Post on 22-Apr-2020
Introduction toOperational Amplifiers
Phyllis R. Nelson
prnelson@csupomona.edu
Professor, Department of Electrical and Computer Engineering
California State Polytechnic University, Pomona
P. R. Nelson – ECE 322 – Fall 2012 – p. 1/50
Ideal Op Amp
Op amp symbol:−
++ input (Vp)- input (Vm)
output (Vo)
Ideal op amp model:1. If there is a branch connecting the output to
the “-” input, Vp − Vm = 0
2. There are no currents flowing into or out ofthe input terminals.
P. R. Nelson – ECE 322 – Fall 2012 – p. 2/50
How It Works
If there is a branch connecting theoutput to the “-” input,
the op amp adjusts Vo so that
Vp = Vm
. . . unless it can’t. (Thinking ahead, what couldgo wrong?)
P. R. Nelson – ECE 322 – Fall 2012 – p. 3/50
Inverting Amp
−
+
R1
input (Vi)
R2
output (Vo)
P. R. Nelson – ECE 322 – Fall 2012 – p. 4/50
Inverting Amp
−
+
R1
input (Vi)
R2
output (Vo)
Vm = Vp = 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 4/50
Inverting Amp
−
+
R1
input (Vi)
R2
output (Vo)
Vm = Vp = 0
Vi − Vm
R1
+Vo − Vm
R2
= 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 4/50
Inverting Amp
−
+
R1
input (Vi)
R2
output (Vo)
Vm = Vp = 0
Vi − Vm
R1
+Vo − Vm
R2
= 0
Vo
Vi
= −R2
R1
P. R. Nelson – ECE 322 – Fall 2012 – p. 4/50
Noninverting Amp
−
+input (Vi)
R1
R2
output (Vo)
P. R. Nelson – ECE 322 – Fall 2012 – p. 5/50
Noninverting Amp
−
+input (Vi)
R1
R2
output (Vo)
Vm = Vp = Vi
P. R. Nelson – ECE 322 – Fall 2012 – p. 5/50
Noninverting Amp
−
+input (Vi)
R1
R2
output (Vo)
Vm = Vp = Vi
Vm − 0
R1
+Vm − Vo
R2
= 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 5/50
Noninverting Amp
−
+input (Vi)
R1
R2
output (Vo)
Vm = Vp = Vi
Vm − 0
R1
+Vm − Vo
R2
= 0
Vo
Vi
= 1 +R2
R1
P. R. Nelson – ECE 322 – Fall 2012 – p. 5/50
Questions?
???
P. R. Nelson – ECE 322 – Fall 2012 – p. 6/50
Example 1
−
+
−+Vref
R1
Vi
R2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 7/50
Example 1
−
+
−+Vref
R1
Vi
R2
Vo
Vm = Vp = Vref
Vi − Vm
R1
+Vo − Vm
R2
= 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 7/50
Example 1 cont’d
−
+
−+Vref
R1
Vi
R2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 8/50
Example 1 cont’d
−
+
−+Vref
R1
Vi
R2
Vo
Vo = −
(
R2
R1
)
Vi +
(
1 +R2
R1
)
Vref
P. R. Nelson – ECE 322 – Fall 2012 – p. 8/50
Example 1 cont’d
−
+
−+Vref
R1
Vi
R2
Vo
Vo = −
(
R2
R1
)
Vi +
(
1 +R2
R1
)
Vref
SUPERPOSITION!
P. R. Nelson – ECE 322 – Fall 2012 – p. 8/50
Example 2
−
+
Ii
R1
RL IL
RSVS
VO
P. R. Nelson – ECE 322 – Fall 2012 – p. 9/50
Example 2
−
+
Ii
R1
RL IL
RSVS
VO
IL =VS − VO
RL
Ii = −VS
R1
= IL +VS
RS
P. R. Nelson – ECE 322 – Fall 2012 – p. 9/50
−
+
Ii
R1
RL IL
RSVS
VO
IL
Ii
=
(
1 +RS
R1
)
P. R. Nelson – ECE 322 – Fall 2012 – p. 10/50
−
+
Ii
R1
RL IL
RSVS
VO
IL
Ii
=
(
1 +RS
R1
)
This is a current-controlled currentsource (current amplifier).
