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AN1353Op Amp Rectifiers, Peak Detectors and Clamps
INTRODUCTIONThis application note covers a wide range
ofapplications, such as half-wave rectifiers, full-waverectifiers,
peak detectors and clamps. Many of thecircuits are simple in terms
of component count, butthey play important roles in overall systems
design,such as:
• AC to DC Power Conversion• Automatic Gain Control Loops• Power
Monitoring Applications• AM Demodulator
BASIC RECTIFIERSThe basic rectifiers have been designed with
diodes.Figure 1 shows such a simple series circuit, driven byan AC
source. When the diode is reverse-biased, itacts as a very high
impedance device. Figure 1showsa negative half wave rectifier. It
outputs nearly the fullinput voltage across the diode when reverse
biased. Asimilar circuit in Figure 2 shows a positive
half-waverectifier. If a full-wave rectifier is desired, more
diodesmust be used to configure a bridge, as shown inFigure 3. The
input signal must be larger than thevoltage across the diode to
ensure that the diode isforward biased.
FIGURE 1: Negative Half-Wave Rectifier.
FIGURE 2: Positive Half-Wave Rectifier.
FIGURE 3: Full-Wave Rectifier.
Choosing the Components
SELECTING THE DIODEWhen choosing the diode, the most
importantparameters are the maximum forward current (IF), andthe
peak inverse voltage rating (PIV) of the diode. Thepeak inverse
voltage is the maximum voltage the diodecan withstand when it is
reverse-biased. If this voltageis exceeded, the diode may be
destroyed. The diodemust have a peak inverse voltage rating that is
higherthan the maximum voltage applied to it in anapplication. In
many diode data sheets, PIV is referredto as peak reverse voltage
(PRV).
Author: Dragos Ducu,Microchip Technology Inc.
AC
RL
D1 VOUT
t
V
VOUTVIN
ACD1
VOUT
RL
VOUTVIN
V
t
VOUT
RL
t
V
VOUTVIN
© 2011 Microchip Technology Inc. DS01353A-page 1
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AN1353
D
The peak inverse voltage of the diode will be equal to:
EQUATION 1:
Every diode has a parasitic capacitance and, bydefault, has a
time charge storage. This charge storagemechanism is nonlinear,
leading to a nonlinearcapacitance. This effect is very important
because thenonlinearity of the diode can generate harmonics.
Forexample, the output voltage becomes negative for ashort time.
This period is called reverse recovery time.During the transition,
the diode’s parasitic capacitancewill interact with the circuit
resistors to modify thecircuit’s behavior.For most general purpose
applications, low powersignal diodes such as 1N4148, are adequate.
For highaccuracy applications, where offset errors and reversediode
leakage current are critical, a low leakage FETtransistor can be
used as a diode (short Drain andSource together), such as 2N4117A.
In applicationswhere speed is important, silicon Schottky
barrierdiodes are worth considering, since they have a lowforward
ON voltage of only 0.4V and are fast.
SELECTING THE RESISTORThe resistor is selected based on the load
current.One limitation is the value of load resistor. The value
ofthe load resistor must be less than the diode resistancewhen in
reverse bias. The parasitic capacitance of thediode interacts with
the load resistor causing a timeconstant. If this constant is
large, the output voltage willhave a delayed recovery.
Advantages and DisadvantagesThe major disadvantage of these
circuits is thenonlinearity of the diodes. If the input signal is
smallerthan the threshold voltage of the diode, the signalcannot be
recovered. To reduce the threshold voltageof the diode and improve
linearity, we need to includethe diode into the feedback loop of
the operationalamplifier.
Practical ExamplesFigures 4 – 6 show practical samples when
using the1N4001 diode and RL = 1 kΩ. The frequency isf = 1 kHz.
FIGURE 4: Negative Half-Wave Rectifier Sample.
FIGURE 5: Positive Half-Wave Rectifier Sample.
FIGURE 6: Full-Wave Rectifier Sample.
VPIV rating( ) VPK max( ) VD on( )+≥Where:
VPIV = Peak inverse voltageVPK(max) = Maximum peak amplitude
VD(on) = Diode voltage on when in
TABLE 1: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Uses few compo-nents
- Poor accuracy
- Simple design - The rectified voltage depends on the diode
voltage threshold
-1.5
-1
-0.5
0
0.5
1
1.5
Time (1 ms/div)
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de (V
)
VOUTVIN
-1.5
-1
-0.5
0
0.5
1
1.5
Time (1 ms/div)
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nitu
de (V
)
VOUTVIN
-1.5
-1
-0.5
0
0.5
1
1.5
Time (1 ms/div)
Mag
nitu
de(V
)
VOUTVIN
S01353A-page 2 © 2011 Microchip Technology Inc.
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AN1353
ACTIVE HALF-WAVE RECTIFIERThe simplest op amp half-wave
rectifier is shown inFigure 7. When the VIN is positive, the diode
is forwardbiased; the signal can be found on the RL load. Whenthe
VIN is negative, the diode is non-conductive, andthe output signal
is ground (0V).
