Op amp needs bias?

[YurkshireLad]

I replaced R4 in the diagram in this attachment - https://www.eevblog.com/forum/beginners/op-amp-needs-bias/msg3797660/#msg3797660

[HB9EVI]

ok, with a 100k pot you have a G between 1 and 11; that won't be in the range, where the opamp starts to clip the signal with a plain passive guitar on the input.

you can either place 2 anti-parallel diodes in the feedback path over the pot, or clip the output by adding a resistor before the diodes

[Terry Bites]

Here are some more friends https://www.diystompboxes.com/smfforum/index.php?topic=110272.0

[YurkshireLad]

ok, with a 100k pot you have a G between 1 and 11; that won't be in the range, where the opamp starts to clip the signal with a plain passive guitar on the input.

you can either place 2 anti-parallel diodes in the feedback path over the pot, or clip the output by adding a resistor before the diodes

I don't have any diodes; can I add a resistor to the output, before the capacitor?

[HB9EVI]

it's not a bad idea to decouple opamp outputs from possible capacitive loads (e.g. long, coaxial audio cables); common value would be 100R
but for clipping that won't do it. the diodes 'cut off' the signal above their forward voltage, so ~0.7V, so the signal turns from sinusoidal to squarewave-like, it adds odd and even harmonics to the signal, which are responsible for the distorted sound

check here once:
https://www.electrosmash.com/big-muff-pi-analysis

they are based on transistors, but that doesn't matter; watch how the diodes in circuits are places - this is what a circuit makes a distortion

[YurkshireLad]

Thanks all. I need to understand the amplitude of the output signal from a typical pedal. Likely mine was too high as it was post-gain. The guitar input signal is about 200mV for low E, so it makes sense that a pedal would output a similar voltage (in my naive opinion) for the next pedal in the chain.

[YurkshireLad]

I thought I'd post my latest diagram for any feedback. I managed to simulate it in LTSpice, but values may be wrong.

[Zero999]

That looks sensible enough. Presumably this is going to be powered off a 9V battery.

Yes, the values are wrong. R3 is 1m which is 1 milliohm, or 0.001Ohms. It should be 1M, which is a million Ohms. SPICE is not case sensitive. It would treat 1M, as 1m, so to get round this, you need to set the value to 1Meg, instead of 1M.

I believe guitar pickups have a fairly high impedance. I'd up the values of R1 & R2 to 1M and move R3 to in-between the op-amp's +input and the 4.5V node.

R6 should be a much lower value, like 100R.

[YurkshireLad]

Yes, I realized it was taking 1m as milli ohms, but I didn't realize I could use 1meg, so thanks! I'll, update that and I'll add your other suggestions too. Many thanks.

[TimFox]

SPICE started out as Fortran on Hollerith cards, and was case-insensitive.
Therefore, "m" or "M" always means "milli" and "meg" or "MEG" is required for "mega".
This is required for backwards-compatibility with earlier SPICE models.

[YurkshireLad]

This is what I have so far. I wired it up on the breadboard, but the signal is only being biased with 20mV and not 4.5V (measured where R1 is connected to the signal input).

[TimFox]

The indicated polarity on the input capacitor is backwards:  you expect +4.5 V on the right side and 0 on the left side for DC voltages.
However, C2 should not be a polarized capacitor at only 100 nF.  The other two capacitors C3 and C4 should be polarized.
What did you use to measure the DC voltage at the junction of C2 and R1?

[YurkshireLad]

Damn, sorry. It isn't polarized - I forgot to change the circuit diagram.

I used my multimeter.

