BJT BIAS CIRCUITS EMT116 –ELECTRONIC …portal.unimap.edu.my/portal/page/portal30/Lecture Notes...The analysis for pnptransistor biasing circuits is the same as that for npntransistor

BJT BIAS CIRCUITS EMT116 – ELECTRONIC DEVICESDR. NUR SYAKIMAH ISMAIL

CHAPTER OUTLINE

§ DC operating point

§ Base Bias

§ Emitter-Feedback Bias

§Voltage Divider Bias

§ Collector-Feedback Bias

§ Emitter Bias

§ Other Bias Configuration

DC OPERATING POINT

§ A transistor must be properly biased with a direct current (DC) voltage in order to operate as a linear amplifier.

§ A DC operating point must be set so that signal variations at the input terminal are amplified and accurately reproduce at the output terminal.

§ When bias a transistor à establish DC voltage and current values.

§ At DC operating point, !" and #"$ have specified values à referred as Q-point (quiescent point) à !"% and #"$%

DC BIAS

§ Bias establishes DC operating point (Q-point) for linear operation of an amplifier.

§ If not bias with correct DC voltages on the input and output, it can go into saturation and cutoff when input signal is applied.

§ In (a), proper biasing produce output signal which amplified input signal except it is inverted. Output signal swings equally above and below DC bias level of the output !"#(%&')§ improper biasing cause distortion of in output signal (b) & (c)

§ (b) illustrate limiting of positive portion of output voltage àQ-point being too close to cutoff.

§ (c) illustrate limiting of negative portion of output voltage àQ-point being too close to saturation.

DC BIAS

§ Transistor is biased with !"" and !## to obtain certain values of $", $#, $&, !#& .

§ The collector characteristic curve for transistor in (a) are shown in (b)

DC BIAS§ Assign !" values and observe what happen to !#and $#%.

§ !" = 200 )*à !# = +!" = 20 ,*$#% = $## − !#.# = 10 − 20 ,* 220 = 5.6 $Q-point shown as Q1

§ !" = 300 )*à !# = +!" = 30 ,*$#% = $## − !#.# = 10 − 30 ,* 220 = 3.4 $Q-point shown as Q2

§ !" = 400 )*à !# = +!" = 40 ,*$#% = $## − !#.# = 10 − 40 ,* 220 = 1.2 $Q-point shown as Q3

DC LOAD LINE§ DC operation of a transistor circuit can be described graphically using a DC load line.

§ A straight line drawn on the characteristic curves from the saturation value where !" = !"(%&') on y-axis to the cutoff value where )"* =)"" on the x-axis.

§ The load line is determine by the external circuit ()"" and +"), not the transistor itself.

§ From !" equation:

!" =)"" − )"*

+"= )""+"

− )"*+"= − 1

+")"* +

)""+"

Compare to general straight line equation / = 01 + 2 where slope is − 34 56, an x intercept of )"* = )"", and y intercept of 78858

which is !"(%&').

The point at which load line intersect with characteristic curve represent Q-point

LINEAR OPERATION§ Region along the load line between cutoff and saturation à linear region of transistor’s operation.

§ in this region à output voltage is ideally a linear reproduction of the input.

§ Assume a sinusoid voltage, !"# is superimposed on !$$ à %$ vary 100()sinusoidally above & below Q point

§ cause %* vary 10+) sinusoidally above & below Q point

§ As a result, !*, vary 2.2 ! sinusoidally above & below Q point

§ Point A, B and Q correspond to positive peak, negative peak and zero value of sinusoid input voltage.

§ !*,/, %*/, %$/ are DC Q-point values with no input sinusoidal voltage applied.

WAVEFORM DISTORTION

EXAMPLE

§ Determine Q-point for the circuit

§ Draw the DC load line.

§ Find the maximum peak value of base current for linear operation.

Assume !"# = 200.

SOLUTION

EXAMPLE

§ Determine Q-point for the circuit

§ Draw the DC load line.

§ Find the maximum peak value of base current for linear operation.

Assume !"# = 100. 24 V

1 kΩ

BASE BIAS CONFIGURATION§ Simplest transistor DC bias configuration.

§ For DC analysis, capacitors can be replaced with open-circuit equivalent.

§ DC supply, !"" can be separated into 2 supplies to permit a separation of input and output circuit.

BASE BIAS CONFIGURATION

§ Consider the B-E circuit loop:

!"#" + %"& − %(( = 0

!" =%(( − %"&

#"

!( = +!"

BASE BIAS CONFIGURATION

§ Consider the C-E circuit loop:!"#" + %"& − %"" = 0

%"& = %"" − !"#"

%"& = %" − %&%*& = %* − %&

§ Since %& = 0%* = %*&%" = %"&

BASE BIAS CONFIGURATION

§ DC Load Line analysis:!"# = !"" − &"'"

§ During saturation, &( maximum à !"# ≈ 0 !∴ &"(-./) =

!""'"

§ If !"# is small (not 0 V):

∴ &"(-./) =!"" − !"#(-./)

'"§ During cutoff, &1 = 0, &" ≈ 0

∴ !"# = !""

