Op-Amp & Negative Feedback

Build a non-inverting or an inverting stage from two resistors, give the op-amp a finite gain and a finite bandwidth, and play a sine wave through it. Step through the cycle to read the voltage at every node, including the small gap between the two inputs.

Worked example — Op-Amp & Negative Feedback

Fraction of the output fed back	0.100000
Gain if A were infinite	10.0000
Loop gain	10,000
Gain with this A, at DC	9.9990
Short of the ideal by	0.0100%
Gain if A were halved	9.9980
Bandwidth (−3 dB)	100,010 Hz
Gain × bandwidth	1,000,000 Hz

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Lesson

The theory — Op-Amp & Negative Feedback

An operational amplifier multiplies the difference between its two inputs by a very large open-loop gain A, V_out = A(V₊ − V₋), until the output reaches a supply rail. Negative feedback returns the fraction β = Rg/(Rg + Rf) of the output to the inverting input, and the non-inverting stage's gain becomes G = (1/β)/(1 + 1/(Aβ)), which tends to 1 + Rf/Rg as A grows; the inverting stage's tends to −Rf/Rg. The product Aβ is the loop gain. With a single pole, A falls with frequency as A(f) = A₀/(1 + j f A₀/GBW), where the gain-bandwidth product GBW is A₀ times the open-loop pole frequency, very nearly the frequency at which A has fallen to 1.

What each symbol means

A: the op-amp's open-loop gain: what it multiplies V₊ − V₋ by. A₀ is its value at DC, and it falls with frequency
β: the feedback fraction, Rg/(Rg + Rf): the share of the output voltage returned to the inverting input
Aβ: the loop gain. The stage misses its ideal gain by exactly 1/(1 + Aβ)
GBW: the gain-bandwidth product, in hertz: A₀ times the open-loop pole frequency, very nearly where the open-loop gain has fallen to 1

Assumes: One pole and nothing else: no slew-rate limit, no input offset voltage or bias current, no second pole, resistors of exact value, and an output that swings all the way to the rails. The clipped waveform is the unclipped steady state cut off at the rails, which assumes the op-amp leaves saturation the instant it is allowed to; a real one takes time to recover. Swapped inputs are drawn as an instant jump from one rail to the other.

Halve the op-amp's gain and the stage moves in the fourth figure 🖖

The opening stage is meant to multiply by ten. With an open-loop gain of 100,000 it gives 9.9990, and the results table works out what it would give if A were halved: 9.9980. A 50% change in the amplifier's own gain moves the stage by 0.01%. The reason is one fraction. The stage falls short of 1 + Rf/Rg by exactly 1/(1 + Aβ). With Aβ = 10,000 that is a hundredth of a percent, and halving or doubling A only doubles or halves a hundredth of a percent. So the gain belongs to the two resistors, which can be bought to 0.1%, and hardly at all to the op-amp. Harold Black made this argument for the amplifiers of telephone carrier systems in 1934: his feedback amplifiers held their gain within 0.01 dB while the plate supply swung from 240 to 260 V, where an ordinary amplifier moved by 0.7 dB. The price is the gain thrown away. Ask the same op-amp for ×1000 and Aβ is down to 100: it gives 990.10, a whole 1% short, in the last row of the second table.

The virtual short is a hundred microvolts, at one hertz 🖖

The usual rule says the two inputs of an op-amp with negative feedback sit at the same voltage. Their gap peaks at |V_out| / |A(f)|, and the circuit prints it as you step through the cycle: 100.49 µV at the peak of the opening preset. Now press Same stage, 1 kHz. The output has hardly changed and the gap is 9.999 mV, a hundred times wider. The A in V_out / A is the open-loop gain at the signal's frequency, and an op-amp with a DC gain of 100,000 and a gain-bandwidth product of 1 MHz is already 3 dB down at 10 Hz and has about 1,000 left at 1 kHz; the dashed line in the frequency-response figure is that fall. Then press Driven into the rail. Asked for 19.998 V, the output stops at 15 V, and with the output pinned nothing holds the inputs together: 0.5000 V apart. Feedback keeps the inputs together only while the output is free to move, and an analysis that assumes a virtual short on a clipped stage is wrong by volts.

