a timer ?
Depending on the timers application, it is needed to read a
number in the timer registers or to register a moment they have
been reset.
- Everything is extremely simple when it is needed to read a
value of the timer which uses only one register for counting (mode
2 or Mode 3) . It is sufficient to read its state at any moment and
it is it!
- It is a bit complicated to read a timers value when it
operates in mode 2. Assuming that the state of the lower byte is
read first (TL0) and the state of the higher byte (TH0) afterwards,
the result is:
TH0 = 15 TL0 = 255
Everything seems to be in order at first sight, but the current
state of register at the moment of reading was:
TH0 = 14 TL0 = 255
In case of negligence, this error in counting ( 255 pulses ) may
occur for not so obvious but quite logical reason. Reading the
lower byte is correct ( 255 ), but at the same time the program
counter was taking a new instruction for the TH0 state reading, an
overflow occurred and both registers have changed their contents (
TH0: 1415, TL0: 2550). The problem has simple solution: the state
of the higher byte should be read first, then the state of the
lower byte and once again the state of the higher byte. If the
number stored in the higher byte is not the same both times it has
been read then this sequence should be repeated ( this is a mini-
loop consisting of only 3 instructions in a program).
There is another solution too. It is sufficient to simply turn
timer off while reading ( the bit TR0 in the register TCON should
be 0), and turn it on after that.
Detecting Timer 0 Overflow
Usually, there is no need to continuously read timer registers
contents. It is sufficient to register the moment they are reset,
i.e. when counting starts from 0. It is called overflow. When this
has occurred, the bit TF0 from the register TCON will be
automatically set. The microcontroller is waiting for that moment
in a way that program will constantly check the state of this bit.
Furthermore, an interrupt to stop the main program execution can be
enabled. Assuming that it is needed to provide a program pause (
time the program appeared to be stopped) in duration of for example
0.05 seconds ( 50 000 machine cycles ):
First, it is needed to calculate a number that should be written
to the timer registers:
This number should be written to the timer registers TH0 and
TL0:
Once the timer is started it will continue counting from the
written number. Program instruction checks if the bit TF0 is set,
which happens at the moment of overflow, i.e. after exactly 50.000
machine cycles and 0.05 seconds respectively.
How to measure pulses?
Suppose it is needed to measure the duration of an event, for
example how long some device has been turned on? Look at the
picture of the timer and pay attention to the purpose of the bit
GATE0 ( which resides in the TMOD register ). If this bit is
cleared then the state on the pin P3.2 does not affect the timer
operating. If GATE0 = 1 the timer will operate as far as the pin
P3.2 has logic one (1) value. If this pin is supplied with 5V
through some external switch at the moment the device is being
turned on, the timer will measure duration of its operating, which
actually was the aim.
How to count up pulses?
This time, the answer lies in the register TCON, and bit C/T0
respectively. Similar to the previous example, this bit brings into
an external signal. If the bit is cleared everything occurs in the
same way as in the previous examples and the timer counts pulses
from oscillator of defined frequency, i.e. measures the time that
went by. If the bit is set, the timer input is provided with pulses
from the pin P3.4 (T0). Since these pulses do not have some
definite time or order, it is not possible to measure time by
counting them. For that reason, this timer is turned into the
counter. The highest frequency that could be measured by such a
counter is 1/24 frequency of used quartz-crystal.
Timer 1
Referring to its characteristics, this timer is a twin brother
to the Timer 0. This means that they have the same purpose, their
operating is controlled by the same registers TMOD and TCON and
both of them can operate in one of 4 different modes.
INCLUDEPICTURE
"http://www.mikroe.com/en/books/8051book/ch2/images/35.gif" \*
MERGEFORMATINET Exercise:
12. Program the 8051 counters as event counters.
13. Program the 8051 timers to generate time delays.
14. Explain the special function registers associated to the two
16-bit timers.