FIGURE 7: Op Amp Half-Wave Rectifier.The big advantage of this
circuit is represented by thesmall threshold voltage and linearity.
This is moreconvenient than the basic rectifiers, since this
circuit isable to rectify signals smaller than the diode
thresholdvoltage.
FIGURE 8: Circuit Behavior on Low Frequency.This circuit has
limitations. The rectifier’s speed islimited by the op amp
bandwidth. This effect isillustrated in Figure 9, where the
rectified output signaloverlaps the input signal. The maximum
frequencythat can be rectified is determined by the slew rate ofthe
op amp.
FIGURE 9: Output Limitation on High-Frequency Input Signals.
Choosing the Components
SELECTING THE OP AMPWhen selecting the op amp, two
importantcharacteristics must be considered:• Gain Bandwidth
Product• Slew Rate (SR)The minimum gain bandwidth product
requirement canbe estimated in Equation 2.
EQUATION 2:
The next parameter that needs to be considered is theslew rate
(SR). This is the maximum time rate changeat the output of the op
amp; it shows how fast the outputcan follow the input signal. The
SR parameter can befound in the selected op amp’s data sheet.The
full bandwidth product (FPBW) defines the highestfrequency sine
wave that will not be distorted by theslew rate limit.
EQUATION 3:
SELECTING THE DIODE AND THE RESISTORRefer to the sections
Selecting the Diode and SelectingThe Resistor, in the Basic
Rectifiers section, for detailson choosing the appropriate
components.
Advantages and DisadvantagesTable 2 shows the main advantages
anddisadvantages of a half-wave rectifier.
VOUT
RLVIN
-+AO1 D1
-1.50
-1.00
-0.50
0.00
0.50
1.00
1.50
Time (1 ms/div)
Mag
nitu
de (V
)
VOUT
VIN
-1.5
-1
-0.5
0
0.5
1
1.5
Time (50 µs/div)
Mag
nitu
de (V
)
VOUTVIN
TABLE 2: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Uses few components - Load dependant- Good linearity - Limited
op amp bandwidth
fGBWP 10 G× fINPUT×=
Where:
fGBWP = Gain bandwidth productG = DC gain
fINPUT = Maximum input frequency
SR ΔVOUTΔT
-----------------max
=
FPBW SRπ VOUT p p–( )×-------------------------------------=
© 2011 Microchip Technology Inc. DS01353A-page 3
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AN1353
Practical ExampleThis example of a half-wave rectifier uses
anMBRM110LT3 Schottky diode and the MCP661 opamp with different
load resistors. For this example, thevalue of the load resistor is
less than 1 kΩ, to avoidglitches in the negative cycle. The
Schottky diode ischosen for higher speed than a small signal
silicondiode. Figures 10 and 11 below are examples of a1 kHz input
signal and different load resistors. Note thatfor the small values
of the resistor (i.e. 100Ω), the glitchis smaller.
FIGURE 10: Half-Wave Rectifier with RL = 100Ω.
FIGURE 11: Half-Wave Rectifier with RL = 1 kΩ.
Improved Op Amp Half-Wave RectifierFigure 12 shows a half-wave
rectifier circuit withimproved performance. The additional diode
preventsthe op amp's output from swinging to the negativesupply
rail. The low level linearity is also improved.Although the op amp
still operates in open-loop at thepoint where the input swings from
positive to negativeor vice versa, the range is limited by the
diode and theload resistor.When the input signal is positive, D1 is
open and D2conducts. The output signal is zero because one sideof
R2 is connected to the virtual ground, with no currentthrough it.
When the input is negative, D1 conducts andD2 is open. The output
follows the positive input cyclewith a gain of G = -R2/R1.
FIGURE 12: Have-Wave Rectifier Circuit Improvement.This type of
circuit also has limitations. The inputimpedance is determined by
the input resistor. It mustbe driven from a low-impedance source.
Likewise, theinput resistor R3 shown in Figure 12 is also
optional,and is needed only if there is no DC path to ground.
Choosing the ComponentsRefer to the section Selecting the Op Amp
in the ActiveHalf-Wave Rectifier section, and to the
sectionSelecting the Diode in the section Basic Rectifiers,
fordetails on choosing the appropriate components.
SELECTING THE RESISTORSThe DC gain is determined in Equation
4:
EQUATION 4:
-0.2-0.15
-0.1-0.05
00.05
0.10.15
0.2
Time (500 µs/div)
Mag
nitu
de (V
)
VOUT VIN
RL = 100 Ohm
-0.2-0.15
-0.1-0.05
00.05
0.10.15
0.2
Time (500 µs/div)
Mag
nitu
de (V
)
RL = 1 k�
��������VOUT VIN
VOUT
RL
VIN R1
R2
D1
D2-+AO1
R3
R2R1------
where G = DC gain
G = –
DS01353A-page 4 © 2011 Microchip Technology Inc.
