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Operational Amplifiers : Chapter 2
- An operational amplifier (called op-amp) is a specially-designed amplifier in
bipolar or CMOS (or BiCMOS) with the following typical characteristics:
- Very high gain (10,000 to 1,000,000)
- Differential input
- Very high (assumed infinite) input impedance
- Single ended output
- Very low output impedance
- Linear behavior (within the range of VNEG < vout < VPOS
- Op-amps are used as generic “black box” building blocks in much analog
electronic design
- Amplification
- Analog filtering
- Buffering
- Threshold detection
- Chapter 2 treats the op-amp as a black box; Chapters 8-12 cover details of op-
amp design
- Do not really need to know all the details of the op-amp circuitry in order to use it
Generic View of Op-amp Internal Structure
- An op-amp is usually comprised of at least three different amplifier stages (see figure)
- Differential amplifier input stage with gain a 1 ( v+ - v- ) having inverting & non-inverting inputs
- Stage 2 is a “Gain” stage with gain a 2 and differential or singled ended input and output
- Output stage is an emitter follower (or source follower) stage with a gain = ~1 and single- ended output with a large current driving capability
- Simple Op-Amp Model (lower right figure):
- Two supplies V (^) POS and VNEG are utilized and always assumed (even if not explicitly shown)
- An input resistance r (^) in (very high)
- An output resistance rout (very low) in series with output voltage source v (^) o
- Linear Transfer function is vo = a 1 a 2 (v+ - v- ) = A (^) o (v+ - v- ) where Ao is open-loop gain
- vo is clamped at V (^) POS or VNEG if Ao (v (^) + - v (^) - ) > VPOS or < VNEG, respectively
Linear Op-amp Operation: Non-Inverting Use
- An op-amp can use negative feedback to set
the closed-loop gain as a function of the
circuit external elements (resistors),
independent of the op-amp gain, as long as
the internal op-amp gain is very high
- Shown at left is an ideal op-amp in a non-
inverting configuration with negative
feedback provided by voltage divider R1, R
- Determination of closed-loop gain:
- Since the input current is assumed zero, we can write v- = R1/(R1 + R2)vOUT
- But, since v (^) + =~ v (^) - for the opamp operation in its linear region, we can write v (^) - = v (^) IN = R1/(R1 + R2)v (^) OUT or, vOUT = ((R1 + R2)/R1)vIN
- We can derive the same expression by writing v (^) OUT = A(v (^) + - v (^) - ) = A{v (^) IN – [R1/(R1 + R2)] v (^) OUT } and solving for v (^) OUT with A>> Look at Example 2.1 and plot transfer curve.
The Concept of the Virtual Short
- The op-amp with negative feedback forces the two inputs v+ and v- to have the same
voltage, even though no current flows into either input.
- This is sometimes called a “virtual short”
- As long as the op-amp stays in its linear region, the output will change up or down until v- is almost equal to v+
- If v (^) IN is raised, v (^) OUT will increase just enough so that v (^) - (tapped from the voltage divider) increases to be equal to v (^) + (= v (^) IN) - In vIN is lowered, vOUT lowers just enough to make v- = v+
- The negative feedback forces the “virtual short” condition to occur
- Look at Exercise 2.4 and 2.
- For consideration:
- What would the op-amp do if the feedback connection were connected to the v+ input and v (^) IN were connected to the v- input?
- Hint: This connection is a positive feedback connection!
Input Resistance for Inverting and Non-inverting Op-amps
- The non-inverting op-amp configuration of slide 2-4 has an apparent input resistance
of infinity, since iIN = 0 and RIN = v IN /i IN = vIN /0 = infinity
- The inverting op-amp configuration, however, has an apparent input resistance of R
- since RIN = vIN/i (^) IN = vIN/[(vIN – 0)/R1] = R
Op-amp Voltage Follower Configuration
- The op-amp configuration shown at left is a
voltage-follower often used as a buffer amplifier
- Output is connected directly to negative input (negative feedback)
- Since v+ = v- = v (^) IN, and v (^) OUT = v-, we can see by inspection that the closed-loop gain A (^) o = 1
- We can obtain the same result by writing v (^) OUT = A (v (^) IN – v (^) OUT ) or vOUT/vIN = A/(1 + A) = 1 for A >> 1
- A typical voltage-follower transfer curve is
shown in the left-bottom figure for the case VPOS
= +15V and VNEG = -10V
- For v (^) IN between –10 and +15 volts, v (^) OUT = v (^) IN
- If v (^) IN exceeds +15V, the output saturates at V (^) POS
- If v (^) IN < -10V, the output saturates at V (^) NEG
- Since the input current is zero giving zero input
power, the voltage follower can provide a large
power gain
Ex. Difference Amplifier with a Resistance Bridge
- The example of Fig’s 2.14 and 2.15 in the text
shows a difference amplifier used with a
bridge circuit and strain gauge to measure
strain.
- Operation:
- The amplifier measures a difference in potential between v1 and v2.
- By choosing R (^) A = R (^) B = Rg (unstressed resistance of Rg1 and Rg2), it is possible to obtain an approx linear relationship between v (^) OUT and ∆L, where ∆L is proportional to the strain across the gauge.