15. What is the significance of GATE signal in a counter.
UART (Universal Asynchronous Receiver and Transmitter)
One of the features that makes this microcontroller so powerful
is an integrated UART, better known as a serial port. It is a
duplex port, which means that it can transmit and receive data
simultaneously. Without it, serial data sending and receiving would
be endlessly complicated part of the program where the pin state
continuously is being changed and checked according to strictly
determined rhythm. Naturally, it does not happen here because the
UART resolves it in a very elegant manner. All the programmer needs
to do is to simply select serial port mode and baud rate. When the
programmer is such configured, serial data sending is done by
writing to the register SBUF while data receiving is done by
reading the same register. The microcontroller takes care of all
issues necessary for not making any error during data exchange.
Serial port should be configured prior to being used. That
determines how many bits one serial word contains, what the baud
rate is and what the pulse source for synchronization is. All bits
controlling this are stored in the SFR Register SCON (Serial
Control).
SCON Register (Serial Port Control Register) SM0 - bit selects
mode
SM1 - bit selects mode
SM2 - bit is used in case that several microcontrollers share
the same interface. In normal circumstances this bit must be
cleared in order to enable connection to function normally.
REN - bit enables data receiving via serial communication and
must be set in order to enable it.
TB8 - Since all registers in microcontroller are 8-bit
registers, this bit solves the problem of sending the 9th bit in
modes 2 and 3. Simply, bits content is sent as the 9th bit.
RB8 - bit has the same purpose as the bit TB8 but this time on
the receiver side. This means that on receiving data in 9-bit
format , the value of the last ( ninth) appears on its
location.
TI - bit is automatically set at the moment the last bit of one
byte is sent when the USART operates as a transmitter. In that way
processor knows that the line is available for sending a new byte.
Bit must be clear from within the program!
RI - bit is automatically set once one byte has been received.
Everything functions in the similar way as in the previous case but
on the receive side. This is line a doorbell which announces that a
byte has been received via serial communication. It should be read
quickly prior to a new data takes its place. This bit must also be
also cleared from within the program!
As seen, serial port mode is selected by combining the bits SM0
and SM2 :
SM0SM1ModeDescriptionBaud Rate
0008-bit Shift Register1/12 the quartz frequency
0118-bit UARTDetermined by the timer 1
1029-bit UART1/32 the quartz frequency (1/64 the quartz
frequency)
1139-bit UARTDetermined by the timer 1
In mode 0, the data are transferred through the RXD pin, while
clock pulses appear on the TXD pin. The bout rate is fixed at 1/12
the quartz oscillator frequency. On transmit, the least significant
bit (LSB bit) is being sent/received first. (received).
TRANSMIT - Data transmission in form of pulse train
automatically starts on the pin RXD at the moment the data has been
written to the SBUF register.In fact, this process starts after any
instruction being performed on this register. Upon all 8 bits have
been sent, the bit TI in the SCON register is automatically
set.
RECEIVE - Starts data receiving through the pin RXD once two
necessary conditions are met: bit REN=1 and RI=0 (both bits reside
in the SCON register). Upon 8 bits have been received, the bit RI
(register SCON) is automatically set, which indicates that one byte
is received.
Since, there are no START and STOP bits or any other bit except
data from the SBUF register, this mode is mainly used on shorter
distance where the noise level is minimal and where operating rate
is important. A typical example for this is I/O port extension by
adding cheap IC circuit ( shift registers 74HC595, 74HC597 and
similar).
Mode 1
In Mode1 10 bits are transmitted through TXD or received through
RXD in the following manner: a START bit (always 0), 8 data bits
(LSB first) and a STOP bit (always 1) last. The START bit is not
registered in this pulse train. Its purpose is to start data
receiving mechanism. On receive the STOP bit is automatically
written to the RB8 bit in the SCON register.
TRANSMIT - A sequence for data transmission via serial
communication is automatically started upon the data has been
written to the SBUF register. End of 1 byte transmission is
indicated by setting the TI bit in the SCON register.
RECEIVE - Receiving starts as soon as the START bit (logic zero
(0)) appears on the pin RXD. The condition is that bit REN=1and bit
RI=0. Both of them are stored in the SCON register. The RI bit is
automatically set upon receiving has been completed.