P. R. Nelson – ECE 322 – Fall 2012 – p. 10/50
Example 3
−
+
Vi
R1
R2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 11/50
Example 3
−
+
Vi
R1
R2
Vo
Rule 1 doesn’t apply.
P. R. Nelson – ECE 322 – Fall 2012 – p. 11/50
Example 3
−
+
Vi
R1
R2
Vo
Rule 1 doesn’t apply.Now what?
P. R. Nelson – ECE 322 – Fall 2012 – p. 11/50
A Comparison
−
+
Vi
R1
R2
Vo−
+Vi
R1
R2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 12/50
A Comparison
−
+
Vi
R1
R2
Vo−
+Vi
R1
R2
Vo
The same external circuit, but different opampinput connections.
P. R. Nelson – ECE 322 – Fall 2012 – p. 12/50
A Comparison
−
+
Vi
R1
R2
Vo−
+Vi
R1
R2
Vo
The same external circuit, but different opampinput connections.
Look again at a familiar example. . .P. R. Nelson – ECE 322 – Fall 2012 – p. 12/50
Inverting Amp Voltage
R1 = 1 kΩ
R2 = 100 kΩ−
+
R1
Vi
R2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 13/50
Inverting Amp Voltage
R1 = 1 kΩ
R2 = 100 kΩ−
+
R1
Vi
R2
Vo
Vi = 0.5 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 13/50
Inverting Amp Voltage
R1 = 1 kΩ
R2 = 100 kΩ−
+
R1
Vi
R2
Vo
Vi = 0.5 V
Vo
Vi
= −R2
R1
⇒ Vo = −50 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 13/50
Inverting Amp Voltage
R1 = 1 kΩ
R2 = 100 kΩ−
+
R1
Vi
R2
Vo
Vi = 0.5 V
Vo
Vi
= −R2
R1
⇒ Vo = −50 V
ALWAYS?
P. R. Nelson – ECE 322 – Fall 2012 – p. 13/50
Inverting Amp Power
−
+
R1
Vi
R2
Vo
RL
R1 = RL = 1 kΩ
R2 = 100 kΩ
Vi = 0.5 V
Vo = −50 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 14/50
Inverting Amp Power
−
+
R1
Vi
R2
Vo
RL
R1 = RL = 1 kΩ
R2 = 100 kΩ
Vi = 0.5 V
Vo = −50 V
Pin =V 2
i
R1
= 0.25 mWPout =
[Vo]2
RL
= 2.5 W = 104Pin
P. R. Nelson – ECE 322 – Fall 2012 – p. 14/50
Inverting Amp Power
−
+
R1
Vi
R2
Vo
RL
R1 = RL = 1 kΩ
R2 = 100 kΩ
Vi = 0.5 V
Vo = −50 V
Pin =V 2
i
R1
= 0.25 mWPout =
[Vo]2
RL
= 2.5 W = 104Pin
What supplies this output power?P. R. Nelson – ECE 322 – Fall 2012 – p. 14/50
Ideal op amp model rule 1 (Vp − Vm = 0 if theoutput is connected to the “-” input) meansthat the op amp adjusts Vo.
P. R. Nelson – ECE 322 – Fall 2012 – p. 15/50
Ideal op amp model rule 1 (Vp − Vm = 0 if theoutput is connected to the “-” input) meansthat the op amp adjusts Vo.
The ideal op amp model thus assumes adependent voltage source inside the opamp.
P. R. Nelson – ECE 322 – Fall 2012 – p. 15/50
Ideal op amp model rule 1 (Vp − Vm = 0 if theoutput is connected to the “-” input) meansthat the op amp adjusts Vo.
The ideal op amp model thus assumes adependent voltage source inside the opamp.
This source must be able to supply power.
P. R. Nelson – ECE 322 – Fall 2012 – p. 15/50
Ideal op amp model rule 1 (Vp − Vm = 0 if theoutput is connected to the “-” input) meansthat the op amp adjusts Vo.
The ideal op amp model thus assumes adependent voltage source inside the opamp.
This source must be able to supply power.
The op amp requires an externalpower supply.
P. R. Nelson – ECE 322 – Fall 2012 – p. 15/50
Ideal op amp model rule 1 (Vp − Vm = 0 if theoutput is connected to the “-” input) meansthat the op amp adjusts Vo.