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AN1353
Resistors R1 and R2 are selected based on theapplication
design:• For a general purpose application, the resistor’s
value should be between 1 kΩ and 100 kΩ.• For a high speed
application, the resistor’s value
should be between 100 Ω and 1 kΩ (consume more power)
• For portable applications between 1 MΩ and 10 MΩ.
The R3 is added to minimize the error caused by theinput bias
current.
EQUATION 5:
Advantages and DisadvantagesTable 3 shows the main advantages
anddisadvantages of an improved half-wave rectifier.
Practical ExampleThe example in Figure 13 is based on the
circuit inFigure 12, and uses the MCP661 op amp, twoMBRM110LT3
Schottky diodes, RL = 1 kΩ, R2 =10 kΩand R1 = 1 kΩ. The input
frequency is 1 kHz.
FIGURE 13: Improved Half-Wave Rectifier with RL = 1 KΩ.
For an input frequency under 600 kHz, the circuitperforms
properly. For frequencies larger than thisvalue, the output signal
is distorted.
FIGURE 14: Circuit Behavior with 600 kHz Input Frequency.To
design a negative half-wave rectifier using the samecomponents, we
only have to invert the diodes, asshown in the circuit in Figure
15.
FIGURE 15: Negative Half-Wave Rectifier.
FIGURE 16: Negative Cycle Rectifier Sample.
TABLE 3: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Good linearity - Uses more components- The second diode
prevents the op amp from swinging into the negative cycle
- Low impedance because of R1
R3R1 R2×
R1 R2+--------------------=
-0.2-0.15-0.1
-0.050
0.050.1
0.150.2
Time (1 ms/div)
Mag
nitu
de (V
)
RL = 1 k�
VOUT VIN
-1-0.8-0.6-0.4-0.2
00.20.40.60.8
1
Time (0.2 µs/div)
Mag
nitu
de (V
)
VOUTVIN
VOUT
RL
VIN R1
R2
D1D2-+AO1
R3
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
Time (1 ms/div)
Mag
nitu
de (V
)
VOUT VIN
© 2011 Microchip Technology Inc. DS01353A-page 5
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AN1353
ACTIVE FULL-WAVE RECTIFIERFull-wave rectifiers are more complex,
compared to thehalf-wave circuits. Full-wave rectifiers output
onepolarity of the input signal and invert the other. A circuitfor
a full-wave rectifier is illustrated in Figure 17.
FIGURE 17: Full-Wave Rectifier Circuit.When in the negative
cycle of the input signal, diode D1is forward biased, and the
output voltage follows theinput. When the input signal (VIN) is
positive, D1 is non-conductive and the input signal passes through
thefeedback resistor (R2), which forms a voltage dividerwith R1 and
RL. Equation 6 shows the calculation forthe output voltage:
EQUATION 6:
When -GM = GP, the full-wave output is symmetric.Note that the
output is not buffered, so it should beconnected only to a circuit
with high impedance, muchhigher than RL.
Choosing the Components
Refer to the section Selecting the Diode in the section Basic
Rectifiers, and to the section Selecting the Op Amp in the section
Active Half-Wave Rectifier, for details on choosing the appropriate
components.
SELECTING THE RESISTORSWhen selecting the resistors for the
circuit in Figure 17,-GM must be equal to GP. The result is shown
inEquation 7:
EQUATION 7:
R3 is added to minimize the error caused by the inputbias
current. Refer to the section Selecting theResistors, in the
section Improved Op Amp Half-WaveRectifier, for details on the
selection of the resistor.
Advantages and Disadvantages
Practical ExampleThis design uses an MCP661 and a general
purposediode rectifier 1N4148. The input frequency is 1 kHz.Table 5
shows the resistor values recommended toobtain the same amplitude
with each input cycle:
The values of the resistors can be scaled depending onthe
application: high speed, portable or generalpurpose. For more
details, refer to the sectionSelecting the Resistors, in the
section Improved OpAmp Half-Wave Rectifier. Figure 18 shows the
result ofthe full-wave rectifier circuit simulation.
VOUT
R3
+-
VIN R1
R2
V+
D1RL
AO1
VOUT VIN GM× VIN 0;=
GPRL
R1 R2 RL+ +-----------------------------------=
TABLE 4: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Uses only one op amp - Low input resistance- Uses a small
number of external components
- The source and load resistance affect rectifying
- Uses a single supply - A reactive load (capacitor or coil)
cannot be tolerated without a buffer- Has a low impedance because
of R1
TABLE 5: VALUES FOR RECTIFIED, EQUAL AMPLITUDE
Resistor Value (kΩ)
R1 2R2 1RL 3
R2 R1 R2 RL+ +( )× R1 RL×=
DS01353A-page 6 © 2011 Microchip Technology Inc.
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AN1353
FIGURE 18: Full-Wave Rectifier Circuit Simulation with the
Recommended Values of the Resistors.
