Inverting Op-Amp Minimum Resistor Values

[D_Ex_Patria]

I've breadboarded a small dual op-amp distortion, which was an experiment in inverting a Rat distortion. The pedal sounds fine, but the resistor values, which I copied verbatim from the Rat schematic, look very different from what I've seen in other inverting op amp designs (like the Op Amp Muff).

This is the part of the schematic I'm concerned with:



The input resistor to the second op amp is 47 Ohms. Most other designs stick to 10k or so between op-amp stages. Are there any concerns I should have in having picked this value? Is this going to run OK but turn to smoke the first chance it gets?

I could scale this up, but then I believe that the larger feedback potentiometer will start becoming a limiting factor.

[ElectricDruid]

In the Rat, the value is used in a non-inverting circuit, which is a bit different. Specifically, it has very little voltage across it because it's the lower part of a voltage divider where the upper part is many KOhms.

In the inverting arrangement, it's taking the input voltage directly and it terminates at a virtual ground node (the op-amps -ve input) so it could have quite a lot of voltage across it. Say the first op-amp stage is driven to clipping and is putting out 8V. The Vref is 4.5V, so that's what the -ve input is at, so there's 3.5V across the 47R.

I = V/R = 3.5 / 47 = 0.074 = 74mA

It's not going to smoke, but that's quite a lot of juice for a pedal.

If I were you, I'd think about gain staging the design a bit more. Rather than having x2100 gain in one stage, put a gain of x10 or x20 in that first inverting stage, and then you can reduce the gain in the second stage (e.g. increase the value of the 47R to something larger) and still have the same gain overall. Say we have x20 in the first stage. Now we only need x100 in the second stage, so 47R can become 1K.

HTH,
Tom

[R.G.]

It will help clarify things to think about what happens at the inverting input of an opamp. The inverting input pin of an opamp is said to be a "virtual ground". It is held by feedback action to the same voltage as the non-inverting input.

The voltage across the inverting input resistor lets a current into the inverting input pin's node. This current must be cancelled by an equal and opposite current from the output pin through th feedback resistor for the amp to be working properly. So the inverting input resistor can be any value at all, from a short circuit up to an open circuit. All that matters is whether the output voltage can drive an equal-and-opposite current through the feedback resistor to keep the inverting input pin from accumulating charge.

You can mentally cut the opamp circuit apart at the inverting input pin and think of the inverting input pin as a floating ground ("virtual ground"). The input voltage sends a current equal to the difference between the input voltage and the non-inverting input's voltage through the input resistor. The output pin sucks and equal and opposite current away from the inverting input through the feedback resistor. You can calculate these separately for their resistor values.

All this works as long as the output voltage and current of the opamp can swing enough to "cancel" the current at the inverting input. This means that the current magnitude at the inverting pin can only be as big as the current limit on the opamp output, and this is further limited by the opamp power supply and output swing.

If the feedback resistor is zero, the opamp merely follows the non-inverting input, and removes current from the inverting input to do so. The feedback resistor at zero ohms merely follows the noninverting input as long as the opamp output current can remove the current from the inverting input. As the feedback resistor increases, the output voltage stays at
Vout = -Rfb/Iin.

For a bit deeper dive:

Opamp inputs are deliberately very high impedance. They can't "eat" any significant current. The inverting input is "magically" held at the same voltage as the positive input by feedback. [There is another story there, but go with me on this.]

The circuit's inverting input resistor has a voltage fed to its outer terminal, and so the resistor sees the difference between that voltage and the input voltage. A current flows in the input resistor equal to the voltage across it divided by the resistance. That current tries to go into/come out of the inverting input pin. But it can't, because the pin is high impedance. The output has been driven the opposite direction to the inverting input by the amplifier gain, so it makes a voltage difference appear across the feedback resistor.

The output of the amplifier goes to any voltage within its available power supply range to prevent a build up of current (charge) at the inverting input, which means that the output causes an equal and opposite current to flow from the inverting input.

