Op-Amps 4: Divided Negative Feedback |
Simple negative feedback, connecting the output to the inverting input, makes an op-amp into a unity gain amplifier. In that setup, all of the output goes to the inverting input. If instead the amount of feedback is reduced, through a voltage divider, then the gain of the op-amp circuit becomes greater than one. This may be the most common way op-amps are used for amplification in stompbox circuits. In addition, by adding some capacitors to the voltage divider, the gain also gets some tonal character that is the foundation of dozens of famous distortion and overdrive pedals (think Boss, Fulltone, Ibanez, Marshall, Proco, Voodoo Labs, …).
Because it keeps the op-amp circuit simpler, let’s work with a bipolar supply. Single-supply versions require only straight-forward additions.
We are going to replace the negative feedback loop from the previous tutorial with a resistive voltage divider. This arrangement is called divided negative feedback. Here is the basic circuit:

Resistors R1 and R2 make up the voltage divider. They have equal resistances, both 10K, so that the voltage at their junction will be half the voltage across R1 and R2 in series to ground. That means that the inverting input will see the output signal with half of its amplitude. Starting with our usual source with amplitude 1V and frequency 800Hz, here is what we get:

The result is that the output signal is the input signal with twice its amplitude. Therefore, by reducing the amount of negative feedback (through the voltage divider), the op-amp amplifies the input signal by a factor greater than one (or unity).
In Op-Amps 2: Hitting the Rails, I describe how an op-amp with no feedback produces a square output signal because the output is on one supply rail or the other. In Op-Amps 3: Between the Rails, I describe how feeding the entire output signal back to the inverting input reduced the effective gain of the ciruit to unity. Divided negative feedback is between these two cases. Or said more carefully, divided negative feedback has both of these as special cases. If R1 has infinite resistance, then the signal will hit the rails. And if R1 has no resistance, then signal is buffered.
If R1 has enough resistance, this circuit will produce some output clipping because part of the output signal hits the supply rails. For example, if we increase R1 in the circuit above from 10K up to 100K then the output looks like this:

The peaks of the output sine wave are clipped, or trimmed, by the limits of the supply voltage. This clipping is heard as distortion and it is one of the sources of distortion in the stompboxes like the DOD Overdrive 250.
Analysis
Before talking more about actual stompboxes, here is an analytical description of how this circuit works. If the op-amp’s gain is 200,000 then for small voltage levels so that output is not hitting the rails,
| Vout = 200,000 × (Vin+ − Vin−) |
where Vout is the output voltage, Vin+ is the non-inverting input voltage, and Vin− is the inverting input voltage. The voltage divider in the feedback loop makes
| Vin− = |
|
Vout |
where R1 and R2 are the values of resistors R1 and R2, respectively. Putting these two equations together gives a relationship between the source signal and the output signal:
|
= Vin+ − |
|
Vout |
In typical applications, 1/200,000 is negligible compared to R2/(R1 + R2). This means that we can treat the left-hand side as approximately zero, giving the approximate solution
| Vout ≅ |
|
Vin+ = (1 + R1/R2) Vin+ |
so that Vout is a multiple bigger than one of Vin+. The larger the ratio R1/R2 is the greater the gain is.
We can place a pot, wired as a variable resistor, in place of either R1 or R2 to make the gain variable. Note that decreasing R2 increases the gain so that if you want gain to increase for a clockwise turn you should have resistance decrease in that direction. R1 would be wired in the opposite way. Also, because the effect of R2 is nonlinear you might use a reverse log pot to compensate.
When R2 is made variable, it is common practice to add a fixed stop resistor in series with the pot to prevent the total R2 resistance from reaching zero ohms. Why this is a good idea is left as an exercise for the reader.
Adding Caps to Shape Tone
Often, in stompbox circuits that use divided negative feedback, a capacitor appears in parallel with R1 and another capacitor appears in series with R2. These capacitors are shaping the tone of this amplifier circuit. Except for single instead of split supply, here is the circuit for Shaka Boost by Aron Nelson (owner and founder of diystompboxes.com):