TWO STAGE OP AMP FULL-WAVE RECTIFIERAnother full-wave rectifier
can be obtained by includingan adder to the single-wave rectifier,
which subtractsVIN from the rectified signal. The rectifier stage
consistsof AO1, R1, R2, D1 and D2, while the adder stageconsists of
AO2, R3, R4 and R5.
FIGURE 19: Two Stage Op Amp Full-Wave Rectifier Circuit.When VIN
is positive, D1 is forward-biased and D2 isreverse-biased, while
when VIN is negative, D2 isforward-biased and D1 is
reversed-biased. The secondstage adds VIN and VO1 and inverts the
polarity of theresulting signal. The output voltage for the
positivecycle of the input voltage is calculated in Equation 8.For
the negative cycle of the input voltage (VIN), D1blocks the signal,
while D2 conducts the whole currentcoming from the input. In this
case, the output voltagefor the first stage is VO1 = 0V. For the
positive cycle of the input signal, VO1 isnegative and, in this
case, the adder stage combinesthe input signals with equal
amplitudes, one positiveand one negative.
EQUATION 8:
Equation 9 calculates the output voltage:
EQUATION 9:
Choosing the ComponentsTo obtain a good performance for the two
stage circuit,the tolerance of resistors R1 to R5 should be 1%,
orbetter; this makes the gains (for negative and positiveVIN) match
well. The circuit in Figure 19 has a goodlinearity, down to a
couple of mV at low frequencies, butthe high-frequency response is
limited by the op ampbandwidth.Refer to the section Selecting the
Diode in the sectionBasic Rectifiers, and to the section Selecting
the OpAmp in the section Active Half-Wave Rectifier, fordetails on
choosing the appropriate components.
SELECTING THE RESISTORSR1 and R2 give the gain for the first
stage; R3 and R5for the second stage.To get the same amplitude for
both cycles, chooseR1 = R3 = R4 and R2 = R5 = 2 x R1.
EQUATION 10:
R6 is added to minimize the error caused by the inputbias
current. Refer to the section Selecting theResistors, in the
section Improved Op Amp Half-WaveRectifier, for details on choosing
the appropriatecomponents.If a greater sensitivity and high
frequency is desired, itis recommended to use lower resistance
value, highspeed diodes and faster op amps.
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
Time (1ms/div)
Mag
nitu
de (V
)
VOUT VIN
VOUTR4
VIN R1 R3
R2
R5D1
D2
VO1AO1
AO2
-+
-+
R6
VO1 VIN G when ,× VIN > 0=
Where:
GR1–
R2---------=
VO1 0 when VIN 0
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AN1353
Advantages and DisadvantagesTable 6 shows the advantages and
disadvantages of atwo stage op amp full-wave rectifier.
Practical ExampleThis example uses the MCP6021 device, two
1N4148diodes, R1 = 1 kΩ, R2 = 2 kΩ, R3 = 1 kΩ, R4 = 1 kΩ,and R5 = 2
kΩ. The input signal frequency is f = 1 kHz.Figure 20 shows the
result of the simulation for the full-wave rectifier shown in
Figure 19:
FIGURE 20: Full-Wave Rectifier Circuit Simulation.For more
topologies of the full-wave rectifier, refer tothe Appendix
section.Figure 21 shows the behavior of the circuit at themaximum
frequency tolerated.
FIGURE 21: Circuit Behavior when Input Frequency = 100 kHz.
BASIC PEAK DETECTORSThe purpose of this circuit is to detect the
maximummagnitude of a signal over a period of time. Theoperation of
a peak detector can be illustrated using asimple diode and
capacitor, as shown in Figure 22.
FIGURE 22: Basic Peak Detector Operation.
Choosing the ComponentsWhen choosing the resistor, the limits
must beconsidered: rdf τ2 >> 1/fc, where fm is the
modulationfrequency and fc is the carrier frequency.
TABLE 6: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Very good performance - Uses two op amps- Low output impedance
- Low input resistance
- Multiple passive components
-1-0.8-0.6-0.4-0.2
00.20.40.60.8
1
Time (1 ms/div)
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)
VOUT VIN
-1-0.8-0.6-0.4-0.2
00.20.40.60.8
1
Time (10 µs/div)
Mag
nitu
de (V
)
VOUT VIN
C1
D1VIN VOUT
R1
ΔVVPEAK
f τ2×-------------------=
Where:
VPEAK = Amplitude maximum value f = Input signal frequency
τ2 = Discharge time constant
VDROP VPEAKt
τ2-----–⎝ ⎠
⎛ ⎞exp×=
Where: τ2 = time constant
DS01353A-page 8 © 2011 Microchip Technology Inc.
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AN1353
Advantages and DisadvantagesTable 7 identifies some of the
advantages anddisadvantages of the peak detectors.
