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[SOLVED] Relation of PSRR at DC to loop gain

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diarmuid

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Hello All,

Just wondering if anyone has an explanation for this:

I am looking at the PSRR at DC of a V2I converter, basic schematic attached.

The loop gain at DC is 90dB. However, my PSRR at DC (taken at the top of the resistor) is 80dB.

Does this make sense? I would have thought the PSRR at DC should equal the loop gain at DC.

Any ideas where I am going wrong here?

Thanks,

Diarmuid
 

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diarmuid said:
The loop gain at DC is 90dB. However, my PSRR at DC (taken at the top of the resistor) is 80dB.

Does this make sense? I would have thought the PSRR at DC should equal the loop gain at DC.
Click to expand...

There's no direct connection between open loop DC gain and PSRR at DC (better: at low frequencies: DC is static, no change, so no influence on the output).

You can say the PSRR at any frequency - in a first order approximation - will be approved by the (neg.) feedback size, i.e. by the ratio open-loop_gain/actual_gain. As your amplifier has full feedback and runs with gain=1 , measured low-frequency PSRR of 80 dB at gain=1 means a 80-90=-10dB low-frequency PSRR at open loop.

PSRR taken as 20*log(power_supply_interference_voltage/output_reaction_voltage).
 
Yes, as described in Willy Sansens Analog Design Essentials: PSRR = Avol/Avdd. In my case, my Avol was 90dB but my Avdd was 10dB (output stage of opamp was gaining up the noise seen
at the output). This is where the 10dB was going.
 

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