[TimFox]

"Multimeter":  do you mean an analog "VOM" or a modern digital "DMM"?
The reason for that question is how much current you are pulling through the 1 megohm resistor from +4.5 V.
Unless your opamp is blown, or you have an undocumented short circuit, or you have mislabeled the pins in your diagram (inverting the power connections), you should have very close to +4.5 V at that junction, assuming you have +4.5 V at the voltage divider connected to the top end of the 1 megohm resistor R1.  Check that node (junction of R7 and R8) first, then disconnect the opamp and see if the +4.5 V reappears.
A typical DMM has an input resistance on DC of 10 megohms, but some cheap ones have only 1 megohm input resistance.  Assuming 10 meg, you have a Thevenin equivalent of +4.5 V in series with 1.5 megohm driving the voltmeter, so the theoretical voltage is about 3.9 V with no load besides the voltmeter.  However, if a polarized C6 is in backwards, it will short out the +4.5 V.

[YurkshireLad]

I'm using a modern digital multimeter to measure the bias voltage at R1/C2.

The junction of R7/R8 is 4.5V. This is what I'm seeing.

[TimFox]

Double-check the pinout on your opamp (none shown in your drawing) and look for accidental short from +input to -power.
Then replace the opamp with a new one, after measuring the voltage when you removed the first one.

[rstofer]

I don't understand C3.  It doesn't provide a DC path to ground so the gain is probably not correct as shown on the most recent schematic.
I would eliminate it completely.

I'm not convinced about the bias circuit either.  See Figure 3 Non-inverting here:

https://mil.ufl.edu/4924/docs/TI_SingleSupply_OpAmp.pdf

Google term:  non-inverting amplifier single supply

Or, cut to the chase, second schematic here:

https://electronics.stackexchange.com/questions/437278/single-supply-non-inverting-amplifier-using-op-amp

I stand corrected on C3, it is useful.

[TimFox]

C3 is a typical circuit to get a low DC gain (+1, since the output goes almost directly to the - input at DC) but a high AC gain at audio frequencies.  If his circuit were working, the + input would be at +4.5 V, and the output and - input would follow, except for small deviation due to small bias currents through the highish resistances.
If you eliminate C3, the - input would be far from the bias on the + input.

[YurkshireLad]

I don't understand C3.  It doesn't provide a DC path to ground so the gain is probably not correct as shown on the most recent schematic.
I would eliminate it completely.

I'm not convinced about the bias circuit either.  See Figure 3 Non-inverting here:

https://mil.ufl.edu/4924/docs/TI_SingleSupply_OpAmp.pdf

Google term:  non-inverting amplifier single supply

Or, cut to the chase, second schematic here:

https://electronics.stackexchange.com/questions/437278/single-supply-non-inverting-amplifier-using-op-amp

I stand corrected on C3, it is useful.

I started with the 2nd schematic but everything I read said that voltage divider can noise from the supply to the input audio signal. So I changed to what I'm using now.

[rstofer]

I don't understand C3.  It doesn't provide a DC path to ground so the gain is probably not correct as shown on the most recent schematic.
I would eliminate it completely.

I'm not convinced about the bias circuit either.  See Figure 3 Non-inverting here:

https://mil.ufl.edu/4924/docs/TI_SingleSupply_OpAmp.pdf

Google term:  non-inverting amplifier single supply

Or, cut to the chase, second schematic here:

https://electronics.stackexchange.com/questions/437278/single-supply-non-inverting-amplifier-using-op-amp

I stand corrected on C3, it is useful.

I started with the 2nd schematic but everything I read said that voltage divider can noise from the supply to the input audio signal. So I changed to what I'm using now.

Yes but with that 1 Mohm resistor, even the slightest bit of bias current will drop the voltage substantially.  Look at the values used in other circuits.  It is more on the order of 10k top and bottom.

[YurkshireLad]

I don't understand C3.  It doesn't provide a DC path to ground so the gain is probably not correct as shown on the most recent schematic.
I would eliminate it completely.

I'm not convinced about the bias circuit either.  See Figure 3 Non-inverting here:

https://mil.ufl.edu/4924/docs/TI_SingleSupply_OpAmp.pdf

Google term:  non-inverting amplifier single supply

Or, cut to the chase, second schematic here:

https://electronics.stackexchange.com/questions/437278/single-supply-non-inverting-amplifier-using-op-amp

I stand corrected on C3, it is useful.