Practical ExampleThe simulation in Figure 23 shows that this
circuit doesnot reach the peak amplitude of the input signal, but
isgood for quickly following sudden changes in thesignal's
amplitude. However, it has significant ripple.This example uses a
1N4148 diode, C1 = 1 µF,R1 = 100 kΩ and f = 1 kHz.
FIGURE 23: Peak Detector Simulation.
Two-Stage Active Peak DetectorIn many applications, the voltage
drop is not desired.To avoid this, we need to include a diode into
the loopof the op amp, as shown in Figure 24.A two-stage peak
detector is shown in Figure 24. In thiscircuit, AO1, R1, R2, D1, R5
and C1 represent the firststage, while AO2, R3 and R4 is the second
stage. AO1charges the capacitor up to the peak value, and AO2acts
as an output buffer. AO1 removes the variability ofthe input
impedance, while AO2 removes the variabilityof the output
impedance.
FIGURE 24: Two-Stage Peak Detector Circuit.The time constant for
charging C1 is very short, andprimarily consists of the C1 and the
forward resistanceof the diode. Thus, C1 charges almost instantly
to thepeak output of the input signal (VIN). When VIN goesbelow the
output signal (VOUT), diode D1 becomesreverse-biased. The only
discharge path for C1 isthrough R5, via leakage or op amp bias
currents. Thedischarge time constant is much longer than the
chargetime constant, so C1 holds its charge and presents asteady
input voltage to AO2 that is equal to the peakamplitude of the
input signal. AO2 is a buffer amplifierthat prevents unintentional
discharging of the C1,caused by the loading impedance of the
followingcircuit. If the R5C1 time constant is too short, then
thevoltage on C1 will not be constant, and will have a highvalue of
ripple. On the other hand, if the R5C1 timeconstant is too long,
the circuit cannot respond quicklyto the changes in the input
amplitude. The lower frequency limit is the frequency that
causesthe ripple voltage to exceed the maximum allowablelevel. It
can be estimated by applying the basicdischarge equation for
capacitors (Equation 13):
EQUATION 13:
The response time describes how quickly C1 canrespond to the
decreases in the magnitude of the inputsignal. This can be computed
from the basic dischargeequation. However, if we assume that the
capacitor ischarged to peak and discharges towards an eventualvalue
of 0, we can use the simplified form(Equation 14).
TABLE 7: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Uses few components
- The output voltage is one diode drop below the actual
out-put
- Very low cost - The input impedance is variable due to the
input characteristics of the diode- The discharge is very slow due
to the leakage current
02468
1012141618
Time (20 ms/div)
Mag
nitu
de(V
)
VOUTVIN
VOUT
R5
VIN
C1R1
R2
D1
R3
R4
AO1AO2
-+
-+
fO1
R5 C1× ln×V Vo–
V Vc–----------------
----------------------------------------------------------=
Where:
V = Capacitor’s discharge voltage VC = Minimum allowable voltage
on the
capacitorVO = Initial charge of the capacitor
© 2011 Microchip Technology Inc. DS01353A-page 9
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AN1353
EQUATION 14:
Choosing the ComponentsRefer to the section Selecting the Diode
in the sectionBasic Rectifiers, and to the section Selecting the
OpAmp, in the section Active Half-Wave Rectifier, fordetails on
choosing the appropriate components.
SELECTING THE RESISTORSR3 limits the current into the positive
input of the AO2when power is disconnected from the circuit.
Withoutthis resistor, the AO2 may be damaged as C1discharges
through it. For capacitors smaller than 1 µF,resistor R3 can
normally be omitted. Resistor R4minimizes the effects of the bias
currents in AO2.Resistor R2 limits the current into the negative
input ofAO1 when power is removed from the circuit. There are two
conflicting circuit parameters that affectthe choice of the values
for R5 and C1: allowable ripplevoltage across C1 and response time.
In general, afaster response time leads to greater ripple.Refer to
the section Selecting the Resistors, in thesection Improved Op Amp
Half-Wave Rectifier, fordetails on choosing the appropriate
components.
Practical ExampleFigure 25 illustrates the simulation result for
one peakdetector, realized with MCP661 device, diode 1N4148,R5 =
100 kΩ and C1 = 1 µF. Input signal has thefrequency equal to 1
kHz.
FIGURE 25: One Peak Detector Simulation Results.
For more topologies on the peak detectors, refer to
theAppendix.
BASIC CLAMPA clamp is used to shift the DC reference level of
theinput signal. Figure 26 shows a basic diode clamp. Itspurpose is
to shift the average or the DC level of theinput signal without
altering the wave shape.When VOUT > VREF and the input signal is
fast, D1 is off,C1 acts like a short circuit, and VOUT looks like
theinput. With slow signals, C1 acts like an open circuit andVOUT
will exponentially decay towards VREF.When VOUT < VREF, VOUT
becomes VREF - VD(on), D1turns on and C1 is forced to accept a new
voltage thatshifts the input to the desired minimum VOUT.For
low-amplitude signals, the diode drop becomessignificant. In fact,
the circuit cannot be used at all if thepeak input signal is below
the diode threshold, sincethe diode cannot be forward-biased. An
active clamp isneeded for signals with an amplitude of
millivolts.Figure 26 shows a negative clamp; it clamps thenegative
extreme of the signal to (near) VREF.Reversing the diode creates a
positive clamp.