The Baud rate in this mode is determined by the timer 1 overflow
time.
Mode 2
In mode 2, 11 bits are sent through TXD or received through RXD:
a START bit (always 0), 8 data bits (LSB first), additional 9th
data bit and a STOP bit (always 1) last. On transmit, the 9th data
bit is actually the TB8 bit from the SCON register. This bit
commonly has the purpose of parity bit. Upon transmission, the 9th
data bit is copied to the RB8 bit in the same register ( SCON).The
baud rate is either 1/32 or 1/64 the quartz oscillator
frequency.
TRANSMIT - A sequence for data transmission via serial
communication is automatically started upon the data has been
written to the SBUF register. End of 1 byte transmission is
indicated by setting the TI bit in the SCON register.
RECEIVE - Receiving starts as soon as the START bit (logic zero
(0)) appears on the pin RXD. The condition is that bit REN=1and bit
RI=0. Both of them are stored in the SCON register. The RI bit is
automatically set upon receiving has been completed.
Mode 3
Mode 3 is the same as Mode 2 except the baud rate. In Mode 3 is
variable and can be selected.
The parity bit is the bit P in the PSW register. The simplest
way to check correctness of the received byte is to add this parity
bit to the transmit side as additional bit. Simply, immediately
before transmit, the message is stored in the accumulator and the
bit P goes into the TB8 bit in order to be a part of the message.
On the receive side is the opposite : received byte is stored in
the accumulator and the bit P is compared with the bit RB8 (
additional bit in the message). If they are the same- everything is
OK!
Baud Rate
Baud Rate is defined as a number of send/received bits per
second. In case the UART is used, baud rate depends on: selected
mode, oscillator frequency and in some cases on the state of the
bit SMOD stored in the SCON register. All necessary formulas are
specified in the table :
Baud RateBitSMOD
Mode 0Fosc. / 12
Mode 11 Fosc.16 12 (256-TH1) BitSMOD
Mode 2Fosc. / 32Fosc. / 6410
Mode 31 Fosc.16 12 (256-TH1)
Timer 1 as a baud rate generator
Timer 1 is usually used as a baud rate generator because it is
easy to adjust various baud rate by the means of this timer. The
whole procedure is simple:
First, Timer 1 overflow interrupt should be disabled
Timer T1 should be set in auto-reload mode
Depending on necessary baud rate, in order to obtain some of the
standard values one of the numbers from the table should be
selected. That number should be written to the TH1 register. That's
all.
Baud RateFosc. (MHz)Bit SMOD
11.05921214.74561620
15040 h30 h00 h0
300A0 h98 h80 h75 h52 h0
600D0 hCC hC0 hBB hA9 h0
1200E8 hE6 hE0 hDE hD5 h0
2400F4 hF3 hF0 hEF hEA h0
4800F3 hEF hEF h1
4800FA hF8 hF5 h0
9600FD hFC h0
9600F5 h1
19200FD hFC h1
38400FE h1
76800FF h1
Multiprocessor Communication
As described in the previous text, modes 2 and 3 enable the
additional 9th data bit to be part of message. It can be used for
checking data via parity bit. Another useful application of this
bit is in communication between two microcontrollers, i.e.
multiprocessor communication. This feature is enabled by setting
the SM2 bit in the SCON register. The consequence is the following:
when the STOP bit is ready, indicating end of message, the serial
port interrupt will be requested only in case the bit RB8 = 1 (the
9th bit).
The whole procedure will be performed as follows:
Suppose that there are several connected microcontrollers having
to exchange data. That means that each of them must have its
address. The point is that each address sent via serial
communication has the 9th bit set (1), while data has it cleared
(0). If the microcontroller A should send data to the
microcontroller C then it at will place first send address of C and
the 9th bit set to 1. That will generate interrupt and all
microcontrollers will check whether they are called.