The ideal op amp model thus assumes adependent voltage source inside the opamp.
This source must be able to supply power.
The op amp requires an externalpower supply.The output voltage is limited by thepower supply voltage(s)!
P. R. Nelson – ECE 322 – Fall 2012 – p. 15/50
Output Voltage Limits
−
+Vp
Vm Vo
Vsp
Vsm Vo =
Vp lim Vp > Vm
0 Vp = Vm
Vm lim Vp < Vm
P. R. Nelson – ECE 322 – Fall 2012 – p. 16/50
Output Voltage Limits
−
+Vp
Vm Vo
Vsp
Vsm Vo =
Vp lim Vp > Vm
0 Vp = Vm
Vm lim Vp < VmVp lim . Vsp
Vm lim & Vsm
P. R. Nelson – ECE 322 – Fall 2012 – p. 16/50
Ideal Op Amp VTC
Vp − Vm
Vo
Vp lim
Vm lim
P. R. Nelson – ECE 322 – Fall 2012 – p. 17/50
Example 3 Continued
−
+
Vi
R1
R2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 18/50
Example 3 Continued
−
+
Vi
R1
R2
VoVp =
(
R1
R1 + R2
)
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 18/50
Example 3 Continued
−
+
Vi
R1
R2
VoVp =
(
R1
R1 + R2
)
Vo
Vi = Vp ⇒ Vo = 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 18/50
Example 3 Continued
−
+
Vi
R1
R2
VoVp =
(
R1
R1 + R2
)
Vo
Vi = Vp ⇒ Vo = 0
If Vi 6= Vp, Vo equals either Vp lim or Vm lim.
P. R. Nelson – ECE 322 – Fall 2012 – p. 18/50
Vout =
Vp lim Vi < Vp
0 Vi = Vp
Vm lim Vi > Vp
Vp =
(
R1
R1 + R2
)
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 19/50
Vout =
Vp lim Vi < Vp
0 Vi = Vp
Vm lim Vi > Vp
Vp =
(
R1
R1 + R2
)
Vo
These equations can’t be solved forVo given Vi,
P. R. Nelson – ECE 322 – Fall 2012 – p. 19/50
Vout =
Vp lim Vi < Vp
0 Vi = Vp
Vm lim Vi > Vp
Vp =
(
R1
R1 + R2
)
Vo
These equations can’t be solved forVo given Vi, but there is a range of Vi
for a given value of Vo.P. R. Nelson – ECE 322 – Fall 2012 – p. 19/50
Choose some values to make the calculationsconcrete:
R1 = 1 kΩ R2 = 5 kΩ V± lim = ±6.0 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 20/50
Choose some values to make the calculationsconcrete:
R1 = 1 kΩ R2 = 5 kΩ V± lim = ±6.0 V
If Vo = Vp lim = 6.0 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 20/50
Choose some values to make the calculationsconcrete:
R1 = 1 kΩ R2 = 5 kΩ V± lim = ±6.0 V
If Vo = Vp lim = 6.0 V
Vi ≤ Vp =
(
1 kΩ
1 kΩ + 5 kΩ
)
6 V = 1 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 20/50
Choose some values to make the calculationsconcrete:
R1 = 1 kΩ R2 = 5 kΩ V± lim = ±6.0 V
If Vo = Vp lim = 6.0 V
Vi ≤ Vp =
(
1 kΩ
1 kΩ + 5 kΩ
)
6 V = 1 V
If Vo = Vm lim = −6 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 20/50
Choose some values to make the calculationsconcrete:
R1 = 1 kΩ R2 = 5 kΩ V± lim = ±6.0 V
If Vo = Vp lim = 6.0 V
Vi ≤ Vp =
(
1 kΩ
1 kΩ + 5 kΩ
)
6 V = 1 V
If Vo = Vm lim = −6 V
Vi ≥ Vp =
(
1 kΩ
1 kΩ + 5 kΩ
)
(−6 V) = −1 VP. R. Nelson – ECE 322 – Fall 2012 – p. 20/50
Vo
Vi
−1 V < Vi < 1 V:Two values of Vo
Vi > 1 V: Vo = −6 V
Vi < −1 V: Vo = 6 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 21/50
Schmitt TriggerVo
Vi
Vp lim
Vm lim
Stable point
Stable pointSwitch point
Switch point
For Vi betweenthe switch points,the circuit hasmemory .