BASE BIAS CONFIGURATION – DC LOAD LINE

If !" is changed by varying #"

If $%% is held fixed and #% increased

If #% is held fixed and $%% decreased

EXAMPLEGiven the DC load line and the defined Q-point.

Determine the required value of !"", $" and $% for a fixed-bias configuration.

Answer:!"" = 20 !$" = 2 )Ω$% = 772 )Ω

EMITTER FEEDBACK BIAS CONFIGURATION

§ Contains emitter resistor to improve stability level over fixed-biased configuration.

EMITTER FEEDBACK BIAS CONFIGURATION

§ Consider the B-E circuit loop:

!"#" + %"& + !&#& − %(( = 0

where !& = !( + !" = +!" + !" = + + 1 !"!"#" + %"& + + + 1 !"#& − %(( = 0!" #" + + + 1 #& = %(( − %"&

!" =%(( − %"&

#" + + + 1 #&!( = +!"

EMITTER FEEDBACK BIAS CONFIGURATION

§ Consider the C-E circuit loop:!"#" + %"& + !&#& − %"" = 0

§ Substitute !& ≅ !"%"& = %"" − !"#" − !"#&%"& = %"" − !"(#" + #&)

Since %& = !&#&%"& = %" − %&

%" = %"& + %& or %" = %"" − !"#"

%-& = %- − %&%- = %-& + %&

EMITTER FEEDBACK BIAS CONFIGURATION

From:

!" =$%% − $"'

(" + * + 1 ('$%' = $%% − !%((% + (')

§ Aside from $"' , resistor (' is reflected back to input base circuit by factor * + 1 .

§ (' is part of C-E loop appears as * + 1 (' in B-E loop.

§ Because * is large value, (' appears great deal larger in base circuit.

(. = * + 1 ('

EMITTER FEEDBACK BIAS CONFIGURATION

§ DC Load Line analysis:!"# = !"" − &"((" + (#)

§ During saturation, &+ maximum à !"# ≈ 0 !∴ &"(/01) =

!""(" + (#

§ If !"# is small (not 0 V):

∴ &"(/01) =!"" − !"#(/01)(" + (#

§ During cutoff, &2 = 0, &" ≈ 0∴ !"# = !""

EXAMPLE

Determine:

a) #$b) #&c) (&)d) (&e) ()f) ($g) ($&

Answer:#$ = 40.125 56#& = 2.006 86(&) = 13.98 ((& = 15.98 (() = 2.006 (($ = 2.706 (

($& = −13.274 (

VOLTAGE-DIVIDER BIAS CONFIGURATION

§ Previous configuration !"# and $"%# were function of &.

§ But & is temperature sensitive and actual value is usually not well defined.

§ Desirable to develop a bias circuit that is less dependent or independent of &à voltage-divider bias configuration.

§ 2 method to analyze – exact method and approximate method.

VOLTAGE-DIVIDER BIAS CONFIGURATION –EXACT ANALYSIS§ Consider the B-E circuit loop using Thevenin equivalent network:

!"# = !% ∥ !' =!%!'!% + !'

)"# =!'*++!% + !'

VOLTAGE-DIVIDER BIAS CONFIGURATION –EXACT ANALYSIS§ Consider the B-E circuit loop:

!"#$% + '"( + !(#( − '$% = 0

where !( = !, + !" = -!" + !" = - + 1 !"!"#$% + '"( + - + 1 !"#( − '$% = 0!" #$% + - + 1 #( = '$% − '"(

!" ='$% − '"(

#$% + - + 1 #(!, = -!"

VOLTAGE-DIVIDER BIAS CONFIGURATION –EXACT ANALYSIS§ Consider the C-E circuit loop:

!"#" + %"& + !&#& − %"" = 0§ Substitute !& ≅ !"

%"& = %"" − !"#" − !"#&%"& = %"" − !"(#" + #&)

Since %& = !&#&%"& = %" − %&

%" = %"& + %& or %" = %"" − !"#"

%-& = %- − %&%- = %-& + %&

VOLTAGE-DIVIDER BIAS CONFIGURATION –EXACT ANALYSIS§ DC Load Line analysis:

!"# = !"" − &"((" + (#)§ During saturation, &+ maximum à !"# ≈ 0 !

∴ &"(/01) =!""

(" + (#§ If !"# is small (not 0 V):

∴ &"(/01) =!"" − !"#(/01)(" + (#

§ During cutoff, &2 = 0, &" ≈ 0∴ !"# = !""

VOLTAGE-DIVIDER BIAS CONFIGURATION –APPROXIMATE ANALYSIS§ Resistance !" is the equivalent resistance between base and ground for the transistor with !#.

§ Resistance between base and emitter: !" = % + 1 !# ≅ %!#§ If !" is much larger than !) à *+ much smaller than *) (current always seeks the path of least resistance) à *) ≈ *-§ If accept this approximation, *+ = 0à *) = *-à !- & !) can be considered series element.

0+ = 012 =!)033!- + !)

