DC biased opamp input resistor function

Thread Starter

rhaggart1

Joined Apr 25, 2020
6
http://www.generalguitargadgets.com/pdf/ggg_dist_plus_sc.pdf

I'm looking to understand this schematic. Currently focussing on the input and power stages.
Here's where I'm at:
INPUT
R1 is a pulldown resistor. It provides a path for the leakage current through C2 when the switch is in the 'bypass' position so that un-bypassing doesn't cause a 'pop' noise.
C3 acts as a low pass filter in combo with the source impedance to filter HF noise.
C2 blocks the DC from the supply. Allows proper DC biasing of the input signal.
R5 is a current limiting resistor to protect the op-amp from ESD

POWER
R2 and R3 form a voltage divider so we have V_s / 2 = 4V5 at the output of the divider (our DC bias)
C1 is our power supply smoothing

... what on earth is R4??!

I've asked this question elsewhere & also searched all over the place for answers. I've had very limited success. Often just 'it's there because we need it' type things... I've been simulating these stages (idealised). The non-inverting input value is very notably attenuated without this resistor.
I've heard some things about an input bias current and impedance matching. But I don't think that's so important here as I'm looking at the ideal case (for now) and it clearly has a significant effect on this case.

Any help greatly appreciated!
 

dl324

Joined Mar 30, 2015
11,000
what on earth is R4??!
It adds a DC offset to the input; likely to workaround the limited common mode input range when operating an LM741 with a single sided supply since the output is AC coupled.

Schematic in question:
clipimage.jpg
Personally, I wouldn't trust a schematic that uses connection dots and "humps". That's a clue that the author is a novice.
 

Thread Starter

rhaggart1

Joined Apr 25, 2020
6
It adds a DC offset to the input; likely to workaround the limited common mode input range when operating an LM741 with a single sided supply since the output is AC coupled.
Sorry - should have cleared this. I understand the need for the DC bias using a single supply through the voltage divider etc. I'm just curious why R4 is needed at all... Surely just connecting the junction of the divider to the non-inverting input. Example:
1593285670560.png

I've noticed that removing C1 & R4 together results in a similar output to having them both in. Removing one at a time results in a very different output...
Perhaps R4 is just there to counter some behaviour introduced by the smoothing cap?
 
Sorry - should have cleared this. I understand the need for the DC bias using a single supply through the voltage divider etc. I'm just curious why R4 is needed at all... Surely just connecting the junction of the divider to the non-inverting input. Example:
View attachment 210854

I've noticed that removing C1 & R4 together results in a similar output to having them both in. Removing one at a time results in a very different output...
Perhaps R4 is just there to counter some behaviour introduced by the smoothing cap?
It's simple. R4 is there to separate the bias voltage from the signal. You are absolutely correct on all other points. Also removing BOTH C1 and R4 will result in the same (almost) behaviour as having both in. But then, with BOTH out, the bias voltage will be an unfiltered version of the supply voltage. C1 and R2, R3 provide filtering of the source voltage to come up with a clean bias voltage = 4.5V. The price for that is having to wait several time constants (3 sec approx) for the filter to settle. With ONLY R4 out (shorted) the signal would be entirely killed by C1.
BTW: The high values of resistors here R4, R8, Gain pot R7, mean that there is more thermal noise in the circuit than needs be. But if you're happy with its sound then that's what matters. On second thought the noise level is not so bad.
 

Thread Starter

rhaggart1

Joined Apr 25, 2020
6
R4 is there to separate the bias voltage from the signal.
Just as an opportunity for me to talk through my thinking here:
Because of the large R4 and the coupling cap., the DC flow from the voltage divider junction to IN+ is very small.
Because of the small current, the voltage drop across R4 is also small (and varies as the AC input does).
If we measure voltage across R4 (+ive at 'top' port), then as the input is sourcing current through R4, the measured voltage across R4 will be -500mV (for a 1Vp-p input).
With the 'top' port of R4 held at 4V5 because of the divider, then the 'bottom' port (IN+) of R4 must be at (4V5 - 500mV) = 4V.
The same would be true for the input sinking current, with IN+ being at 5V

Without R4, there is nothing to drop this +/-500mV across, and so IN+ would be held at 4V5 (no variation).

With this in mind, then yes I see where you're coming from by saying R4 'separates' the DC bias from the AC input.

Let me know if I'm off here...
 
Just as an opportunity for me to talk through my thinking here:
Because of the large R4 and the coupling cap., the DC flow from the voltage divider junction to IN+ is very small.
Because of the small current, the voltage drop across R4 is also small (and varies as the AC input does).
If we measure voltage across R4 (+ive at 'top' port), then as the input is sourcing current through R4, the measured voltage across R4 will be -500mV (for a 1Vp-p input).
With the 'top' port of R4 held at 4V5 because of the divider, then the 'bottom' port (IN+) of R4 must be at (4V5 - 500mV) = 4V.
The same would be true for the input sinking current, with IN+ being at 5V

Without R4, there is nothing to drop this +/-500mV across, and so IN+ would be held at 4V5 (no variation).

With this in mind, then yes I see where you're coming from by saying R4 'separates' the DC bias from the AC input.

Let me know if I'm off here...
Nope, that's essentially correct. The "top" end of R4 is at +4.5V with no signal on it and the "bottom" end of R4 has the full amount of input signal on it biased up to +4.5V.
 

Veracohr

Joined Jan 3, 2011
715
R4 increases the input impedance, and reduces the high-pass filter cutoff by about an octave (from 30.8Hz (Vout) to 15.9Hz (Vout2)), reference designators are different in the images below:

Screen Shot 2020-06-27 at 7.39.08 PM.pngScreen Shot 2020-06-27 at 7.39.23 PM.png


This assumes the opamp input impdeance is infinity. I simulated it with an OPA196 model (max input bias current 20pA) and it was the same.
 
The usual methodology is to first pick an op-amp with desireable characteristics: high input impedance, large GBW, low noise, low input bias voltage, for instance. Then choose the impedance levels of the R's to suit those OA caracteristics (and hopefully the noise levels) and then design their values to give the gain desired. Lastly to choose the cap values for frequency rolloff points, adequate decoupling, etc. The 741 has an input impedance of 1 M Ohm so surrounding it with 1 M resistors is not the best strategy to begin with due to voltage offset considerations. But that aside, the caps are always chosen afterwards to set the corner frequencies so lowering the resistance values immediately means that the caps have to be recalculated. But I didn't want to tear this simple circuit to pieces and completely confuse the guy.
Anyway, as I said before, if the bias cap (C3 in your circuit above) and the 1M bias resistor (R5 above) are left out then the circuit behaves pretty much as before, except that the bias voltage at the junction of R1, R2 in your circuit above, would have 1/2 all the noise present on the Vcc rail, which may not be all that quiet. This noise would be added right into the audio and gained up by whatever the gain knob is set for.
 

Veracohr

Joined Jan 3, 2011
715
Attach files
Anyway, as I said before, if the bias cap (C3 in your circuit above) and the 1M bias resistor (R5 above) are left out then the circuit behaves pretty much as before, except that the bias voltage at the junction of R1, R2 in your circuit above, would have 1/2 all the noise present on the Vcc rail, which may not be all that quiet. This noise would be added right into the audio and gained up by whatever the gain knob is set for.
I was responding to the OP, but I agree with you. The addition of the bias filtering cap indeed removes noise from the DC bias, but then the resistor (R4 in the OP's original schematic) is necessary, otherwise the overall response would be a high pass filter formed from the source's output reisistance (assuming no reactive component) and the C2+C1 capacitance (dominated by C2).
 
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