Op Amps
Reading: Nilsson & Riedel Chapter 5
Sedra & Smith Chapter 2/5
Signal Amplification
• Signal
• a time-varying quantity
• Carries information
• Extract info by signal processing, with
help of transducer
• Amplification
• Linearity
• Saturation
• Gain in db
Amplifier Gain
If the power delivered
to the load is greater
than the power drawn
from the source, where
does the additional
power come from?
Transfer Function
Express gain in dB
• Use absolute figures. -20dB means attenuation, but 20db is
amplification
• Eg Vin= 1V, Vout= 0.1V → attenuation, -20dB
Power gain
voltage gain in decibels = power gain only if the impedances at input and
output are equal.
Op Amp
• Terminals
• +, -, out V+ V• Offset pin used to compensate for degradation in performance
• Voltages & currents @ terminals
• All Currents into the terminals
TI A741 op amp data sheet
• Top row, left to right:
• 14-pin plastic dual in-line
package (DIP),
• 8-pin plastic DIP (“mini-DIP”).
• Middle row:
• 14-pin thin-shrink small-outline
package (TSSOP),
• 8-pin small-outline package
(SO- 8),
• 8-pin TSSOP (where height matters)
• Bottom row:
Src: Art of electronics
• 5-pin small-outline transistor
package (SOT23),
• 6-ball chip-scale package (CSP
– top and bottom views),
• 5-pin SC-70
LM741
LF412
LF347
Voltage transfer characteristics
• Constraints for linear operation
• Linear operation if input voltage difference is small
• Amp saturates → non linear operation
• In linear region, Vo= gain A * diff bt Vp and Vn
• Typically A>10K
Model for voltage amplifier.
Check Sedra&Smith Fig 1.17
Ideal OpAmp
At the input, the op amp does not load the circuit it is
connected to. Its input terminal draws little to no
power.
• infinite A, infinite input impedance, zero output impedance
The voltage between output terminal and ground
always equal to 𝐴(𝑣2 − 𝑣1 ), independent of current
drawn from output terminal into a load impedance
Notes – for linear, ideal operation
• Ideal OpAmp: infinite A, infinite input R, zero output R
• Input Voltage constraint
• Typically, Vcc<20V. Thus if A= 104 ,then |(Vp-Vn)| < 20/104 = 2mV
• But usual input voltages >> 2mV. Voltage difference less than 2mV
→ the two voltages essentially equal
• Linear region only if input voltages essentially equal
(called virtual short condition)
• Input Constraint Vp=Vn
• Input Current constraint
• Large input resistance > 1MΩ
• ip = in =0
• In linear operation, Vo < +Vcc or > -Vcc
Notes on op amp operation
1. Output current is generally not known (it is provided by
external power supplies)
2. KCL at input nodes is generally a good starting point in
op amp circuit analysis
3. (vp-vn) is multiplied by a large number to get Vout
1. Vout is limited by the external power supplies
Eg 1a – LINEAR OPERATION
• Calculate Vo if Va is 1V and Vb is 0V
Eg 1a – LINEAR OPERATION
• Calculate Vo if Va is 1V and Vb is 0V
• Write node voltage eqn at inverting input.
in=0
• (Va-Vn)/25 +(Vo-Vn)/100 = 0
• Vn = Vb
• → 1 / 2 5 + Vo/100 =0
(less than 10V, linear region of operation)
• → Vo = -4V
• Note: -ve feedback, inverted output
• Note: avoid writing node voltage eqn at output, Vo
Avoid doing KCL
at output
terminal!
It is possible to specify values of va and vb in
the circuit below such that vo = −12 V
A.
B.
True
False
It is possible to specify values of va and vb in
the circuit below such that vo = −12 V
A.
B.
True
False
Try: Eg 1b
• Find Vo if Va=1V, Vb=2V
in=0
Try: Eg 1b
• Find Vo if Va=1V, Vb=2V
(Va-Vn)/25 +(Vo-Vn)/100 = 0
-1/25+(Vo-2)/100=0
V=6V
(linear)
in=0
Eg
• If Va= 1.5 V, what Vb avoids
saturation?
Eg
• If Va= 1.5, what Vb avoids saturation?