This schematic is drawn the way you will usually see this circuit. It does not emphasize that R1 and C1 are both in the voltage divider in parallel, though that is clear once it is pointed out. Also, note that while C2 usually appears at the junction of the voltage divider, the order of R2 and C2 could be reversed. Finally, the presence of C2 allows us to replace the ground connection at one lead of R2 with either V+ or V−. C2 will block any DC differences between the feedback loop and the voltage to which R2 connects. What is critical is that the voltage be constant. Capacitors respond only to the rate of change of voltage and that must be zero.
Consider the effect of C1, placed in parallel with R1. Capacitors attenuate AC signals more as frequency falls. C1 has the effect of making the R1 half of the voltage divider have more resistance for low frequencies. According to the analysis above, such resistance means more amplification. Therefore, C1 has the effect of boosting lows relative to highs.
C2, placed in series with R2, increases the resistance in the R2 half of the voltage divider as frequency falls and has the opposite effect of C1. With C1 removed, C2 emphasizes low frequencies at the inverting input and, therefore, causes the op-amp to attenuate low frequencies relative to high frequencies.
Fortunately, the effects of C1 and C2 do not cancel each other out. Their nonlinearity causes the combined effect to be a mid-frequency hump. Extreme lows and highs are both attenuated relative to the frequencies in between. Here is a plot of each effect alone, showing the amplitude of the output for each frequency.

The green line plots amplitude when no C1 capacitor is present and the blue one when no C2 is present. Around a frequency of 1KHz, both amplitudes reach their peak at 11V. This is the gain we would expect from the circuit without capacitors. Using the formula for gain given above,
| gain ≅ |
|
= |
|
= 11 |
So we see that around 1KHz, neither capacitor is having much effect. But at low frequencies C2 attenuates and at high frequencies C1 attenuates. With both C1 and C2 present, we get the combined effect shown in red below.