Vo depends onhistory as wellas the currentvalue of Vi.
P. R. Nelson – ECE 322 – Fall 2012 – p. 22/50
Example 4
−
+
R1
Vi
R2
Vo
Now it’s your turn.
Sketch Vo vs. Vi
P. R. Nelson – ECE 322 – Fall 2012 – p. 23/50
Hint 1
Vp =
(
R1
R1 + R2
)
Vo +
(
R2
R1 + R2
)
Vi
P. R. Nelson – ECE 322 – Fall 2012 – p. 24/50
Hint 1
Vp =
(
R1
R1 + R2
)
Vo +
(
R2
R1 + R2
)
Vi
WHY?
P. R. Nelson – ECE 322 – Fall 2012 – p. 24/50
Hint 1
Vp =
(
R1
R1 + R2
)
Vo +
(
R2
R1 + R2
)
Vi
WHY?
superposition and the voltage divider formula,or
KCL at the “+” op amp terminal plus algebra
P. R. Nelson – ECE 322 – Fall 2012 – p. 24/50
Hint 1
Vp =
(
R1
R1 + R2
)
Vo +
(
R2
R1 + R2
)
Vi
WHY?
superposition and the voltage divider formula,or
KCL at the “+” op amp terminal plus algebra
Voltage dividers, superposition, current dividers,and the Thevénin and Norton equivalent circuitscan save a lot of algebra, and potentially errors!
P. R. Nelson – ECE 322 – Fall 2012 – p. 24/50
Hints 2 & 3
Hint 2: The switch points are at Vp = 0 V.
P. R. Nelson – ECE 322 – Fall 2012 – p. 25/50
Hints 2 & 3
Hint 2: The switch points are at Vp = 0 V.
Hint 3:Viswitch
= −
(
R1
R2
)
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 25/50
Another SchmittTrigger
Vo
Vi
V lim
−Vm lim
− R1
R2
Vm lim
R1
R2
Vp lim
P. R. Nelson – ECE 322 – Fall 2012 – p. 26/50
Questions?
???
P. R. Nelson – ECE 322 – Fall 2012 – p. 27/50
Finite Open Loop Gain
−
++ input (Vp)- input (Vm)
output (Vo)
Rule 1 becomesVo = Avo (Vp − Vm)
P. R. Nelson – ECE 322 – Fall 2012 – p. 28/50
Finite Open Loop Gain
−
++ input (Vp)- input (Vm)
output (Vo)
Rule 1 becomesVo = Avo (Vp − Vm)
Vo
Vp − Vm
Avo = ∞
Avo finite
P. R. Nelson – ECE 322 – Fall 2012 – p. 28/50
Finite Open Loop Gain
−
++ input (Vp)- input (Vm)
output (Vo)
Rule 1 becomesVo = Avo (Vp − Vm)
Vo
Vp − Vm
Avo = ∞
Avo finite
Vp − Vm 6= 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 28/50
Finite Gain CircuitModel
If Vm lim < Vo < Vp lim the op amp can be modeledas a voltage-controlled voltage source:
Vm
Vp Vo
+− A(Vp − Vm)
P. R. Nelson – ECE 322 – Fall 2012 – p. 29/50
Inverting Amp, FiniteGain
−
+
R1
Vi
R2
Vo
Rule 1:
Vo = Avo (0 V − Vm)
Rule 2:Vi − Vm
R1
+Vo − Vm
R2
= 0
Vo
Vi
=− R2
R1
1 + 1
Avo
(
1 + R2
R1
)
P. R. Nelson – ECE 322 – Fall 2012 – p. 30/50
Differential InputOffset Voltage
The internal components that connect to the “+”and “-” terminals of an op amp are not perfectlyidentical due to process variations.
A small voltage difference VIO must be used atthe input of a real op amp to make the outputvoltage zero.
−
+
−+VIO
Vo = 0 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 31/50
The differential input offset voltage is modeled bya voltage source in series with the “+” terminal ofan ideal op amp.
−
+
Vm
−+
VIO
VpVo
The value of VIO can be either positive ornegative.