FIGURE 26: Basic Diode Negative Clamp.
Practical ExampleFigure 27 shows a simulation for the above
schematicwith VREF = 2V, C1 = 1 nF and diode 1N4148. Theinput
signal has the frequency equal to 500 Hz.
FIGURE 27: Basic Diode Clamp – Circuit Simulation.
tR R5 C× 1
VPK old( )VPK new( )---------------------------ln×=
Where:
VPK(old) = Peak input signal amplitude before the decrease
VPK(new) = Peak input signal amplitude after the decrease
0.10.110.120.130.140.150.160.170.180.19
0.2
Time (1 ms/div)
Mag
nitu
de (V
) VOUT VIN
AC
C1
D1
VOUT
VREF
R1
-1-0.5
00.5
11.5
22.5
33.5
4
Time (10 ms/div)
Mag
nitu
de (V
)
VOUTVIN
DS01353A-page 10 © 2011 Microchip Technology Inc.
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Active ClampTo reduce the threshold voltage of the diode, and
forlinearization, the circuit needs a diode in the feedbackloop of
the operational amplifier. Figure 28 shows an op amp clamp where
the inputsignal is positive and D1 is forward-biased. The
diodeconverts the circuit into a voltage follower withreference to
the positive input. This means that theoutput of the op amp has
approximately the samevoltage as the reference voltage. When the
input signalis negative, the diode is reversed-biased. The op
ampwill also be at the reference voltage level. Capacitor C1is
charged with the difference of potential between VINand VREF. This
effectively disconnects the op amp fromthe circuit so the output
will be the same as VIN plusC1’s voltage. The capacitor has no
rapid discharge pathand will act as a DC source, providing the
clampingaction.
FIGURE 28: Op Amp Clamp.
Choosing the ComponentsRefer to the section Selecting the Diode,
in the sectionBasic Rectifiers, and to the section Selecting the
OpAmp, in the section Active Half-Wave Rectifier, fordetails on
choosing the appropriate components.
SELECTING THE RESISTANCE AND THE CAPACITORThe input impedance of
the circuit varies with the inputfrequency and with the state of
the circuit. As frequencyincreases, the reactance of C1 decreases
and lowersthe input impedance.Usually, R1 gives the input
impedance, so the chosenresistance should be the minimum of the
desiredimpedance.The value of C1 is shown in Equation 15:
EQUATION 15:
R(-) and R(+) are calculated as a ratio between themaximum
voltage allowed by the circuit on the inputterminal and the maximum
input bias current.
Advantages and DisadvantagesTable 8 shows the main advantages
anddisadvantages of the clamp circuit.
Practical ExampleThis example uses the MCP6021 device, a
1N4148diode, C1 = 150 nF, R1 = 1.2 kΩ, R2 = 43 kΩ,R3 = 47 kΩ, VREF
= 0.7V. The input frequency is10 kHz. Figure 29 shows the
simulation result for thisexample.
VOUTR1VIN
VREF
C1
D1
-+ -
+
AO1AO2
TABLE 8: ADVANTAGES AND DISADVANTAGES OF THE CIRCUIT
Advantages Disadvantages
- Uses only one op amp - The input impedance var-ies with the
input frequency
- Few external components
- The output impedance varies with the input frequency
- Adjustable level for voltage reference
- Uses a potentiometer
- Uses a positive and a negative voltage reference
C116.7
R flow×-------------------=
Where:
flow = minimum frequency desired
R = rdr || R (-) || R (+)Where:
rdr = Reversed diode resistanceR(-) = Input resistance on the
negative
terminal of AO1R(+) = Input resistance on the positive ter-
minal of AO1 for voltage follower
© 2011 Microchip Technology Inc. DS01353A-page 11
-
AN1353
FIGURE 29: Op Amp Clamp Circuit Simulation Result.
CONCLUSIONThis application note examined the circuits that
canrectify the amplitude signal, detect the peak signal andchange
the DC level of waveforms. The op amp-basedsolutions bring
improvements to the basic solutions,such as operating with
millivolt signals or isolating theoutput and input impedance. The
applicationsproposed are based on low cost op amps, and
offercircuits with few peripheral components, givingdesigners
simple, but effective solutions to theirproblems.
-1-0.5
00.5
11.5
22.5
33.5
4
Time (100 ms/div)
Mag
nitu
de (V
)
VOUTVIN
DS01353A-page 12 © 2011 Microchip Technology Inc.
-
AN1353
APPENDIXThis Appendix includes schematics for additional halfand
full-wave rectifiers, peak detectors and clamps.Each of these can
be implemented using the rulespresented in this application
note.