I started with the 2nd schematic but everything I read said that voltage divider can noise from the supply to the input audio signal. So I changed to what I'm using now.

Yes but with that 1 Mohm resistor, even the slightest bit of bias current will drop the voltage substantially.  Look at the values used in other circuits.  It is more on the order of 10k top and bottom.

Will do, thanks.

I've been referencing the ProCo cat - https://www.electrosmash.com/images/tech/pro-co-rat/pro-co-rat-schematic-parts.jpg

[TimFox]

I don't understand C3.  It doesn't provide a DC path to ground so the gain is probably not correct as shown on the most recent schematic.
I would eliminate it completely.

I'm not convinced about the bias circuit either.  See Figure 3 Non-inverting here:

https://mil.ufl.edu/4924/docs/TI_SingleSupply_OpAmp.pdf

Google term:  non-inverting amplifier single supply

Or, cut to the chase, second schematic here:

https://electronics.stackexchange.com/questions/437278/single-supply-non-inverting-amplifier-using-op-amp

I stand corrected on C3, it is useful.

I started with the 2nd schematic but everything I read said that voltage divider can noise from the supply to the input audio signal. So I changed to what I'm using now.

Yes but with that 1 Mohm resistor, even the slightest bit of bias current will drop the voltage substantially.  Look at the values used in other circuits.  It is more on the order of 10k top and bottom.

Yes, 1 uA of bias current (far larger than should happen with a TL072) would give 1 V change in the voltage from 1 megohm.  However, at room temperature, the TI spec (TL072H, similar for other grades) for maximum bias current is 120 pA.  Over temperature extremes (-40 to +125 C), the maximum bias current is 5 nA.  5 nA x 1 megohm = 5 mV.
The OP's choice of resistor and capacitor values may not be optimal, but the huge error in the DC voltage at the + input indicates either a wiring error, a component failure, or an accidental huge difference between the circuit diagram value and the actual resistor value.

[rstofer]

Yes, 1 uA of bias current (far larger than should happen with a TL072) would give 1 V change in the voltage from 1 megohm.  However, at room temperature, the TI spec (TL072H, similar for other grades) for maximum bias current is 120 pA.  Over temperature extremes (-40 to +125 C), the maximum bias current is 5 nA.  5 nA x 1 megohm = 5 mV.
The OP's choice of resistor and capacitor values may not be optimal, but the huge error in the DC voltage at the + input indicates either a wiring error, a component failure, or an accidental huge difference between the circuit diagram value and the actual resistor value.

I lost track of the device, it doesn't show on the latest schematic.  Nevertheless, 4.48 uA is going somewhere.  The thing is, that's not a lot of current.  Or, maybe the resistor is a lot larger - like 10M ohm.

[YurkshireLad]

Yes, 1 uA of bias current (far larger than should happen with a TL072) would give 1 V change in the voltage from 1 megohm.  However, at room temperature, the TI spec (TL072H, similar for other grades) for maximum bias current is 120 pA.  Over temperature extremes (-40 to +125 C), the maximum bias current is 5 nA.  5 nA x 1 megohm = 5 mV.
The OP's choice of resistor and capacitor values may not be optimal, but the huge error in the DC voltage at the + input indicates either a wiring error, a component failure, or an accidental huge difference between the circuit diagram value and the actual resistor value.

I lost track of the device, it doesn't show on the latest schematic.  Nevertheless, 4.8 uA is going somewhere.  The thing is, that's not a lot of current.

The output signal (GAIN_OUTPUT) for now is going to my guitar amp.

[Jan Audio]

You always need a resistor at the input of your opamp.
If you want 5volt you could take a microchip rail to rail opamp.
Are you overdriving your signal with this opamp ?, it sound nicer if you use germanium diodes.