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Old 26th March 2011, 10:37 PM   #1
yakideo is offline yakideo  Poland
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Default Cmoy resistor to ground

Hello,

I've been looking at and trying to understand the cmoy op-amp circuit. One thing that I couldn't figure out the use for was the R2 resistor connected from the input to ground. If C2 is used to block dc voltage what is the use of R2?

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Old 26th March 2011, 10:41 PM   #2
Atilla is offline Atilla  Norway
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C2 and R2 together form the high-pass filter that blocks the DC on the input.
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Old 26th March 2011, 10:45 PM   #3
yakideo is offline yakideo  Poland
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Quote:
Originally Posted by Atilla View Post
C2 and R2 together form the high-pass filter that blocks the DC on the input.
Would removing R2 still block the DC current? I mean why the high-pass filter, unless it would damage the headphones?
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Old 27th March 2011, 06:59 AM   #4
Mooly is online now Mooly  United Kingdom
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R2 is vital and the circuit will not work and will quite possibly damage the headpnones without it.

It defines the DC voltage on the + input of the opamp

On a practical level, the value of 100k in combination with 0.1uf is not ideal for good low frequency response. I would suggest at least a 0.47uf cap.

Filter Circuits with Capacitors

[Also... from a theoretical point, the resistor modifies the "law" of the pot and adding a resistor of the appropriate value can be used with a linear pot to give a pseudo log response. You would need a much lower value than 100k though.]

Edit... and why use C2 at all ?
Firstly because you can not guarantee that the source feeding the amp will have no DC present. Second... any DC current that flows in the wiper of the pot tends to make it very crackly and noisy as it is turned. Thirdly... depending on the opamp, the DC offset will change as the pot is rotated. FET opamps will not exhibit that behaviour in this circuit.
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Last edited by Mooly; 27th March 2011 at 07:03 AM.
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Old 27th March 2011, 02:02 PM   #5
yakideo is offline yakideo  Poland
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Quote:
Originally Posted by Mooly View Post
It defines the DC voltage on the + input of the opamp
I though that C2 was there to block any DC?

Quote:
Originally Posted by Mooly View Post
On a practical level, the value of 100k in combination with 0.1uf is not ideal for good low frequency response. I would suggest at least a 0.47uf cap.
Right, a 0.47uF cap would lower to cutoff frequency to 3.4Hz.

Quote:
Originally Posted by Mooly View Post
Second... any DC current that flows in the wiper of the pot tends to make it very crackly and noisy as it is turned.
Wouldn't it be better then to have C2 (possibly R2) before the wiper?
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Old 27th March 2011, 05:26 PM   #6
Mooly is online now Mooly  United Kingdom
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Not easy to explain quickly

C2 is to block any DC from the source and to stop it affecting the opamp in any way.

The two inputs of the opamp must have their DC conditions exactly defined... and that means tying or referencing the + input to ground. The - input has it's DC conditions set by the feedback network.

I can understand your confusion.
If it were AC coupled only the + input would just quickly drift to some ill defined level. The opamp inputs don't assume zero volts when not connected to anything.
The golden rule of design for 99% of opamp applications is,

"The output will do whatever is neccessary to keep the voltage difference between the inputs at zero

Try and understand that, put some numbers in and have a go at calculating the output and feedback voltages and you'll see how it makes sense.

Have a read at this,
http://users.ece.gatech.edu/mleach/e...4/OpAmps01.pdf

Location of C2 ? Good question. Where it is now stops any DC current in the wiper of the pot and that's essential to keep it quiet.
Some opamps such as bjt types (NE5532 etc) have a significant current flowing from the inputs (it's in the data sheets) and that is enough to make the pot noisy. If you used a FET opamp then there is essentially zero current and so you could couple directly to the wiper and move the cap but that is not considered good practice... but yes you could
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