Of course, only one of them will recognize this address and
immediately clear the bit SM2 in the SCON register. All following
data will be normally received by that microcontroller and ignored
by other microcontrollers.
Exercise:
16. Explain the special function registers associated to
USART?
17. Define Baud rate?
18. Write a Assembly program to transfer alternate alphabets
using RS232?
8051 MICROCONTROLLER INTERRUPTSThere are five interrupt sources
for the 8051, which means that they can recognize 5 different event
that can interrupt regular program execution. Each interrupt can be
enabled or disabled by setting bits in the IE register. Also, as
seen from the picture below the whole interrupt system can be
disabled by clearing bit EA from the same register.
Now, one detail should be explained which is not completely
obvious but refers to external interrupts- INT0 and INT1. Namely,
if the bits IT0 and IT1 stored in the TCON register are set,
program interrupt will occur on changing logic state from 1 to 0,
(only at the moment). If these bits are cleared, the same signal
will generate interrupt request and it will be continuously
executed as far as the pins are held low.
IE Register (Interrupt Enable) EA - bit enables or disables all
other interrupt sources (globally)
0 - (when cleared) any interrupt request is ignored (even if it
is enabled)
1 - (when set to 1) enables all interrupts requests which are
individually enabled
ES - bit enables or disables serial communication interrupt
(UART)
0 - UART System can not generate interrupt
1 - UART System enables interrupt
ET1 - bit enables or disables Timer 1 interrupt
0 - Timer 1 can not generate interrupt
1 - Timer 1 enables interrupt
EX1 - bit enables or disables INT 0 pin external interrupt
0 - change of the pin INT0 logic state can not generate
interrupt
1 - enables external interrupt at the moment of changing the pin
INT0 state
ET0 - bit enables or disables timer 0 interrupt
0 - Timer 0 can not generate interrupt
1 - enables timer 0 interrupt
EX0 - bit enables or disables INT1 pin external interrupt
0 - change of the INT1 pin logic state can not cause
interrupt
1 - enables external interrupt at the moment of changing the pin
INT1 state
Interrupt Priorities
It is not possible to predict when an interrupt will be
required. For that reason, if several interrupts are enabled. It
can easily occur that while one of them is in progress, another one
is requested. In such situation, there is a priority list making
the microcontroller know whether to continue operating or meet a
new interrupt request.
The priority list cosists of 3 levels:
1. Reset! The apsolute master of the situation. If an request
for Reset omits, everything is stopped and the microcontroller
starts operating from the beginning.
2. Interrupt priority 1 can be stopped by Reset only.
3. Interrupt priority 0 can be stopped by both Reset and
interrupt priority 1.
Which one of these existing interrupt sources have higher and
which one has lower priority is defined in the IP Register (
Interrupt Priority Register). It is usually done at the beginning
of the program. According to that, there are several
possibilities:
Once an interrupt service begins. It cannot be interrupted by
another inter rupt at the same or lower priority level, but only by
a higher priority interrupt.
If two interrupt requests, at different priority levels, arrive
at the same time then the higher priority interrupt is serviced
first.
If the both interrupt requests, at the same priority level,
occur one after another , the one who came later has to wait until
routine being in progress ends.
If two interrupts of equal priority requests arrive at the same
time then the interrupt to be serviced is selected according to the
following priority list :
1. External interrupt INT0
2. Timer 0 interrupt
3. External Interrupt INT1
4. Timer 1 interrupt
5. Serial Communication Interrupt
IP Register (Interrupt Priority)
The IP register bits specify the priority level of each
interrupt (high or low priority).
