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Op amp current noise calculations

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ElecDesigner

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Hi,

I am designing a preamp for a type of transducer element (mostly capacitive) and want to make sure that I am using the best (lowest noise) OpAmp for the job so I need to find out if Voltage noise or Current noise is the dominating factor.

See attached. I have attempted to reduce the circuit for analysis as much as possible and have used En x gain to find the output referred voltage noise but am unsure how to find the current noise. Not sure if I its valid to make it output referred by multiplying by the gain. Not sure if I need to combine the effect from the feedback resistors. I think I do but not sure how.

I realize for a full analysis I should include the resistor Johnson noise but I am not worried about that right now.
 

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KlausST

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Hi,

input voltage noise is given in datasheet.

input current noise is given in datasheet. This has to be multiplied with the input impedance. (To make it more simple you can choose a single frequency (noise) of interest) to give a corresponding voltage noise.

you need to add them with square and root

e(t) = sqrt( e_volt^2 + e_curr^2) = input referred

for two opamps: you can multiply it with 1.414

for output referred you need to multiply the input referred noise with the OPAMP gain. (simplified)

****
a simple rule:
your values: en =3.9nV/sqrt(Hz), in = 0.4pA(sqrt(Hz) --> r = U/I --> 3.9nV/0.4pA = 9750 Ohms
if your input impedance is > 9750 Ohms, then current noise dominates
if your input impedance is < 9750 Ohms, then voltage noise dominates

Although this means that this OPAMP has best noise performance with about 10kOhms source impedance but to compare two OPAMPs you allways need to calculate both input noise values.

*****
for more information i recommend to read:
<Noise Calculations in Op Amp Circuits – Design Note 15 / Alan Rich>
https://www.linear.com/docs/4201
(linear technology, www.linear.com)

Klaus
 

ElecDesigner

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Hi,

input voltage noise is given in datasheet.

input current noise is given in datasheet. This has to be multiplied with the input impedance. (To make it more simple you can choose a single frequency (noise) of interest) to give a corresponding voltage noise.

you need to add them with square and root

e(t) = sqrt( e_volt^2 + e_curr^2) = input referred

for two opamps: you can multiply it with 1.414

for output referred you need to multiply the input referred noise with the OPAMP gain. (simplified)

****
a simple rule:
your values: en =3.9nV/sqrt(Hz), in = 0.4pA(sqrt(Hz) --> r = U/I --> 3.9nV/0.4pA = 9750 Ohms
if your input impedance is > 9750 Ohms, then current noise dominates
if your input impedance is < 9750 Ohms, then voltage noise dominates

Although this means that this OPAMP has best noise performance with about 10kOhms source impedance but to compare two OPAMPs you allways need to calculate both input noise values.

*****
for more information i recommend to read:
<Noise Calculations in Op Amp Circuits – Design Note 15 / Alan Rich>
https://www.linear.com/docs/4201
(linear technology, www.linear.com)

Klaus


Hi,

This makes good sense to me on one hand. However I am still not certain about the -ve pin. Do I need to include some current noise for that.

I am looking at https://www.youtube.com/watch?v=W0vfALQ_n54 where an AD engineer shorts the +ve pin to ground and works out the current noise due to the feedback resistors. Actually I think he should have only done the calculations for the one between the output and -ve input according to the comments below the video.
 

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