Half-Wave RectifiersFigures 30 – 33 show more half-wave
rectifiers withtheir DC transfer functions.
FIGURE 30: Positive Half-Wave Rectifier 1.
FIGURE 31: Negative Half-Wave Rectifier 1.
FIGURE 32: Positive Half-Wave Rectifier 2.
FIGURE 33: Negative Half-Wave Rectifier 2.Every one of these
circuits can be used to design full-wave rectifiers by adding an op
amp adder. Thismethod is illustrated in Figure 19.
Full-Wave RectifiersThe circuits shown in this section are based
on half-wave rectifiers. For example, the circuit in Figure
36contains two half-wave rectifiers, one for the positivecycle, the
other for the negative cycle, and onedifference (or adder)
amplifier.For Figures 34 – 36, VOUT is positive. Reversing
thediodes creates a negative rectifier.
FIGURE 34: Two Stage Full-Wave Rectifier 1.
FIGURE 35: Two Stage Full-Wave Rectifier 2.
R2
D1
D2
R1
VOUT
VIN
VOUT-+
AO1
VIN
VOUT
VIN
R2
D1
D2
R1 VOUT-+
AO1
VIN
R2
D1
D2R1
VOUT
VINVOUT-
+AO1
VIN
VOUT
VIN
R2
D1
D2R1 VOUT-
+
VIN
AO1
VOUT
R4
VIN
R1
R3R2
D1
D2AO1AO2
-+
-+
R4
R3
R2
D1
D2
R1
R4
R5
VOUT
AO1AO2
-+
-+
VIN
© 2011 Microchip Technology Inc. DS01353A-page 13
-
AN1353
FIGURE 36: Three Stage Full-Wave Rectifier.
Peak DetectorsThe circuit in Figure 37 has the capacitor
dischargethrough R2, which causes the output to droop. DiodeD2
provides the local feedback around AO1, once apeak has been
detected. This prevents AO1 fromsaturating during the peak hold
mode and decreasesthe peak acquisition time. You can omit D2, but
thecircuit will be slower when detecting peaks.
FIGURE 37: Peak Detector Rectifier 1.
For Figure 38, VOUT is positive. Reversing the diodecreates a
negative rectifier.
FIGURE 38: Peak Detector Rectifier 2.
For Figure 39, VOUT is positive. Reversing the diodescreates a
negative rectifier. To reset this circuit, we canuse a relay reed,
or a transistor with a low leakagecurrent.
FIGURE 39: Peak Detector Rectifier 3.
ClampFigure 40 shows another positive active clamp wherethe
reference voltage can be adjusted. If the diode isinverted, a
negative active clamp will result.
FIGURE 40: Active Clamp Sample.
VOUT
VIN R1
R2
R1
R3
R3
R2
D1D2
D1D2
R5
R4
R6
R7
AO1
AO2
AO3
-+
-+
-+
VOUTVIN R1
R2
R3
D1D2
AO1-+ AO2
-+
C1
D1 VOUT-+
AO1
-+
AO2
VIN
VOUTVIN
R2
D1D2
AO1 AO2-+
-+
C2
-VC
Run
0VReset
C1
VOUTR1
+-
VIN
P1 R3R2
C1
C2
D1
+V -V
AO1
VREF
DS01353A-page 14 © 2011 Microchip Technology Inc.
-
Note the following details of the code protection feature on
Microchip devices:• Microchip products meet the specification
contained in their particular Microchip Data Sheet.
• Microchip believes that its family of products is one of the
most secure families of its kind on the market today, when used in
the intended manner and under normal conditions.
• There are dishonest and possibly illegal methods used to
breach the code protection feature. All of these methods, to our
knowledge, require using the Microchip products in a manner outside
the operating specifications contained in Microchip’s Data Sheets.
Most likely, the person doing so is engaged in theft of
intellectual property.
• Microchip is willing to work with the customer who is
concerned about the integrity of their code.
• Neither Microchip nor any other semiconductor manufacturer can
guarantee the security of their code. Code protection does not mean
that we are guaranteeing the product as “unbreakable.”
Code protection is constantly evolving. We at Microchip are
committed to continuously improving the code protection features of
ourproducts. Attempts to break Microchip’s code protection feature
may be a violation of the Digital Millennium Copyright Act. If such
actsallow unauthorized access to your software or other copyrighted
work, you may have a right to sue for relief under that Act.
Information contained in this publication regarding
deviceapplications and the like is provided only for your
convenienceand may be superseded by updates. It is your
responsibility toensure that your application meets with your
specifications.MICROCHIP MAKES NO REPRESENTATIONS ORWARRANTIES OF
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Microchipintellectual property rights.
© 2011 Microchip Technology Inc.
Trademarks
The Microchip name and logo, the Microchip logo, dsPIC, KEELOQ,
KEELOQ logo, MPLAB, PIC, PICmicro, PICSTART, PIC32 logo, rfPIC and
UNI/O are registered trademarks of Microchip Technology
Incorporated in the U.S.A. and other countries.