PS - Serial Port Interrupt priority bit
Priority 0
Priority 1
PT1 - Timer 1 interrupt priority
Priority 0
Priority 1
PX1 - External Interrupt INT1 priority
Priority 0
Priority 1
PT0 - Timer 0 Interrupt Priority
Priority 0
Priority 1
PX0 - External Interrupt INT0 Priority
Priority 0
Priority 1
Handling Interrupt
Once some of interrupt requests arrives, everything occurs
according to the following order:
1. Instruction in progress is ended
2. The address of the next instruction to execute is pushed on
the stack
3. Depending on which interrupt is requested, one of 5 vectors
(addresses) is written to the program counter in accordance to the
following table:
Interrupt SourceVector (address)
IE03 h
TF0B h
TF11B h
RI, TI23 h
All addresses are in hexadecimal format
4. The appropriate subroutines processing interrupts should be
located at these addresses. Instead of them, there are usually jump
instructions indicating the location where the subroutines
reside.
5. When interrupt routine is executed, the address of the next
instruction to execute is poped from the stack to the program
counter and interrupted program continues operating from where it
left off.
From the moment an interrupt is enabled, the microcontroller is
on alert all the time. When interrupt request arrives, the program
execution is interrupted, electronics recognizes the cause and the
program jumps to the appropriate address (see the table above ).
Usually, there is a jump instruction already prepared subroutine
prepared in advance. The subroutine is executed which exactly the
aim- to do something when something else has happened. After that,
the program continues operating from where it left off
Reset
Reset occurs when the RS pin is supplied with a positive pulse
in duration of at least 2 machine cycles ( 24 clock cycles of
crystal oscillator). After that, the microcontroller generates
internal reset signal during which all SFRs, excluding SBUF
registers, Stack Pointer and ports are reset ( the state of the
first two ports is indefinite while FF value is being written to
the ports configuring all pins as inputs). Depending on device
purpose and environment it is in, on power-on reset it is usually
push button or circuit or both connected to the RS pin. One of the
most simple circuit providing secure reset at the moment of turning
power on is shown on the picture.
Everything functions rather simply: upon the power is on,
electrical condenser is being charged for several milliseconds
through resistor connected to the ground and during this process
the pin voltage supply is on. When the condenser is charged, power
supply voltage is stable and the pin keeps being connected to the
ground providing normal operating in that way. If later on, during
the operation, manual reset button is pushed, the condenser is
being temporarily discharged and the microcontroller is being
reset. Upon the button release, the whole process is repeated
Through the program- step by step...
The microcontrollers normally operate at very high speed. The
use of 12 Mhz quartz crystal enables 1.000.000 instructions per
second to be executed! In principle, there is no need for higher
operating rate. In case it is needed, it is easy to built-in
crystal for high frequency. The problem comes up when it is
necessary to slow down. For example, when during testing in real
operating environment, several instructions should be executed step
by step in order to check for logic state of I/O pins.
Interrupt system applied on the 8051 microcontrollers
practically stops operating and enables instructions to be executed
one at a time by pushing button. Two interrupt features enable
that:
Interrupt request is ignored if an interrupt of the same
priority level is being in progress.
Upon interrupt routine has been executed, a new interrupt is not
executed until at least one instruction from the main program is
executed.
In order to apply this in practice, the following steps should
be done:
1. External interrupt sensitive to the signal level should be
enabled (for example INT0).
2. Three following instructions should be entered into the
program (start from address 03hex.):
What is going on? Once the pin P3.2 is set to 0 (for example, by
pushing button), the microcontroller will interrupt program
execution jump to the address 03hex, will be executed a
mini-interrupt routine consisting of 3 instructions is located at
that address.
The first instruction is being executed until the push button is
pressed ( logic one (1) on the pin P3.2). The second instruction is
being executed until the push button is released. Immediately after
that, the instruction RETI is executed and processor continues
executing the main program. After each executed instruction, the
interrupt INT0 is generated and the whole procedure is repeated (
push button is still pressed). Button Press = One Instruction.
