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Old 2nd March 2006, 08:17 AM   #1
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Default Opamp confusion about gain

Hello,

I'm quite a newbie in audio DIY design but I already successfully "designed" a preamp based on a opamp input stage, a PGA2310 und uController.

For the next project I again design a preamplifier and looked at many great pages (mhennessy, sound.westhost.com and a bunch of opamp PDFs) which give me the necessary hint for the classic non-inverting gain circuit:

Click the image to open in full size.

I do understand that:
1. Input impedance is defined by Rimp in parallel to the input impedance of the opamp

2. Gain is defined by Av = (Rf + Rg) / Rg

3. C1-C5 are used for decoupling

4. Cin forms a high pass filter to prevent DC voltage at the input of the opamp.

But there are several questions open to me which I couldn't find an answer for yet:
1. What are the "dimension rules" for the resistor values for Rf and Rg? The formula allows mathematically infinite possibilities. I saw on other posts that "big" values can cause higher noise. So should I go as low in the values as I can?

2. What is the purpose of Rin? I saw in one post the recommendation that this value should be the parallel value of Rf and Rg. Why?

3. Why is Rd able to "drive longer cables" as stated in many posts and what are the dimensioning rules for this one?

I'm sorry if these questions are too dumb but I would like to understand

Best regards,
Sergiusz
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Old 2nd March 2006, 08:34 AM   #2
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Rf and Rg should be dimensioned to the application and chip used. Too high impedance and noise and stray capacitance come into play, too low value and they load the op-amp output too much, or you exceed their power rating.

Rin is used to make sure the inverting and non-inverting inputs have the same impedance to ground to minimise offset voltage due to input bias current. I never bother with it as it's a negligible effect.

Rd is used to swamp out the capacitance of long cables by making a highly resistive component in comparison to the capacitive reactance. Driving reactive loads reduces the stability margin in feedback systems due to phase shift. Make the value too high and you attenuate the output and have poor damping, make it too low and it's effect is not enough. Again you must choose it to suit your application.
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Old 2nd March 2006, 09:31 AM   #3
AndrewT is offline AndrewT  Scotland
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Hi,
Rd forms a low pass filter with any capacitance to ground that follows it. Cable often has about 50pF to 100pF /m so you get a different lowpass filter depending on cable length and type.

A low value Rd allows higher capacitance but then stresses the opamp both in current output at high frequency and phase margin. Your 240r is a good compromise for cables upto about 30m long (3nF).

The resistive load to ground (or zero impedance) from the input pins should match.
your non-inverting pin has 1k+100k =101k to ground.
your inverting pin has 10k//10k =5k to ground.
for lowest output offset (no DC blocking cap) these should match. Unfortunately input offset current and input offset voltage complicate this matching and for precision, trial & error is probably required.

If ALL your following poweramps have DC block at their inputs then your DC coupled opamp is OK.

I have not experimented with noise but for an opamp driven to near maximum levels at peak output (+20dbu) then feedback resistors approaching 50K would probably not be noticeable. But if your peak output was only 775mV(0dbu) then 50ks would probably be noisy and you might want lower values, 10ks maybe. OPA2134 can drive down to 2k (check the data sheet) with ease but this is the parallel total of feedback and output load and capacitance equivalent.
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Old 2nd March 2006, 01:19 PM   #4
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Hi,

Now the picture clears up crystal-clear, thank you very much for enlightenment

Especially Rin was not clear to me but now is. The values are taken from Rod Elliots project at http://sound.westhost.com/project88.htm that's why I used these values for the example.

I assume DC-Offset measurement at the opamp output is just as simple as measuring DC across the output to ground (as in regular power amplifiers)?

Thanks again,
Serge
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Old 2nd March 2006, 01:26 PM   #5
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Yes
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Old 2nd March 2006, 06:50 PM   #6
shusha is offline shusha  Croatia
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Quote:
Originally posted by AndrewT
your inverting pin has 10k//10k =5k to ground.
Can you explain this? As I understand there is only 10k from Inv. Input to the ground.
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Old 2nd March 2006, 08:52 PM   #7
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Yes but the output is also at ground potential (at idle) so the other 10k resistor is effectively connected to ground.
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Old 3rd March 2006, 09:42 PM   #8
shusha is offline shusha  Croatia
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Thanks! I just haven't look at this that way. It makes sense.
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Old 4th March 2006, 08:27 AM   #9
AndrewT is offline AndrewT  Scotland
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Hi,
I had made the comparison of power amp to opamps and used ALL the resistors connected to the inv & non inv pins. These parallel combinations control the output offset voltage and in turn current, forgetting that some of the resistors do not carry DC current due to placement of DC blocking caps.
This applies to opamps, filters, power amps and preamps, anywhere that a discrete or IC pair of inv & non-ivn appear, so it's use is universal.
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