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Input Z of single supply opamp circuit

Dan Moos

Oct 5, 2015
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Ok, take a typical 9 volt powered single supply opamp circuit, in this case a simple unity buffer if it matters.

The 4.5v Vref is derived from the nine volt supply from a voltage divider (100k both halves of divider)

The Vref is is applied to the non inverting input of the opamp through a resister. Lets say its a 100k resister, but the crux of my question is here, so maybe its not.

Finally, lets just say the opamp achieves the theoretical infinite input impedance for simplicity's sake. I imagine for our purposes, it might as well be.

Ok, taking the application of Vref to the non inverting through the 100k resistor into account, how does one determine the input resistance of the circuit at that node?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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The impedance you would measure at the non-inverting input would be 150K. This is comprise of the impedance of the divider (50k -- assume both resistors are in parallel) plus the 100k effectively in series with that.

This is in parallel with the input impedance of the op-amp, but that is probably so much greater that the impact of it would be less than that of the tolerance of the resistors.
 

Dan Moos

Oct 5, 2015
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Ok, I thought this might be the case. I had wondered if the bias voltage muddied things, but , correct me if I'm wrong, it being effectively the same as 0v in a split supply, it's just as if the bias network is a group of grounded resisters.

Thanks for the reply!

For a follow up, this needs to be a nice quiet circuit. While none of the resisters in question here are in the signal path, I imagine I still want to keep them small for noise purposes (ignoring current draw for now). Is this the case?
 

(*steve*)

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It depends on what you call quiet. You should probably bypass one (or both) of the resistors in your voltage divider to help prevent noise on the supply feeding back into the input. This will also have the effect of reducing your input impedance (worst case is 100k).
 

Dan Moos

Oct 5, 2015
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Ok, actually I do have both divider resistors bypassed. I didn't think of it, but I suppose that eliminates the divider resisters from the equation as far as AC is concerned. Since I last posted, I figured out how to do noise analysis in LTSpice, and the resister feeding the Vref to the opamp is a very small noise contributer even when I go much larger. This is a for an overdrive pedal for guitar, so while I would like it to be quiet, I suspect my quiet is louder than alot of other applications!
 

AnalogKid

Jun 10, 2015
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Can you share a schematic of your design? It will help greatly with the advice.

ak
 

Nicholas Schlensky

Oct 7, 2015
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If it is inverting amp as you seem to suggest the impedance is Rin ( you dont mention what this is) plus the res on non inverting input to vref =100k plus 100//100k (from Vref divider) so Rin+150K is input impedance.
If you use a 100k to make vref divider you will get problems because input curent and bias current will flow through this and cause dc offsets that will be multiplied by amp gain. Vref should be a low impedance source like a opamp buffer.
If you use a non inverting cct. the impedance will be 50K (100k//100k) plus 100k//Rf(you dont say what it is). you could save an opamp this way becasue the Vref eq resistance could provide one of the resistors in the noninverting amp. 100k//100k= R1 100K=R2 Av= 1+100k/50k. This doesnt include any caps in the cct and if the resistors arnt to big and the caps are not present you wont have to work out anything else other wise you need to add 1/2pi*fC to add cap effect whereever
 
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