Exercise:
19. Progarm 8051 interrupts in assembly language.
20. Explain the purpose of the interrupt vector table.
21. Where are the Interrupt addresses stored in 8051.
8051 Microcontroller Power Consumption Control
Conditionally said microcontroller is the most part of its
lifetime is inactive for some external signal in order to takes its
role in a show. It can make a great problem in case batteries are
used for power supply. In extremely cases, the only solution is to
put the whole electronics to sleep in order to reduce consumption
to the minimum. A typical example of this is remote TV controller:
it can be out of use for months but when used again it takes less
than a second to send a command to TV receiver. While normally
operating, the AT89S53 uses current of approximately 25mA, which
shows that it is not too sparing microcontroller. Anyway, it doesnt
have to be always like this, it can easily switch the operation
mode in order to reduce its total consumption to approximately
40uA. Actually, there are two power-saving modes of operation: Idle
and Power Down.Idle mode
Immediately upon instruction which sets the bit IDL in the PCON
register, the microcontroller turns off the greatest power
consumer- CPU unit while peripheral units serial port, timers and
interrupt system continue operating normally consuming 6.5mA. In
Idle mode, the state of all registers and I/O ports is remains
unchanged.
In order to terminate the Idle mode and make the microcontroller
operate normally, it is necessary to enable and execute any
interrupt or reset.Then, the IDL bit is automatically cleared and
the program continues executing from instruction following that
instruction which has set the IDL bit. It is recommended that three
first following one which set NOP instructions. They do not perform
any operation but keep the microcontroller from undesired changes
on the I/O ports.
Power Down mode
When the bit PD in the register PCON is set from within the
program, the microcontroller is set to Powerdown mode. It and turns
off its internal oscillator reducing drastically consumption in
that way. In power- down mode the microcontroller can operate using
only 2V power supply while the total power consumption is less than
40uA. The only way to get the microcontroller back to normal mode
is reset.
During Power Down mode, the state of all SFR registers and I/O
ports remains unchanged, and after the microcontroller is put get
into the normal mode, the content of the SFR register is lost, but
the content of internal RAM is saved. Reset signal must be long
enough approximately 10mS in order to stabilize quartz oscillator
operating.
PCON register The purpose of the Register PCON bits :
SMOD By setting this bit baud rate is doubled.
GF1 General-purpose bit (available for use).
GF1 General-purpose bit (available for use).
GF0 General-purpose bit (available for use).
PD By setting this bit the microcontroller is set into Power
Down mode.
IDL By setting this bit the microcontroller is set into Idle
mode.
Exercise:
22. Give the two modes of PCON register.EXERCISE
1. Describe the dual role of port 0 in providing both data and
addresses.
2. Explain the memory organization of 8051 with a neat
sketch.
3. List the SFR addresses.
4. Explain the Flag register of 8051.
5. Explain the purpose of each pin of the 8051
microcontroller.
6. List the timers of the 8051 timers and their associated
registers.
7. Describe the various modes of the 8051 timers.
8. Define the Interrupt Priority of the 8051.
9. List the 6 interrupts of the 8051.
10. Compare and contrast interrupts versus polling.
Multiple-choice Questions
1. The port1 pins of 8051 MC are multiplexed with _______
pins.
A. Address B. Data
C. Address/Data D. Address & Data
2. PSEN stands for_________.
A. Processor status enable B. Program status enable
C. Processor start enable D. Program store enable
3. Does all micro controllers have internal memory?
A. Yes B. No
C. It depends on D. None
the configurations
4. What is the capacity of internal ROM of 8051?
A. 4K bytes B. 128 bytes
C. 256 bytes D. 8K bytes
5. What is the capacity of internal RAM of 8051?
A. 4K bytes B. 128 bytes
C. 256 bytes D. 8K bytes
6. How many ports do 8051 have?
A.1 B. 2
C. 3 D. 4
7. Intel 8051 is a ______ bit processor.
A. 8 B. 16
C. 32 D. 64
8. How many timers do 8051 have?
A.one16 bit B. two16-bit
C. Two32 bit D. two 8-bit
9. The following processors have 16-bit program counter.
A. 8085 & 8086 B. 8051 & 8086
C. 8085 & 8051 D. None
10. What is the pin used for serial communication?
A. RXD B. RI
C. TI D. DTE
11. DPTR is used for external memory.
A. True
B. False
12. Do all series of the 8051 micro controllers have flash
memory?