P. R. Nelson – ECE 322 – Fall 2012 – p. 32/50
Example 5: Integrator
Although I used resistors to derive theclosed-loop gain of the inverting amplifier, thesame reasoning works with compleximpedances.
−
+
R1
Vi
C2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 33/50
Example 5: Integrator
Although I used resistors to derive theclosed-loop gain of the inverting amplifier, thesame reasoning works with compleximpedances.
−
+
R1
Vi
C2
Vo
Vo =−Vi
jωR1C2
= −1
R1C2
∫
Vi dt
P. R. Nelson – ECE 322 – Fall 2012 – p. 33/50
−
+−+
VIO
R1
Vi
C2
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 34/50
−
+−+
VIO
R1
Vi
C2
Vo
Vo = −VIO −1
R1C2
∫
(Vi − VIO) dt
P. R. Nelson – ECE 322 – Fall 2012 – p. 34/50
−
+−+
VIO
R1
Vi
C2
Vo
Vo = −VIO −1
R1C2
∫
(Vi − VIO) dt
If Vi = 0, there is still a signal (VIO) to integrate.The steady-state output with Vi = 0 V will be atVp lim or Vm lim.
P. R. Nelson – ECE 322 – Fall 2012 – p. 34/50
Stability of VIO
If the value of VIO could be measured, the error itintroduces could be eliminated by adding avoltage to cancel it.
However, VIO changes with temperature anddrifts over time as the op amp ages.
Approaches to minimizing errors due to VIO willbe discussed later.
P. R. Nelson – ECE 322 – Fall 2012 – p. 35/50
Input Bias and InputOffset Currents
Some current flows into the input terminals of areal op amp.
P. R. Nelson – ECE 322 – Fall 2012 – p. 36/50
Input Bias and InputOffset Currents
Some current flows into the input terminals of areal op amp.
−
+
Vo
Im
Vm
Ip
Vp
These currents can be modeled ascurrent sources Ip and Im externalto an ideal op amp.
P. R. Nelson – ECE 322 – Fall 2012 – p. 36/50
Designer’s Model
−
+
Vo
IB
Vm
IB
Vp
IOS/2
The input bias cur-rent IB and input off-set current IOS are indata sheets.
IB =Im + Ip
2IOS = Im − Ip
Im = IB +IOS
2Ip = IB −
IOS
2P. R. Nelson – ECE 322 – Fall 2012 – p. 37/50
Example 6: InvertingAmp With IB and IOS
−
+
R1
Vi
R2
VoIm
P. R. Nelson – ECE 322 – Fall 2012 – p. 38/50
Example 6: InvertingAmp With IB and IOS
−
+
R1
Vi
R2
VoIm
Vi
R1
+Vo
R2
− Im = 0
P. R. Nelson – ECE 322 – Fall 2012 – p. 38/50
Example 6: InvertingAmp With IB and IOS
−
+
R1
Vi
R2
VoIm
Vi
R1
+Vo
R2
− Im = 0
Vo = −R2
R1
Vi + R2Im Im = IB ±
∣
∣
∣
∣
IOS
2
∣
∣
∣
∣
P. R. Nelson – ECE 322 – Fall 2012 – p. 38/50
Numerical Example
−
+
R1
Vi
R2
VoIm
R1 = 10 kΩ
R2 = 1 MΩ
Vin = 0 V
IB = 100 nA
IOS = 0 nA
P. R. Nelson – ECE 322 – Fall 2012 – p. 39/50
Numerical Example
−
+
R1
Vi
R2
VoIm
R1 = 10 kΩ
R2 = 1 MΩ
Vin = 0 V
IB = 100 nA
IOS = 0 nA
Vo = 0.1 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 39/50
Numerical Example
−
+
R1
Vi
R2
VoIm
R1 = 10 kΩ
R2 = 1 MΩ
Vin = 0 V
IB = 100 nA
IOS = 0 nA
Vo = 0.1 V
How can the circuit be redesigned tominimize this error?
P. R. Nelson – ECE 322 – Fall 2012 – p. 39/50
Redesign Ideas
Idea 1: Smaller R2, constant R2/R1
P. R. Nelson – ECE 322 – Fall 2012 – p. 40/50
Redesign Ideas
Idea 1: Smaller R2, constant R2/R1
. . . but R1 is the input resistance of this circuit,which is often specified.