This plot shows that at the extremes the combined effect equals the attenuation of one capacitor or the other and, of course, in the middle they are all the same. This combined effect is often called a mid-frequency hump. The range of frequencies near the maximum gain is called the passband.
In LTSpice Analysis and the DOD Overdrive 250 you can see similar results for that stompbox. There, you will also see that changing gain is linked to changing a corner frequency.
More Analysis
It is possible to work out formulas for some of the features of this circuit. Divided Negative Feedback and Capacitors provides a detailed description. To close this tutorial, here is a summary of the results.
First, the frequency response plotted above for the Shaka Booster shows several general properties:
- the response is symmetric in log-frequencies around one maximum and
- at the extremes the gain approaches unity.
In addition, if the corner frequencies are far enough apart, then
- the lower and upper corner frequencies are approximately f2 = 1/(2πR2C2) and f1 = 1/(2πR1C1) respectively and
- the maximum gain equals the no-capacitor gain 1 + R1/R2 and occurs at the frequency f0= √(f1f2), the square root of the product of the two corner frequencies.
For low gain levels, more accurate formulas for the corner frequencies are f2′ = af2 and f1′ = f1/a where a2 = 1 − 2/g2 and g = 1 + R1/R2 > 2. The corner frequencies are “far enough apart” if
| (1 − 1/g2) |
|
≤ 1% |
The range of frequencies between these corner frequencies is the passband.
Note that the dependence of the corner frequencies on the values of the resistors may also affect the decision where to place the variable resistor. Making R2 variable makes the lower corner frequency rise with gain and making R1 variable makes the upper corner frequency fall with gain. Aron chose to make R1 variable in the Shaka Booster whereas DOD chose to make R2 variable in the Overdrive 250.
Examples
In Aron’s Shaka Boost,
| C1 = 220pF = 220 × 10-12F, | R1 ≤ 25KΩ = 25 × 103Ω, | |
| C2 = 10uF = 10 × 10-6F, | R2 = 1KΩ = 1 × 103Ω, |
and if we set R1 = 10K then
| g = 1 + R1/R2 = 11 | ||
| f2 = 15.9Hz, | f2′ = 15.8Hz, | |
| f1 = 72.3KHz, | f1′ = 72.9KHz, | |
| f0 = √(f1 f2) = 1073Hz. |
Notice that with g = 11, the differences between f1 and f1′ and between f2 and f2′ are ignorable. Also
| (1 − 1/g2) |
|
= 0.02% < 1% |
so that the corner frequency formulas are accurate for this case.
If we increase the gain pot to 25K, then f2 will not change and f1 = 28.9KHz. I suspect that my hearing would not notice this large difference in the upper corner frequency because it is so high in both cases.
If we turn the gain pot down to 2K, then g = 3, f1 = 362KHz and f1′ = 410KHz. There is a big difference in these frequencies, but they are so far out of the range of human hearing that this does not matter. So theory predicts that Aron’s Shaka Boost leaves the tone of the signal effectively unchanged over its entire range of gain.
The DOD Overdrive 250 is a different story. For this circuit, the variable resistor is in the R2 position in series with a 4.7K stop resistor:
| C1 = 25pF = 25 × 10-12F, | R1 = 1MΩ = 1 × 106Ω, | |
| C2 = 47nF = 47 × 10-9F, | 4.7K ≤ R2 ≤ 505KΩ = 505 × 103Ω, |
and if we set R2 = 500K then we get the minimum gain and
| g = 1 + R1/R2 = 3 | ||
| f2 = 6.8Hz, | f2′ = 6.0Hz, | |
| f1 = 6.4KHz, | f1′ = 7.2KHz, | |
| f0 = √(f1 f2) = 208Hz. |
The difference between f1 and f1′ is appreciable this time, so f1′ is preferred. Also
| (1 − 1/g2) |
|
= 0.09% < 1% |
so that the corner frequency formulas are still accurate for this case. A passband from 6Hz–7.2KHz represents an audible cut in the high frequencies.
Based on separate simulations, we find that the op-amp starts to clip around R2 = 30K, or about a 2/3 rotation, so that is a sensible place to check what happens for higher gains. Beyond this point, clipping of the signal invalidates the formulas for amplitudes. Changing R2 to 30K increases g to 34.3, leaves f1 unchanged, and raises the lower corner frequency to 113Hz. Because the bass E string on a guitar has a frequency of 82Hz, this will also be clearly audible.
These figures are slightly misleading, but they tell the correct qualitative story. When we check
| (1 − 1/g2) |
|
= 2% > 1% |
we just violate the requirement for accuracy. This implies that both capacitors have an influence in for the passband frequencies and that the maximum gain falls short of the no-capacitor gain by 2% which is still small. To obtain more accurate information, one can use SPICE simulations, as illustrated in LTSpice Analysis and the DOD Overdrive 250.
To summarize what these calculations show, the DOD Overdrive 250 cuts highs and cuts lows more as the gain is increased.
Freedreamer said:
very interesting and really good work!
Posted 08.01.2008 at 2:27 pm
Martin said:
I really like the way this is presented – it gives a good understanding of what is really going on in these types of feedback circuits. Much more useful than the old, “change these resistors to set the gain” kind of thing I usually see. Very helpful. Thanks!
Posted 09.01.2008 at 11:22 am
juse said:
Well done! Thanks for taking the time to present this in such a logical & professional manner. Helps with the learning curve!
Posted 22.01.2008 at 9:39 am
Good Charlotte Billy Martin said:
Thanks for the nice read, keep up the interesting posts..
Posted 28.01.2008 at 1:59 pm
Jessie said:
Thanks, this is good stuff. You are spot on.
Posted 29.01.2008 at 1:40 pm
Nick C. said:
Wow, you’ve really shed light on the subject, unlocking some mysteries for me. Great site.
Thanks
Posted 06.03.2008 at 11:08 am
Glen said:
Thanks
This is starting to make sense
Posted 21.04.2009 at 8:21 pm
rob said:
this is brilliant…thank god for guys like you..well explained…at last!!!!!…
Posted 02.11.2009 at 10:28 am
Jon said:
I know this posting is years old, but I must ask, if you added the pot on both locations (where the shaka boost places it, and also where the DOD Overdrive places it), is that how pedals can use both the treble and bass frequency knobs? or does that ruin the circuit.
Also great work, this was an awesome tutorial!
Posted 20.05.2010 at 10:29 am
drknownothing said:
This is the best explanation of opamp feedback circuit I have seen so far, so clear and to-the-point. Thanks for the good work!
Posted 03.05.2011 at 3:54 am