A. Yes B. No
C. It depends on D. None
the configurations
13. If EA pin = 0, which memory is accessed?
A. Internal ROM B. internal RAM
C. External RAM D. External ROM
14. The total number of general purpose registers in RAM
are________.
A. 128 B. 256
C. 64 D. 8
15. What is the address range of special function registers in
RAM?
A. 33h- 80h B. 33h-66h
C. 80h- FFh D. 66h-FFh
16. How do we select register banks of RAM?
A. Through R0 B. through R1
C. Through R2 D. through PSW
17. MOV A, 33h is an example for __________ addressing mode
A. Direct B. Indirect
C. Immediate D. Relative
18. MOV A, @R0 means
A. R0 contains data directly
B. R0 contains address
C. R0 contains address indirectly
D. None
19. How many serial ports do 8051 have?
A.1 B. 2
C. 3 D. 4
20. XTAL2 is used for
A. Providing Clock
B. For calculation of Frequency of clock
C. Vcc
D. Gnd
21. XTAL1 is used for
A. Providing Clock
B.For calculation of Frequency of clock
C.Vcc
D.Gnd
22. Upon Power on RESET, the CPU loads SP to________
address.
A. FFFFH B. 0000H
C. 0007H D. FFFF0H
23. EA pin of 8051 is used to differentiate between ________
memories.
A. Internal ROM & RAM
B. Internal ROM & RAM
C. Internal RAM & External RAM
D. Internal ROM & External ROM
24. Find the machine cycle for XTAL = 11.0592 MHz. A. 1.085 us
B. 0.75us
C. 1.085 ms D. 0.75ms
25. One machine cycle=________ T-states.
A.6 B.12
C. 13 D. 14
26. Write the syntax for setting 7th pin of port1 of 8051.
A.SETB P1 B. SETB P1-7
C.SETB P17 D. SETB P1.7
27. ADD A, 6 means
A. A= A+6 B. A= A+R6
C. A= A+[6] D. Both B & C
28. NOP instruction in 8051 consumes _____ number of machine
cycle.
A.1 B. 2
C. 3 D. 4
29. Rotate left operation can be performed only with following
registers. What are
they?
A. R0-R7 B. A and B
C. A, B, R0-R7 D. only A
30. Give the address locations of bit addressable RAM.
A.00-0FH B. 20-2FH
C.20-80H D. 40-4FH
31. PSW.4 is______
A. Carry B. Parity
C. Bank selection bit D. Auxiliary carry
32. Which of the following is an illegal instruction?
A.MOV DPH, #FCH B. MOV DPTR, A
C.MOV DPTR, # PORT 1 D. All
33. MOVC A, @A+DPTR comes under _______ addressing mode.
A. Immediate B. Register
C. Indirect D. Indexed
34. The result of MUL AB is stored in________.
A. A B. B
C. A & B D. None
35. LJMP is a ____ byte instruction.
A.1 B. 2
C. 3 D. 4
36. What data is to be written into port to read data?
A.0FFH B. 2FH
C. Not necessary D. Depends
37. How many external interrupt sources do 8051 have?
A.1 B. 2
C. 3 D. 4
38. How do we make a timer operated as a counter in 8051?
A. By sending an external clock
B. By using internal clock itself
C. By setting timer run flag
D. By clearing timer flags
39. Timer mode 2 is used for:
A.13-bit timer B. 8-bit timer
C. 16-bit timer D. 8-bit auto reload
40. Which of the following values should be loaded into timer
TH1 register for
Setting a baud rate of 9600?
A. -3 B. -6
C. -12 D. -24Answers:1. C
2. D
3. A
4. A
5. B
6. D
7. A
8. B
9. B
10. A
11. A
12. C
13. A
14. A
15. C
16. D
17. A
18. A
19. A
20. B
21. A
22. C
23. C
24. A
25. B
26. D
27. A
28. A
29. D
30. B
31. C
32. B
33. D
34. C
35. C
36. A
37. B
38. A
39. D
40. A