A follower and an inverter both have unity gain, and one has twice the bandwidth 🖖

Press ×100 at 20 kHz. The bandwidth is 10,010 Hz and the DC gain 99.9001, and their product is 1,000,000 Hz, the op-amp's gain-bandwidth product, to the last digit. With a single pole that is an identity for any non-inverting stage, which is why four rows of the second table end in the same number. The inverting rows do not. The inverter of gain −1 reaches 500,010 Hz, where the follower of gain +1 reaches 1,000,010 Hz. What the op-amp trades its bandwidth against is 1/β, the gain the same two resistors would give with the signal on the non-inverting pin, and for an inverter of gain −1 that is 2. An inverter always pays for one more unit of gain than it delivers: about 1% of the bandwidth at ×−100, and half of it at ×−1.

Problem solved in full

Getting 20 kHz through a ×100 preamplifier 6 steps
A sensor gives 50 mV and has to reach 5 V, at frequencies up to 20 kHz. Your op-amps have A₀ = 100,000 and a gain-bandwidth product of 1 MHz. Will one ×100 non-inverting stage, Rf = 99 kΩ over Rg = 1 kΩ, do it? If not, what will?

Open this problem in the tool
1. β = 1/(1 + 99) = 0.01, so Aβ = 1,000 and the DC gain is 100/(1 + 1/1,000) = 99.9001. At low frequencies 50 mV becomes 4.9950 V, short of 5 V by the 0.1% that 1/(1 + Aβ) predicts.
  
  $β = \frac{1}{1 + 99} = 0.01, A β = 1000, G_{0} = \frac{100}{1 + 1/1 000} = 99.9001$
2. The op-amp's own pole is at GBW/A₀ = 10 Hz. Closing the loop moves it up by the factor 1 + Aβ = 1,001, to 10,010 Hz. That is the stage's bandwidth, and 99.9001 × 10,010 = 1,000,000 Hz.
  
  $f_{p} = \frac{GBW}{A _{0}} = 10 Hz, f_{- 3 dB} = 10 \cdot 1001 = 10010 Hz, 99.9001 \cdot 10010 = 1000000$
3. At 20 kHz, about twice the bandwidth, the gain is 99.9001/√(1 + (20,000/10,010)²) = 44.7124, and the output peaks at 2.2356 V. Less than half of what was asked for.
  
  $∣ G (20 kHz) ∣ = \frac{99 . 9001}{1 + ( 20 000/10 010 ) ^{2}} = 44.7124, 0.05 \cdot 44.7124 = 2.2356 V$
4. Split the gain over two ×10 stages, each with Rf = 9 kΩ and Rg = 1 kΩ. Each has Aβ = 10,000, a DC gain of 9.9990 and a bandwidth of 100,010 Hz, ten times the single stage's. Set the tool to that stage at 20 kHz and it reads 9.8049.
  
  $β = 0.1 : G_{0} = 9.9990, f_{- 3 dB} = 100010 Hz, ∣ G (20 kHz) ∣ = 9.8049$
5. In cascade the gains multiply: 9.8049² ≈ 96.14, so 50 mV becomes 4.8068 V at 20 kHz.
  
  $9.804 9^{2} \approx 96.14, 0.05 \cdot 96.14 = 4.8068 V$
6. The pair has two poles at 100,010 Hz, and its own −3 dB point is lower than either: the squared magnitude has fallen by a factor of √2 when 1 + (f/100,010)² = √2, at f = 100,010 · √(√2 − 1) = 64,366 Hz.
  
  $1 + (f /100 010)^{2} = 2 \Rightarrow f = 100010 2 - 1 = 64366 Hz$
Answer

One ×100 stage gives 2.2356 V at 20 kHz; two ×10 stages give 4.8068 V, 96% of the target, with the pair 3 dB down at 64,366 Hz. Each op-amp spends its own gain-bandwidth product, so each ×10 stage has ten times the bandwidth of the ×100 one, and the pair together 6.4 times. The tool draws one stage, so the cascade is arithmetic on its readouts, and the last step shows why the arithmetic is needed: two identical stages do not keep the bandwidth of one.

$2.2356 V (one stage) vs 4.8068 V (two)$

Learning path

What is not a resistor

References (1)

The first insight's history, and the result the whole page rests on: with |μβ| ≫ 1 the output is −e/β, "independent of μ but does depend upon β". The 0.01 dB against 0.7 dB figures quoted in that block are Black's own: H. S. Black, "Stabilized Feedback Amplifiers." Bell System Technical Journal 13(1), January 1934, pp. 1–18. DOI 10.1002/j.1538-7305.1934.tb00652.x

By the 42Math team · Reviewed against published sources. Corrections welcome.