Feedback: the loop gain is the truth meter
Introducido| English | Significado |
|---|---|
| loop gain | ganancia de bucle |
| transimpedance amplifier | An amplifier that turns an input current into an output voltage. |
| gain-bandwidth product/ɡeɪn ˈbændwɪdθ ˈprɒdʌkt/ | producto ganancia-ancho de banda |
Why do the golden rules work?
- An op-amp's open-loop gain may be 100 000 one day and 50 000 the next, yet a circuit built from it gives a gain of exactly 10. Two resistors set it, not the chip.
- The earlier golden rules said so without proof. This lesson shows when they hold and how wrong they are when they do not.
Feedback and the loop gain
- Negative feedback returns a fraction $\beta_{fb}$ of the output to the input, where it is subtracted. What matters is the loop gain 环路增益 $A_{OL}\beta_{fb}$, the gain all the way round the loop.
- For the inverting stage, $\beta_{fb} = \dfrac{R_{in}}{R_{in} + R_{f}}$. Solving the exact node equation with a finite $A_{OL}$ gives:
An inverting stage has R_in = 10 kΩ and R_f = 90 kΩ. What is β_fb?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$\beta_{fb} = \dfrac{10}{10 + 90} = 0.1$. The ideal gain is $-9$; with $A_{OL} = 10^{5}$ the loop gain is $10^{4}$.
The error factor
- If the loop gain is huge, the denominator is 1 and the golden-rule gain $-\dfrac{R_{f}}{R_{in}}$ is exact. Otherwise it is reduced by the factor $\dfrac{1}{1 + 1/(A_{OL}\beta_{fb})}$.
- So the golden rules are really a statement about loop gain: large loop gain, rules exact; loop gain near 1, rules fail.
Closed-loop gain 100 with A_OL = 10⁵ and β_fb = 0.01 actually gives
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$A_{CL} = \dfrac{100}{1 + 1/(A_{OL}\beta_{fb})} = \dfrac{100}{1.001} = 99.90$. The golden rules err by about $\dfrac{1}{\text{loop gain}}$.
Worked example: how wrong?
- Closed-loop gain 100, so $\beta_{fb} = 0.01$. With $A_{OL} = 10^{5}$: $A_{CL} = \dfrac{100}{1 + 0.001} = 99.90$, a 0.1% error.
- With $A_{OL}$ down to $10^{3}$, as it is at higher frequencies: $A_{CL} = \dfrac{100}{1.1} = 90.9$, a 9.1% error.
Where the open-loop gain has fallen to 10³, the same closed-loop-100 stage reads
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
$A_{CL} = \dfrac{100}{1 + 0.1} = 90.9$. With a loop gain of only 10, the golden rules are visibly off.
Four standard stages
- Inverting: $-\dfrac{R_{f}}{R_{in}}$, input resistance $R_{in}$. Non-inverting: $1 + \dfrac{R_{f}}{R_{in}}$, input resistance nearly infinite. Summing: $-R_{f}\left(\dfrac{v_{1}}{R_{1}} + \dfrac{v_{2}}{R_{2}}\right)$.
- The transimpedance amplifier 跨阻放大器 turns a current into a voltage: $v_{o} = -I_{p}R_{f}$. A photodiode giving 2 µA into 1 MΩ produces 2 V.
Match each op-amp stage to its output.
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
The follower is the non-inverting stage with $R_{f} = 0$ y $R_{in}$ removed, giving gain exactly 1.
A photodiode gives 5 µA into a transimpedance amplifier with R_f = 200 kΩ. What is the size of the output, in volts?
Dificultad 2/5 · 140 monedas + un pequeño bono en el primer intento de hoy
$|v_{o}| = I R_{f} = 5 \times 10^{-6} \times 200 \times 10^{3} = 1$ V, negative for current flowing into the inverting input.
Gain-bandwidth product
- An op-amp's open-loop gain falls at 20 dB per decade. Along that line, gain × frequency is constant: the gain-bandwidth product 增益带宽积 (GBW).
- With GBW = 1 MHz, a closed-loop gain of 100 gets 10 kHz of bandwidth and a gain of 10 gets 100 kHz. At 10 kHz the open-loop gain is $\dfrac{10^{6}}{10^{4}} = 100$: loop gain 1, which is exactly the corner.
A 1 MHz GBW op-amp at closed-loop gain 10 has bandwidth
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Gain × bandwidth stays at 1 MHz: $\dfrac{1\,\text{MHz}}{10} = 100$ kHz. At gain 100 it would be 10 kHz.
Three common slips
- Inverting formula for a non-inverting stage. $1 + \dfrac{R_{f}}{R_{in}}$ differs by 1, which matters most at low gain.
- Ignoring frequency. A gain that is exact at 100 Hz can be 10% low at 10 kHz as $A_{OL}$ falls.
- Asking one stage for too much. High gain and wide bandwidth compete through GBW; use two stages.
Two cascaded gain-10 stages on a 1 MHz GBW op-amp give gain 100 with more bandwidth than one gain-100 stage.
Dificultad 1/5 · 100 monedas + un pequeño bono en el primer intento de hoy
Each stage has 100 kHz. Cascading shrinks that to about 64 kHz, still far above the 10 kHz of a single gain-100 stage.
Recap
- Loop gain $A_{OL}\beta_{fb}$ decides how exact the golden rules are.
- Error factor: $\dfrac{1}{1 + 1/(A_{OL}\beta_{fb})}$.
- Closed-loop gain × bandwidth ≈ GBW.