>Current goes into (out of) the inverting input pin node, and is canceled by an equal and opposite current pulled out of (into) the pin by the output pin through the feedback resistor.<

It's this action that makes for the voltage gain. The output voltage is any voltage that the amplifier can possibly do that moves the cancelling current off the input pin. The feedback resistor converts that voltage difference to an output voltage to keep the inverting input pin from building up charge.

The bottom line is that the inverting input resistor can be anything at all, from open to a short circuit. All that matters is what amount of current the input signal puts into/out of the inverting input pin. The inverse of that current has to flow from the output to cancel at the inverting input.

[PRR]

> turn to smoke the first chance it gets?

Wire it up. If it survives an hour of power-chords, it is OK.

FIRST: my pointer won't reach your whiteboard. Put numbers on parts so I can yell "C46!" instead of playing "not that one, no to the left, now up..."


Agree with Tom: I think you have incompletely copied the plan, drew something not-the-same.

Why is A1 biased both to Vref AND to Gnd? While that can be valid, it seems far more likely you want C1 moved over.

If you look at it with Ideal Opamp Theory, it has "problems". Tom computes 74mA. Where does that flow? Out of A1 (which may be a 20-40mA part) through R6 and R7, from A2 output. But 74mA in 100k requires 7,400 Volts at A2 output. Clearly not going to happen on 9V or even 36V power. Even if R7 is turned-down to a minimum ~~1k (need 7.4V peak so not-happening for V+ under 15V).

Assuming R7=1k (min) and 4V swing at A2 output, the actual max current is 4mA. All audio opamps will deliver this much, safely.

And what a lot of gain!! 100k/47 is over 2,000. Plus the gain of A1(?). Input overload is under 1mV, very low relative to typical guitar signals. This will distort ALL THE TIME.

Extending R.G.'s thoughts: When distorting hard, all opamp theory goes out the window. The "-" input of A2, which we thought was a virtual earth (= "+" input), is slamming around at 1/2000th of A2's output swing, a few mV. In an extended ideal opamp theory, this is OK/safe. Real opamps have some limit how far you can force the + and - input pins apart. Most take 7V by default. Many take 30V, essentially all you can do with typical 30V-36V maximum supply. I think this will be safe for 99% of opamps you can buy. But it is a point to consider. (If it was more than mV, you can try two diodes across the input to limit to 0.6V. But there's usually a better plan.)

[PRR]

Also the "R6 leg" is not just R6 but also reactance of C2. Unless C2 > 200uFd it will be significant in the audio band. If C2 is 1uFd it will dominate ALL through the guitar range to 3.5kHz. In the "body zone" like 350Hz a 1uFd C2 acts like 500 Ohms not 50 Ohms.

For "clean work" you want most NFB parts larger than the opamp's minimum load. For jellybean opamps, 2K is the nominal minimum. If you run low voltage (like 9V) and don't need point-oh THD, you can go 1K. These are the impedances "seen" by the output. A heavy voltage divider can use a 100r lower resistor with a 1k upper resistor since that is a 1.1k total load.

[D_Ex_Patria]

Thanks everyone! This is a lot of information to go through. I'm still wrapping my head around the science and I'm going to try my best to make sense of it.

I apologize for not marking the parts; I can see how that would have made this easier. I also left some parts out, like the feedback diodes PRR mentioned. I incorrectly assumed that limiting the complexity of the diagram would make my question easier to answer.

Biasing A1 against ground and Vref was a mix-up when I was sketching. Bad me. That capacitor must be moved over.

The amount of gain is very intentional; I do want this to clip. Furthermore, I want it to push up against the available bandwidth of the op-amp and start to limit the frequency range. From my understanding, that's part of the Rat's sound, and what I'm interested in hearing here. So gain staging wouldn't get me there.

The large current draw pretty much tells me I should rethink this. The specs for the MC1458 I'm using for A1 won't even give me the time of day for a 47Ohm load, so I'm probably clipping way earlier than I want. I'm looking into a T network to see if I can get to that level of gain from A2 with a larger R6. More to come.

Thanks again, everyone. You're all fantastic.