- Design:
- In order for the bridge to be accurate, the input resistances of the difference op-amp must be large compared to R (^) A , R (^) B,, & Rg - Input resistance at v1 (with v2 grounded) is R + R2 =~ 10 Mohm - Input resistance at v2 (with v1 grounded) is just R1 = 12 K due to the v1-v2 virtual short
Instrumentation Amplifier
- Some applications, such as an
oscilloscope input, require differential
amplification with extremely high
input resistance
- Such a circuit is shown at the left
- A3 is a standard difference op-amp with differential gain R2/R
- A1 and A2 are additional op-amps with extremely high input resistances at v1 and v2 (input currents = 0)
- Differential gain of input section:
- Due to the virtual shorts at the input of A1 and A2, we can write iA = (v2 – v1) /RA
- Also, iA flows through the two R (^) B resistors, allowing us to write v 02 – v 01 = iA (R (^) A + 2 RB )
- Combining these two equations with the gain of the A3 stage, we can obtain v (^) OUT = (R2/R1)(1 + [2R (^) B /R (^) A ])(v1 – v2)
- By adjusting the resistor RA, we can adjust the gain of this instrumentation amplifier
Op-amp with T-bridge Feedback Network
- To build an op-amp with high closed-loop gain may require a high value resistor R
which may not be easily obtained in integrated circuits due to its large size
- A compromise to eliminate the high value resistor is the op-amp with T-bridge feedback
network, shown below
- R (^) A and R (^) B comprise a voltage divider generating node voltage v (^) B = v (^) OUT RB /(R (^) A + R (^) B ), assuming that R2 >> R (^) A||RB
- Since v (^) B is now fed back to v-, an apparent gain v (^) B /v (^) IN = -(R2/R1) can be written
- Combining these two equations allows us to write v (^) OUT = - (R2/R1)([R (^) A + RB ]/RB )v (^) IN
- Fairly large values of closed-loop gain can be realized with this network without using
extremely large IC resistors
Op-amp Integrator Network
- Shown below is an op-amp integrator network
- The output will be equal to the integral of the input, as long as the op-amp remains in its linear region
- Due to the virtual short property of the op-amp input, we can write i 1 = v (^) IN/R 1
- This current i 1 starts charging the capacitor C according to the relation i 1 = C(dv (^) C /dt)
- Since v- remains at GND, the output drops below GND as C charges and the time
derivative of vOUT becomes the negative of the time derivative of vC
- since v (^) C = 0 - v (^) OUT
- Combining the above equations, we obtain
- dv (^) OUT/dt = -i 1 /C = -v (^) IN/R 1 C
- Solving for v (^) OUT(t) and assuming C is initially uncharged, we obtain
- v (^) OUT (t) = (-1/R 1 C) ⁄ v (^) IN dt where the integral is from 0 to t
Op-amp Integrator Example with Long Pulse
- Consider a case with an infinitely long 4V pulse
- The capacitor will continue to charge linearly in time, but will eventually reach 10V which will force v (^) OUT to –10V (= V (^) NEG) and saturate the op-amp (at 12.5 ms)
- After this time, the op-amp will no longer be able to maintain v- at 0 volts
- Since v (^) OUT is clamped at –10V, the capacitor will continue to charge exponentially with time constant R 1 C until v- = +4V - During this time the capacitor voltage will be given by vC (t) = 10 + 4[1 – exp(t 1 – t)/R 1 C] where t 1 = 12.5 ms - At t = t 1 , vC = 10 V and at t = infinity, vC = 14 V
- The resulting capacitor and output waveforms are shown below.
Op-amp as a Differentiator
- The two op-amp configurations shown below perform the function of differentiation
- The circuit on the left is the complement of the integrator circuit shown on slide 2-14, simply switching the capacitor and resistor
- The circuit on the right differentiates by replacing the capacitor with an inductor
- For the circuit on the left we can write
- i 1 = C(dv (^) IN/dt) = i 2 = (0 – v (^) OUT )/R2 or vOUT = - R 2 C (dvIN/dt)
- Similarly, for the circuit on the right we can obtain
vOUT = - (L/R 1 ) (dvIN/dt)
- By nature a differentiator is more susceptible to noise in the input than an integrator,
since the slope of the input signal will vary wildly with the introduction of noise spikes.
- Do exercises 2.23 and 2.25.
Open-Loop Comparator (Example 2.8 in text)
- Given the open-loop comparator shown at the
left with VPOS = +12V and VNEG= -12V, plot the
output waveforms for VR = 0, +2V, and –4V,
assuming v IN is a 6V peak triangle wave
- The solution is shown at the left
- In (a) the output switches symmetrically from VPOS rail to V (^) NEG rail as the input moves above or below GND
- In (b) the output switches between the rail voltages as the input goes above or below +2 V
- In (c) the output switches between the rail voltages as the input varies above or below –4 V
- The output becomes a pulse generator with adjustable pulse width
- Do Exercise 2.28.
Schmitt Trigger Op-amp Circuit
- The open-loop comparator from the previous two
slides is very susceptible to noise on the input
- Noise may cause it to jump erratically from + rail to – rail voltages
- The Schmitt Trigger circuit (at the left) solves this
problem by using positive feedback
- It is a comparator circuit in which the reference voltage is derived from a divided fraction of the output voltage, and fed back as positive feedback.
- The output is forced to either V (^) POS or VNEG when the input exceeds the magnitude of the reference voltage
- The circuit will remember its state even if the input comes back to zero (has memory)
- The transfer characteristic of the Schmitt Trigger is
shown at the left
- Note that the circuit functions as an inverter with hysteresis
- Switches from + to – rail when v (^) IN > VPOS (R1/(R1 + R2))
- Switches from – to + rail when v (^) IN< VNEG(R1/(R1 + R2))