P. R. Nelson – ECE 322 – Fall 2012 – p. 40/50
Redesign Ideas
Idea 1: Smaller R2, constant R2/R1
. . . but R1 is the input resistance of this circuit,which is often specified.
Idea 2: Add a voltage at the “+” terminal.
P. R. Nelson – ECE 322 – Fall 2012 – p. 40/50
Redesign Ideas
Idea 1: Smaller R2, constant R2/R1
. . . but R1 is the input resistance of this circuit,which is often specified.
Idea 2: Add a voltage at the “+” terminal.. . . but we don’t know what value to use becausewe don’t know Im.
P. R. Nelson – ECE 322 – Fall 2012 – p. 40/50
Redesign Ideas
Idea 1: Smaller R2, constant R2/R1
. . . but R1 is the input resistance of this circuit,which is often specified.
Idea 2: Add a voltage at the “+” terminal.. . . but we don’t know what value to use becausewe don’t know Im.
Idea 3: IB flows in both op amp terminals . . .
P. R. Nelson – ECE 322 – Fall 2012 – p. 40/50
Redesign Ideas
Idea 1: Smaller R2, constant R2/R1
. . . but R1 is the input resistance of this circuit,which is often specified.
Idea 2: Add a voltage at the “+” terminal.. . . but we don’t know what value to use becausewe don’t know Im.
Idea 3: IB flows in both op amp terminals . . .. . . so we could use a resistor at the “+” terminalto generate a voltage proportional to IB.
P. R. Nelson – ECE 322 – Fall 2012 – p. 40/50
New Design
−
+
IpR3
R1
Vi
R2
VoIm
P. R. Nelson – ECE 322 – Fall 2012 – p. 41/50
New Design
−
+
IpR3
R1
Vi
R2
VoIm
Vo = −
(
R2
R1
)
Vi + R2Im −
(
1 +R2
R1
)
R3Ip
P. R. Nelson – ECE 322 – Fall 2012 – p. 41/50
New Design
−
+
IpR3
R1
Vi
R2
VoIm
Vo = −
(
R2
R1
)
Vi + R2Im −
(
1 +R2
R1
)
R3Ip
R3 = R1 ‖ R2 ⇒ Vo = −
(
R2
R1
)
Vi + R2IOS
P. R. Nelson – ECE 322 – Fall 2012 – p. 41/50
Improvement
R1 = 10 kΩ R2 = 1 MΩ Vin = 0 V
IB = 90 nA IOS = 20 nA ⇒ Im = 100 nA
Vo = 20 mV (100 mV without R3)
P. R. Nelson – ECE 322 – Fall 2012 – p. 42/50
Improvement
R1 = 10 kΩ R2 = 1 MΩ Vin = 0 V
IB = 90 nA IOS = 20 nA ⇒ Im = 100 nA
Vo = 20 mV (100 mV without R3)
A more typical case with the same gain:
R1 = 1 kΩ R2 = 100 kΩ ⇒ Vo = 2 mV
P. R. Nelson – ECE 322 – Fall 2012 – p. 42/50
Improvement
R1 = 10 kΩ R2 = 1 MΩ Vin = 0 V
IB = 90 nA IOS = 20 nA ⇒ Im = 100 nA
Vo = 20 mV (100 mV without R3)
A more typical case with the same gain:
R1 = 1 kΩ R2 = 100 kΩ ⇒ Vo = 2 mV
Small R2 minimizes IOS errors!
P. R. Nelson – ECE 322 – Fall 2012 – p. 42/50
Output Resistance
So far we have neglected the internal impedanceof the voltage-controlled voltage source.
P. R. Nelson – ECE 322 – Fall 2012 – p. 43/50
Output Resistance
So far we have neglected the internal impedanceof the voltage-controlled voltage source.
We need to use a Thevénin equivalent circuitinstead of an ideal source.
P. R. Nelson – ECE 322 – Fall 2012 – p. 43/50
Output Resistance
So far we have neglected the internal impedanceof the voltage-controlled voltage source.
We need to use a Thevénin equivalent circuitinstead of an ideal source.
−+Avo(Vp − Vm)
Vo
−+Avo(Vp − Vm)
RoVo
P. R. Nelson – ECE 322 – Fall 2012 – p. 43/50
Output Resistance
So far we have neglected the internal impedanceof the voltage-controlled voltage source.
