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Old 18th March 2005, 02:16 PM   #1
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Default Balanced resistor network?

Hi

I like to now why Nelson and other use this type of input network.
As an example on the + input we use 10k in series and 10k + 100k to ground.
And the – input use 10k series and 10k to ground and 100k feedback.
With this network you lose the half of the input signal and double the gain of the amp over unbalanced input.
Also you have on the + input 10k in series with the capacitane to ground from the mosfet input and thereby lower the high freq response.
Is it possible to only use the 10k to ground on the + Input and short the – input to ground.
Then you have the same gain from the amp unbalanced or balanced and have a higher roll off point.
Can somone eplain this in a simple way.

Rob
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Old 19th March 2005, 04:01 AM   #2
PRR is offline PRR  United States
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> Is it possible to only use the 10k to ground on the + Input and short the – input to ground.

Then you don't have ANY feedback. What sets your gain?

If I am reading your text-description correctly: you describe the very-standard "Four-R One op-amp differential amplifier", plus some DC return resistors. That is the only way to get a good approximation of differential input feedback-controlled gain with just one amplifier. There are 2-amp and 3-amp differential-inputs with better unbalanced performance, but not necessarily better audio performance in line-level work, and of course possible degradation from the extra amplifying stages.

Yes, it is possible to design a diff-input without overall feedback. And of course to design a power amp without overall feedback, though usually with a lot of local feedback snuck in. But the simple 4-R opamp diff-input works very well.

Those extra 10K resistors do hurt some. Can you point to an example?
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Old 19th March 2005, 09:33 AM   #3
uli is offline uli  Austria
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Default Re: Balanced resistor network?

Quote:
Originally posted by Rob Dingen
As an example on the + input we use 10k in series and 10k + 100k to ground.
And the – input use 10k series and 10k to ground and 100k feedback.
Hi Rob,

for offset minimizing reasons the impedance seen at both inputs should be equal.
Therefore you need to place this 100k parallel to the 10k from +input to gnd,
as the 100k fb is virtually parallel to the 10k to gnd at the -input.

Uli

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Old 19th March 2005, 01:19 PM   #4
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Hi PRR

The input networks are from aleph amps.
I don’t want an amp without feedback I just want to now why to use this type of input network and the reason.
If I should design an input network then on the + input I use only 10k to ground and on the – input I use 10k in series and 100k to the output.
Then both input are 10k input impedance.
Or did I miss something?

Rob
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Old 20th March 2005, 07:00 AM   #5
uli is offline uli  Austria
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Quote:
Originally posted by Rob Dingen

If I should design an input network then on the + input I use only 10k to ground and on the – input I use 10k in series and 100k to the output.
Then both input are 10k input impedance.
Offset depends on the impedance seen BY the input not by the source.
In your arrangement the impedance seen by the + input is nearly zero,
assuming a zero impedance source, the impedance seen by the - input is
about 9k0909 (10k || 100k). In the aleph arrangement both inputs
see the same impedance: 10K || 10k ||100k = 4k762.

Uli

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Old 20th March 2005, 09:41 AM   #6
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Hi Uli

Ok now I understand thanks.

Rob
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