[SOLVED] Relation of PSRR at DC to loop gain

[diarmuid]

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

 

[erikl]

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.

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).

 

[diarmuid]

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.