We need to use a Thevénin equivalent circuitinstead of an ideal source.
−+Avo(Vp − Vm)
Vo
−+Avo(Vp − Vm)
RoVo
Output resistance is important when the opamp supplies significant current.
P. R. Nelson – ECE 322 – Fall 2012 – p. 43/50
Example 7: AudioAmplifier
−
+
1 kΩ
Vi
1 kΩ
24 kΩ
Vo
8 Ω
P. R. Nelson – ECE 322 – Fall 2012 – p. 44/50
Example 7: AudioAmplifier
−
+
1 kΩ
Vi
1 kΩ
24 kΩ
Vo
8 Ω
Want 100 mWRMS
. . . to 8 Ω speaker
P. R. Nelson – ECE 322 – Fall 2012 – p. 44/50
Example 7: AudioAmplifier
−
+
1 kΩ
Vi
1 kΩ
24 kΩ
Vo
8 Ω
Want 100 mWRMS
. . . to 8 Ω speaker
Vo/Vi = 25
P. R. Nelson – ECE 322 – Fall 2012 – p. 44/50
Example 7: AudioAmplifier
−
+
1 kΩ
Vi
1 kΩ
24 kΩ
Vo
8 Ω
Want 100 mWRMS
. . . to 8 Ω speaker
Vo/Vi = 25
Vo = 0.894 VRMS = 1.26 Vpeak Vi = 50.6 mVpeak
P. R. Nelson – ECE 322 – Fall 2012 – p. 44/50
With 15 Ω Ro
−
+ X
15 Ω
1 kΩ
Vi
1 kΩ
24 kΩ
Vo
8 Ω
Vx − Vo
Ro
=Vo
RL
+Vo − Vi
R2
P. R. Nelson – ECE 322 – Fall 2012 – p. 45/50
Vx
Vo
= 1 + Ro
(
1
RL
+1
R2
[
1 −1
Av
])
= 2.88
P. R. Nelson – ECE 322 – Fall 2012 – p. 46/50
Vx
Vo
= 1 + Ro
(
1
RL
+1
R2
[
1 −1
Av
])
= 2.88
Vx = 3.64 Vpeak
P. R. Nelson – ECE 322 – Fall 2012 – p. 46/50
Vx
Vo
= 1 + Ro
(
1
RL
+1
R2
[
1 −1
Av
])
= 2.88
Vx = 3.64 Vpeak
You would not like the sound if you run this circuitfrom a 3 V supply (two AA batteries in series)!
P. R. Nelson – ECE 322 – Fall 2012 – p. 46/50
Finite Input Impedance
Model:
Vm
Rd
Vp
Cd
Rm Cm
Rp CP
P. R. Nelson – ECE 322 – Fall 2012 – p. 47/50
Example 8: Amp forHigh- R Sensor
−
+
1 MΩ0.1 µA
sensor
1 kΩ
24 kΩ
Vo
P. R. Nelson – ECE 322 – Fall 2012 – p. 48/50
Example 8: Amp forHigh- R Sensor
−
+
1 MΩ0.1 µA
sensor
1 kΩ
24 kΩ
Vo
Ideal op amp model →
Vo = 0.1 µA × 1 MΩ ×
(
1 +24 kΩ
1 kΩ
)
= 2.5 V
P. R. Nelson – ECE 322 – Fall 2012 – p. 48/50
With Finite Rp
−
+
1 MΩ0.1 µA
sensor
1 kΩ
24 kΩ
Vo
Rp = 2 MΩ
P. R. Nelson – ECE 322 – Fall 2012 – p. 49/50
With Finite Rp
−
+
1 MΩ0.1 µA
sensor
1 kΩ
24 kΩ
Vo
Rp = 2 MΩ
Vo = 0.1 µA × (1 MΩ ‖ 2 MΩ) ×
(
1 +24 kΩ
1 kΩ
)
= 1.67 VP. R. Nelson – ECE 322 – Fall 2012 – p. 49/50
Other EffectsThis discussion has covered only a few of themost important op amp parameters.
For an extensive listing of op amp parameters,see section 11-2 of Op Amps For Everyone.
P. R. Nelson – ECE 322 – Fall 2012 – p. 50/50