• Solve for -10 ≤Vo ≤ 10V
• Upper limit:
• (1.5-Vb)/25k +(10-Vb)/100k = 0
• Lower Limit
• (1.5-Vb)/25k +(-10-Vb)/100k = 0
• Or Vb= 1/5(6+Vo)
• Ans: -0.8 V ≤ Vb ≤ 3.2 V
Chapter 5 – Op Amps
Example
(a) Find vo for vs = 0.4 V,
2.0 V, 3.5 V, −0.6 V, −1.6 V,
and −2.4 V. Assume the
op amp is ideal.
(b) Find the range of input voltage
vs for which the op amp operates
in its linear region.
𝑣𝑝 = 0 ⇒ 𝑣𝑛 = 𝑣𝑝 = 0
Chapter 5 – Op Amps
Example
Find vo for vs = 0.4 V, 2.0 V,
3.5 V, −0.6 V, −1.6 V, and
−2.4 V. Assume the op
amp is ideal.
𝑣𝑝 = 0 ⇒ 𝑣𝑛 = 𝑣𝑝 = 0
0 − 𝑣𝑠
0 − 𝑣𝑜
KCL at 𝑣𝑛 :
+
+0=0
16,000 80,000
80,000
∴
𝑣𝑜 = −
𝑣 = −5𝑣𝑠
16,000 𝑠
vs [V]
vo = −5vs [V]
OK?
0.4
-2.0
Yes
2.0
-10.0
Yes
3.5
-17.5
NO – saturate at
-15 V
-0.6
3.0
Yes
-1.6
8.0
Yes
-2.4
12.0
NO – saturate at
10 V
Chapter 5 – Op Amps
Example, continued
Find the range of input voltage vs for which the op amp
operates in its linear region.
Chapter 5 – Op Amps
Example, continued
Find the range of input voltage vs for which the op amp
operates in its linear region.
−15 ≤ 𝑣𝑜 ≤ 10
−15 ≤ −5𝑣𝑠 ≤ 10
3 ≥ 𝑣𝑜 ≥ −2
−2 𝑉 ≤ 𝑣𝑜 ≤ 3 𝑉
vo = −5vs = 10

vs = −2 V
vo = −5vs = −15

vs = 3 V

− 2 V  vs  3 V
Inverting Amp Circuit
• (Vs-0)/Rs + (Vo-0)/Rf =0
• True even though infinite open loop gain A
• and infinite input impedance for ideal op amp
• Determine closed loop gain
• But now gain is fixed at = -Rf/Rs
• Using feedback resistor
• Assuming pwr rails are ±Vcc , to avoid saturation, Vo< Vcc
Assume ideal opamp, linear region
• Choice of resistors for design should factor
this in
Can bypass very high open-loop gain, A, of
op amp to get finite, predictable and stable
closed-loop gain.
Standard resistor values (Appendix H)
Characteristics of inverting opamp:
• Input impedance
• Rs
• Output impedance
• Zero (directly connected to OpAmp)
• Gain: -Rf/Rs
Design Inverting Amp
• Design inverting op amp with gain of 12, with ±15V power
supplies.
• Vo= -Rf/Rs X Vin.
• Can use 12k and 1k, (also 18k, 1.5k)
• What range of input voltages results in operation in linear
region?
• Vo=-12k/1k X Vs; |Vo| ≤ 15V
• → Vs = -1/12 X Vo
• → -1.25 V ≤ Vs ≤ 1.25 V
Summing Amp (weighted summer)
in=0
Applications:
Audio Mixer
What do to if sums should include a
difference
Non Inverting Amp
Potential divider at inverting input
No voltage drop across Rg, negligible current. Vn= Vg
Rg
Change of subject of equation
For operation in linear region, Vo< Vcc,
Input impedance high (infinity), output zero
or Vg*gain < Vcc
Eg Design Non –Inverting Amp
• Design non inverting amp with gain of 6
• What minimum power supply values can be used such that
Eg Design Non –Inverting Amp
• Design non inverting amp with gain of 6
• What minimum power supply values can be used such that
• Solve Vo= 6Vg, where Vg is ±1.5V, giving ±9V
• Add a limiting input resistor to + input
What if 𝑅𝑓 = 0 and 𝑅𝑠 = ∞ ?