FilterLab, Hampshire, HI-TECH C, Linear Active Thermistor,
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Analog-for-the-Digital Age, Application Maestro, CodeGuard,
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FanSense, HI-TIDE, In-Circuit Serial Programming, ICSP, Mindi,
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Omniscient Code Generation, PICC, PICC-18, PICDEM, PICDEM.net,
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SQTP is a service mark of Microchip Technology Incorporated in
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All other trademarks mentioned herein are property of their
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© 2011, Microchip Technology Incorporated, Printed in the
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Printed on recycled paper.
ISBN: 978-1-60932-931-0
DS01353A-page 15
Microchip received ISO/TS-16949:2002 certification for its
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DS01353A-page 16 © 2011 Microchip Technology Inc.
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IntroductionBasic RectifiersFIGURE 1: Negative Half-Wave
Rectifier.FIGURE 2: Positive Half-Wave Rectifier.FIGURE 3:
Full-Wave Rectifier.Choosing the ComponentsTABLE 1: Advantages and
Disadvantages of the Circuit
Practical ExamplesFIGURE 4: Negative Half-Wave Rectifier
Sample.FIGURE 5: Positive Half-Wave Rectifier Sample.FIGURE 6:
Full-Wave Rectifier Sample.
Active Half-Wave RectifierFIGURE 7: Op Amp Half-Wave
Rectifier.FIGURE 8: Circuit Behavior on Low Frequency.FIGURE 9:
Output Limitation on High-Frequency Input Signals.Choosing the
ComponentsAdvantages and DisadvantagesTABLE 2: Advantages and
Disadvantages of the Circuit
Practical ExampleFIGURE 10: Half-Wave Rectifier with RL =
100W.FIGURE 11: Half-Wave Rectifier with RL = 1 kW.
Improved Op Amp Half-Wave RectifierFIGURE 12: Have-Wave
Rectifier Circuit Improvement.
Choosing the ComponentsAdvantages and DisadvantagesTABLE 3:
Advantages and Disadvantages of the Circuit
Practical ExampleFIGURE 13: Improved Half-Wave Rectifier with RL
= 1 KW.FIGURE 14: Circuit Behavior with 600 kHz Input
Frequency.FIGURE 15: Negative Half-Wave Rectifier.FIGURE 16:
Negative Cycle Rectifier Sample.
Active Full-Wave RectifierFIGURE 17: Full-Wave Rectifier
Circuit.Choosing the ComponentsAdvantages and DisadvantagesTABLE 4:
Advantages and Disadvantages of the Circuit
Practical ExampleTABLE 5: Values For rectified, equal
amplitudeFIGURE 18: Full-Wave Rectifier Circuit Simulation with the
Recommended Values of the Resistors.
Two Stage Op Amp Full-Wave RectifierFIGURE 19: Two Stage Op Amp
Full- Wave Rectifier Circuit.Choosing the ComponentsAdvantages and
DisadvantagesTABLE 6: Advantages and Disadvantages of the
Circuit
Practical ExampleFIGURE 20: Full-Wave Rectifier Circuit
Simulation.FIGURE 21: Circuit Behavior when Input Frequency = 100
kHz.
Basic Peak DetectorsFIGURE 22: Basic Peak Detector
Operation.Choosing the ComponentsAdvantages and DisadvantagesTABLE
7: Advantages and Disadvantages of the Circuit
Practical ExampleFIGURE 23: Peak Detector Simulation.
Two-Stage Active Peak DetectorFIGURE 24: Two-Stage Peak Detector
Circuit.
Choosing the ComponentsPractical ExampleFIGURE 25: One Peak
Detector Simulation Results.
Basic ClampFIGURE 26: Basic Diode Negative Clamp.Practical
ExampleFIGURE 27: Basic Diode Clamp – Circuit Simulation.
Active ClampFIGURE 28: Op Amp Clamp.
Choosing the ComponentsAdvantages and DisadvantagesTABLE 8:
Advantages and Disadvantages of the Circuit
Practical ExampleFIGURE 29: Op Amp Clamp Circuit Simulation
Result.
ConclusionAppendixHalf-Wave RectifiersFIGURE 30: Positive
Half-Wave Rectifier 1.FIGURE 31: Negative Half-Wave Rectifier
1.FIGURE 32: Positive Half-Wave Rectifier 2.FIGURE 33: Negative
Half-Wave Rectifier 2.
Full-Wave RectifiersFIGURE 34: Two Stage Full-Wave Rectifier
1.FIGURE 35: Two Stage Full-Wave Rectifier 2.FIGURE 36: Three Stage
Full-Wave Rectifier.
Peak DetectorsFIGURE 37: Peak Detector Rectifier 1.FIGURE 38:
Peak Detector Rectifier 2.FIGURE 39: Peak Detector Rectifier 3.
ClampFIGURE 40: Active Clamp Sample.Corporate
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