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enter image description here

Its supposed to be part of a guitar effect pedal, why is there positive feedback besides the negative feedback? I was reading about op amps and from what i understand positive feedback is very rarely used. How would this affect the signal?

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  • \$\begingroup\$ I see no positive feedback there. Although op-amps are more often drawn with the inverting input at the top, this one has it drawn with the inverting input on the bottom... \$\endgroup\$ – Jerry Coffin Jan 28 '18 at 17:07
  • \$\begingroup\$ @Jerry Coffin i dont know much about op amps, but I can see that the output is connected to both the noninverting and inverting input isnt it? Isnt connecting the output to noninverting input positive feedback? \$\endgroup\$ – TwoheadedFetus Jan 28 '18 at 17:14
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    \$\begingroup\$ The input connects to both the non-inverting and inverting inputs (which is somewhat unusual). The output connects only to the inverting input (well, yes, it can also feed back through the pot to the non-inverting input, but that will be pretty heavily attenuated, since it goes through 21K vs. 1K going to the inverting input). \$\endgroup\$ – Jerry Coffin Jan 28 '18 at 17:16
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    \$\begingroup\$ Get a free sim tool and analyse it. In the medium to long term you will see this as good advice. \$\endgroup\$ – Andy aka Jan 28 '18 at 17:18
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Quite often the way to analyse these sorts of things is to ask yourself what happens at the extremes (of control position and frequency). The nice thing about this approach is that at low frequency (whatever that means) caps are approximately open circuits and at high frequency they are approximately short circuits.

Ok the 1k and 220nF at the left are a more or less fixed passive lowpass, so forget them, not that interesting.

Now with the pot slider all the way to the left: It just adds 220nF in series with 220 ohms across the existing 220nF so the thing gets a bit more lowpassy (With a bit of a shelf)...

With the slider all the way to the right we get something more interesting: Now at low frequency we model the cap as open circuit so by the usual rules of the opamp (The output is driven such that the two inputs are pretty much equal) we can clearly see that we have a gain of 1 from the opamp stage.

At high frequency however the caps appear short circuit (ignoring that passive input LPF for the moment!), so the opamp now has a gain of about 1 + (1000/220) = ~5.5 times, we can ignore the 20K pot because it is swamped by the 220R resistor, so we have a variable treble boost of about 14dB preceded by a passive lowpass filter.

As to where the action happens, just figure the timeconstants (1K & 220n, 220R & 220n).

It is amazing how often this sort of grossly simplified analysis is all you really need, and it is a ton quicker then disappearing into a mess of S plane bullshit.

Hope this helps.

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About the positive feedback, there is none. The negative feedback path is the 1k resistor between output and inverting input. This negative feedback implies that inverting and non-inverting inputs are almost equal (they try to be and ideally they are equal).

That means that what you think is positive feedback, actually is part of the input path.

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I recognize it as similar to a Mid-range tone control with flat EQ unity gain in the middle.

The reason being , the source impedance dominates a non-inverting gain of 2 and inverting gain of -1 , nets to an overall gain of 1. Any LPF effect going into Vin(-) is opposed by that impedance ratio now in the the denominator as a source Zf/Zs so it acts as a HPF and balanced in the mid-position.

It's not a huge amount of gain variation , maybe 10~15dB

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