§ This approximation analysis only valid if 456 ≥ 895:

EXAMPLE

Determine the levels of !"# and $"%# for the voltage divider configuration using exact and approximate techniques.

Answer:Exact analysis:

!"# = 1.98 +,$"%# = 4.54 $

Approximate analysis:!"# = 2.59 +,$"%# = 0.388 $

COLLECTOR FEEDBACK CONFIGURATION

An improve level of stability can be obtained by introducing a feedback path from collector to base.

§ Consider the B-E circuit loop:!′#$# + !&$& + '&( + !($( − '## = 0

where !( = !# + !& = ,!& + !& = , + 1 !& ≅ ,!& ≅ !#Assume !′# ≅ !#

!#$# + !&$& + '&( + !#$( − '## = 0,!&$# + !&$& + '&( + ,!&$( − '## = 0!& $& + ,($# + $() = '## − '&(

!& ='## − '&(

$& + ,($# + $()

COLLECTOR FEEDBACK CONFIGURATION

COLLECTOR FEEDBACK CONFIGURATION

§ Consider the C-E circuit loop:!′#$# + &#' + !'$' − &## = 0

§ Substitute !' ≅ !# and !′# ≅ !#&#' = &## − !#$# − !#$'&#' = &## − !#($# + $')

Since &' = !'$'&#' = &# − &'

&# = &#' + &' or &# = &## − !#$#

&.' = &. − &'&. = &.' + &'

COLLECTOR FEEDBACK CONFIGURATION

§ DC Load Line analysis:!"# = !"" − &"((" + (#)

§ During saturation, &+ maximum à !"# ≈ 0 !∴ &"(/01) =

!""(" + (#

§ If !"# is small (not 0 V):

∴ &"(/01) =!"" − !"#(/01)(" + (#

§ During cutoff, &2 = 0, &" ≈ 0∴ !"# = !""

EXAMPLE

Determine the Q-point of !"# and $"%#

Answer:!"# = 1.07 +,$"%# = 3.69 $

EXAMPLE

Determine DC level of !" and #$ for the network.

Answer:!" = 35.5 )*#$ = 9.22 #

EMITTER BIAS CONFIGURATION

§ Emitter bias provides excellent bias stability in spite of changes in ! or temperature.

§ it uses both a positive and a negative supply voltage.

-

EMITTER BIAS CONFIGURATION

§ Consider the B-E circuit loop:

!"#" + %"& + !&#& − %&& = 0

where !& = !* + !" = +!" + !" = + + 1 !" ≅ +!" ≅ !*!"#" + %"& + + + 1 !"#& − %&& = 0!"[#"+ + + 1 #&] = %&& − %"&

!" =%&& − %"&

#" + + + 1 #&

EMITTER BIAS CONFIGURATION

§ Consider the C-E circuit loop:!"#" + %"& + !&#& − %&& − %"" = 0

§ Substitute !& ≅ !"!"#" + %"& + !"#& − %&& − %"" = 0!" #" + #& + %"& − %&& − %"" = 0%"& = %&& + %"" − !" #" + #&

§ Since %& = !&#& − %&&%"& = %" − %&

%" = %"& + %& or %" = %"" − !"#"

%+& = %+ − %&%+ = %+& + %&

EXAMPLE

Determine how much the Q-point (!" and #"$) for the circuit will change if %increases from 100 to 200.

COMMON COLLECTOR (EMITTER BIAS)

Determine !"#$ and %#$

Answer:%#$ = 4.16 +,!"#$ = 11.68 !

COMMON BASE

Determine:

a) #$b) #&c) ()$d) ()&

Answer:#$ = 2.75 01#& = 45.08 51()$ = 4.1 (()& = 3.51 (

PNP TRANSISTOR

§ The analysis for pnp transistor biasing circuits is the same as that for npn transistor circuits.

§ The only difference is that the currents are flowing in the opposite direction.

PNP TRANSISTOR

§ Consider the B-E circuit loop:

−"#$# − %#& − "&$& + %(( = 0

where "& = "( + "# = +"# + "# = + + 1 "#−"#$# − %#& − + + 1 "#$& + %(( = 0

%(( − %#& = "# $# + + + 1 $&

"# =%(( − %#&

$# + + + 1 $&

PNP TRANSISTOR

§Consider the C-E circuit loop:

−"#$# − %#& − "&$& + %## = 0

§ Substitute "& ≅ "#%#& = %## − "#$# − "#$&%#& = %## − "#($# + $&)

SUMMARY

§ The purpose of biasing a circuit is to establish a proper stable DC operating point (Q-point).

§The Q-point of a circuit is defined by specific values for !" and #"$ . These values are coordinates of Q-point.

§A DC load line passes through Q-point on a transistor’s collector curves intersecting the vertical axis at !"(&'() and the horizontal axis at #"$(*+(,--).§ The linear (active) operating region of a transistor lies along the load line below saturation and above cutoff.

§ There are many types of biasing such as voltage divider, emitter bias, base bias, emitter-feedback bias, collector-feedback bias, etc.