• Buffer (Remember this…)
• Voltage follower: (gain is 1)
• Infinite (High) input impedance
• Zero (Low) output impedance
• Primary use: impedance matching
• Used to connect a source with a high impedance to a low
impedance load
When opamp not ideal
• Gain is not infinite
• Input impedance is not infinite.
• Output impedance is not zero
NB: as R o → 0 and R i → ∞,
A → ∞ we get eqn for ideal
Similar analysis if RL exists at output
Realistic model of op amp.
Effect of finite open loop gain …II
Can express Vin(difference)
in terms of Vo
•
When using node voltage
method, cannot assume Vn
is 0 anymore. Vn = -Vo/A
(Assuming A is finite,
therefore Vp not equal to Vn. )
KVL
The inverting configuration suffers
from a low input resistance
To try:
• Find an expression for gain
(Vo/Vi) for the circuit shown,
if opamp is ideal.
• Eg 5.2. SS p521
Check SS p523
Hint: Write node voltage eqns at
inverting input and node x.
Substitute for vx. Solve.
The Difference Amplifier
The difference amp
Cannot use op-amp by itself as a difference amp bc Gain
will be unstable, infinite
• At inverting input: (in=0)
• At non inverting input
• Together:
Steps:
Node voltage analysis at -ve
Express Vn in terms of Vb
Equate ratios of scaling factors
Express in terms of Ra and Rb (-ve)
Use superposition to prove.
The difference amp
Cannot use op-amp by itself as a difference amp bc Gain
will be unstable, infinite
• At inverting input: (in=0)
• At non inverting input
Steps:
Node voltage analysis at -ve
Express Vn in terms of Vb
Equate ratios of scaling factors
Express in terms of Ra and Rb (-ve)
• Together:
• Make scaling factors equal
Redo with Rb/Ra= Rd/Rc
(PTO)
Use superposition to prove.
→
Notes:
• **redo with Rd/Rc=Rb/Ra
• In general: the output voltage is proportional to the difference
between a scaled replica of Va and a scaled replica of Vb
• By making
the output is a scaled replica of the difference
Eg – Designing a difference amp
• A) Design an ideal op amp with gain 8 and power supplies ±8V
• Using standard values of resistors, choose equal
scaling factors
• B) If Va=1V, what range of values for Vb
will result in linear operation?
Eg – Designing a difference amp
• A) Design an ideal op amp with gain 8 and power supplies ±8V
• Using standard values of resistors:
• Choose Ra=1.5k, Rb=12k, same choices for Rc, Rd
• B) If Va=1V, what range of values for Vb
Result in linear operation?
Uses of Difference Amp
• Useful in amplifying desired signal in the presence of noise.
• Useful signal is in the difference
• Noise affects both inputs equally, but the difference is amplified.
• Eg ecg signal from heart weaker than noise from external sources (eg
fluorescent lamp), but latter is rejected/not amplified.
• An ideal difference amp
• amplifies only the difference mode portion of the input
• eliminates the common mode portion (proof: SS $5.4.1 p532)
50/60 Hz common-mode signal between electrodes
and local power-system ground.
Small differential signal is
important and strong
interfering common-mode
signal is present
Another point of view
Differential and
Common-Mode Signals
• Difference mode: Vdm= Vb-Va
• Common mode input: (avg of the inputs): Vcm= ½(Va+Vb)
• Expressing Va and Vb in terms of Vdm and Vcm components:
•
and
Determine output if only Vcm applied (and Rb/Ra= Rd/Rc) → zero
Common Mode Rejection Ratio (CMRR)
Ideally, 𝑣𝑜 = 𝐴𝑑𝑚 𝑣𝑑𝑚 . But not for practical circuits.
→ 𝑐𝑜𝑚𝑚𝑜𝑛 𝑚𝑜𝑑𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦 𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑.
• 𝑣𝑜 = 𝐴𝑐𝑚 𝑣𝑐𝑚 + 𝐴𝑑𝑚 𝑣𝑑𝑚
• 𝐴𝑐𝑚 − common mode gain
• 𝐴𝑑𝑚 − differential mode gain
• CMRR is a measure of quality/nearness to ideal of an opamp
• For ideal opamp and matched components, Acm is zero, leading to
infinite CMRR.
