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Old 17th September 2007, 05:18 PM   #1
hags is offline hags  United States
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Default Ideal feedback resistor vales

I want to know how to determine what the ideal feedback resistor values for a given circuit would be.
I have an op amp based line amp (OPA627) that I use with no buffer. It has a gain of 3 or 9db. I have chosen feedback resistors of 1000 and 2000.
My question is, would higher value resistors at the same ratio be better? Would they raise the output impedance?
These feedback resistors are in parallel with the input of the next stage (amplifier), correct?
By using low values am I causing the op amp to drive to low of an input impedance?
I also have a line amp using the LME49710 with a BUF634 in the feedback loop. I am using the same values for the resistors here for the same amount of gain (or slightly lower with the buffer).
Are the values more important here or less important?
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Old 17th September 2007, 05:36 PM   #2
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Remember that essentially no current flows in the inputs, so the load to ground is just the series sum of the resistors, plus the actual load. Most op-amps drive 2K nicely, and some do well with 600 ohms. At very low signal levels, they can go even lower, thus phono input stages with a few hundred ohms. IMO, your values are fine. If you go a lot lower, you might get more distortion, and if you go (quite a bit) higher, stray capacitance and pickup get involved. There's really a huge range of acceptable values, with little difference between them. I typically use about 10K as a starting point, though with bipolar amps you want to also think about equalizing the values seen by both inputs.
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Old 17th September 2007, 05:40 PM   #3
hags is offline hags  United States
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I should mention that both designs are used in the non-inverting topology.
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Old 17th September 2007, 05:50 PM   #4
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Shouldn't matter, as the rules are about the same. The problem when you go inverting is the input resistance can get too low for the source. OTOH, I've read that op-amp performance is usually a bit better one way over the other. Problem is, I can never remember which way!
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Old 17th September 2007, 06:24 PM   #5
AndrewT is offline AndrewT  Scotland
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Hi,
yes, the Zin of the next stage appears in parallel to the Rfb of this stage.
So take care to not go too low, particularly at maximum output voltage. Have a look at the distortion Vs load impedance and Output voltage Vs load impedance.
Both these will show a variety of closely spaced load curves and then a curve where performance is markedly different. I would suggest that the one that is different has gone too low. Your parallel impedance should be about the same value (or higher) as the lowest of the closely grouped bunch.
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Old 17th September 2007, 08:58 PM   #6
KSTR is offline KSTR  Germany
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With FET-input op-amps in non-inv mode it is good practise to exactly match the impedances at both input pins, Z(+IN)==Z(-IN). That is, not only the resistive part. This keeps distortion low and helps to minimize offsest. Of course one usually strives to use as low an impedance as one can afford (without getting other drawbacks, e.g. overloading the output), from noise and stability standpoints. Walt Jung (www.waltjung.org) has elaborated on this.

For your example, we would get a 1k//2k=667Ohms impedance at the -IN, so the source impedance should also be 667Ohms. With normal sources like a CD player (~100Ohms Z_out) one would need a series resistor in the input, which would result in a neglegible noise penalty for the OPA627 (below 3k total Z there is no improvement). Not a bad idea anyway...

- Klaus
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Old 17th September 2007, 09:23 PM   #7
hags is offline hags  United States
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Quote:
Originally posted by KSTR
With FET-input op-amps in non-inv mode it is good practise to exactly match the impedances at both input pins, Z(+IN)==Z(-IN). That is, not only the resistive part. This keeps distortion low and helps to minimize offsest. Of course one usually strives to use as low an impedance as one can afford (without getting other drawbacks, e.g. overloading the output), from noise and stability standpoints. Walt Jung (www.waltjung.org) has elaborated on this.

For your example, we would get a 1k//2k=667Ohms impedance at the -IN, so the source impedance should also be 667Ohms. With normal sources like a CD player (~100Ohms Z_out) one would need a series resistor in the input, which would result in a neglegible noise penalty for the OPA627 (below 3k total Z there is no improvement). Not a bad idea anyway...

- Klaus


Interesting, I have never given the impedance that the inverting input sees any consideration.
I was more concerned with what load the OPA627 sees due to not using it with a buffer.
The non-inverting input is preceded by a 25K discrete attenuator, how would one adjust for that?.
With an input impedance of 100K on my amplifier that this line amp is driving and the feedback resistors I'm using, the load the op amp sees is 2.913K? I'm I figuring this correctly?
Does that seem low to anyone?
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Old 17th September 2007, 09:33 PM   #8
AndrewT is offline AndrewT  Scotland
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Hi,
an unbuffered or unblocked variable source resistance will behave as a variable output offset adjustment.
If an RF cap is also fitted then the attenuator now becomes a variable filter.
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Old 17th September 2007, 09:40 PM   #9
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Great, the secret's out. Now everybody will want one.
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Old 17th September 2007, 09:41 PM   #10
hags is offline hags  United States
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Quote:
Originally posted by AndrewT
Hi,
an unbuffered or unblocked variable source resistance will behave as a variable output offset adjustment.
If an RF cap is also fitted then the attenuator now becomes a variable filter.

How else would I attenuate the source? What would be a better topology, putting the volume control at the output?
Variable output offset adjustment? Could you explain?
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