• The larger the CMRR the better
• With resistance mismatch or non-ideal (internal components of)
opamp, CMRR is large, but finite.
Downside: Differential Input resistance Rid
•
What if the amplifier is required to have a large
differential gain, R2/R1? Any implications?
What can be done?
Circuit model for voltage
amplifier
Voltage amplifier with input
signal source and load.
At input, amplifier forms a voltage divider with source.
Downside: Differential Input resistance Rid
•
Differential gain is R2/R1, so R1 will be small
Not desirable to have low input resistance
Add buffers before R1 → high input impedance, gain=1
Do better: get some voltage gain in stage 1 as well.
Instrumentation amplifier overcomes drawback
The Instrumentation Amplifier
• Has higher input impedance
• Stage 1 → amplification, Stage
2 → differencing function
• Stage 1: non-inverting configuration
gain G1= (1+R2/R1)
• Stage 2: difference configuration
A1, A2 are buffers with gain.
Flaws:
• Common mode signals are amplified by first stage (as well as
differential);
• Common mode signals are larger, can cause saturation of amp.
• Stage 2 thus deals with larger CM signals
• Two amps in stage 1 must be perfectly matched
• Any differences will be magnified in stage 2
• To change the gain of stage 1, modify both sets of R1 resistors equally
Better design
• Ground at R1 removed
• Gain is still
• (next slide)
• Common mode signals create
same voltage after A1, A2
Hence no current through
2𝑅1 and 𝑅2 resistors
• CM signal not amplified.
• Difference signal amplified
• Gain varied by varying 2R1 only
• Even if 𝑅2 ′𝑠 not matched doesn’t affect operation →
Improved circuit
3: Sum the I x R from Vo2 till Vo1
(three terms)
*Differential voltage
Vld appears across
2R1 bc of virtual
short cct.
*Determine current
through 2R1
*Determine voltage
appearing at A3.
2
0
1
3
Wrt common mode signal: equal voltage across 2R1 → no current → voltage directly passed via R2 to
Vo1&Vo2 A3 does not deal with amplified CM signal
Integrators
• Inverting integrator (Miller integrator )
𝑑𝑣 (𝑡)
•𝑖=𝐶 𝑐
𝑑𝑡
• 𝑉𝑐 − initial voltage on C at t=0
Derivation
Output is inverted scaled replica of integral of the input voltage. (only within linear range)
Differentiators
Applications of opAmps
• Useful especially in instrumentation circuits
• Voltage controlled voltage source
• Vo= (-Rf/Ri)Vin = kVin
• [SS, Fig 5.21]
• Voltage controlled current source
• [SS, fig 5.23] current flows through Rf (which is the
load)
• Io= Vin/Ri → io=k Vin
• Current Controlled Voltage Source
• Current Controlled Current Source
Frequency dependence of open loop gain
(nonideal property)
• Open loop gain is finite,
and decreases with frequency
Open loop gain variation with frequency
Gain x bandwidth = constant (for closed
loop)
Trade-off between gain and bandwidth
Large Signal Operation of OpAmps
• Output voltage saturation can occur
• Clips peaks of output waveform.
• Keep input signal small
• Output current limits
• Eg 741 has 20mA output current limit (includes both load and feedback
currents)
• Output voltage will saturate at the level in spec corresponding to max output
current
• Slew Rate → max rate of voltage change at output
• Causes non linear distortion
In V/𝝁s
Op Amp application circuits
• Voltage follower
• Impedance matching
• Comparator
• Schmitt trigger, comparators with hysteresis
• Oscillator
• Window comparator PE4E p710, 714
(Pract Elect 4 Eng Bk)
• Level comparator
• Sample & hold AoE p254
(Art of Electronics bk)
• Peak detector
Problems to try:
• Nilsson & Riedel
• Q 5.5, 5.11, 5.13, 5.17, 5.20, 5.27, 5.30, 5.34,
5.44, AP5.6
• Sedra & Smith
• Q5.2, 5.8, 5.20, 5.38