OPA192: Output error due to Power Supply ripple

Filipe Lage35

Part Number: OPA192

Hi,

I'm working on the power supply requirements for a signal conditioning system and I'm currently analyzing the impacts of power supply ripple on output error.

I'm doing my analysis based on a previous blog post (e2e.ti.com/.../my-amplifier-has-high-psrr-so-i-don-t-need-to-worry-about-supply-variation-right) but this post doesn't cover what happens when the noise signal frequency is greater than the closed loop bandwidth.

Here is an example:

The circuit above (simplified) is used for measuring the output current of a driver, which is driving an external load (R7). Considering that the power supply presents a 125 mV ripple and a switching frequency of 1 MHz, this would give:

OPA192 PSRR @ 1 MHz: ~23 dB

Vos = Vs/(10^(PSRR/20)) = 8.85 mV

The closed loop noise gain is:

Gcl = 1 + R6/R3 = 34.33

Therefore:

Vout = Vos * Gcl = 304 mV.

Considering that the output signal's valid range (Vout) is +-5V, this 304 mV corresponds to more than 3% of full scale.

However, the noise signal frequency is greater than the closed loop bandwidth, given by:

BWn = GBW/Gcl = 10/34.33 = 291 kHz.

What is the expect output error in this case?

My simulation results (figure below), shows the supply voltage (V(n002)), the load current and Vout. A 290 mV oscillation can be seen in Vout.

Are these results correct? Or there should be some attenuation due to the closed loop bandwidth?

Regards,

Filipe

over 6 years ago

0 kai klaas69 over 6 years ago

Guru 140200 points

Hi Filipe,

I cannot open the link. Can you show us a schematic or your circuit?

Kai

0 Filipe Lage35 over 6 years ago in reply to kai klaas69

Prodigy 80 points

Sorry, Kai. The figures I had attached disappeared when I finished the post.

Following is the circuit:

And the simulation results:

0 kai klaas69 over 6 years ago in reply to Filipe Lage35

Guru 140200 points

Hi Filipe,

an AC ripple of 1MHz on the supply line would appear as input voltage of OPAmp, rejected by a factor of about 23dB. This input referred ripple would appear at output of OPAmp, amplified with the noise gain of 30.7dB. But becasue the open loop gain of OPAmp at 1MHz is only 20dB, the input referred ripple is only amplified by a factor of 20dB. So, the 1MHz supply ripple in your circuit appears at output, amplified by a gain of 20dB - 23dB = -3dB.

Kai

0 Zak Kaye over 6 years ago in reply to Filipe Lage35

TI__Mastermind 23050 points

1232-PSRR ac and dc - 12-11.pptxHi Filipe,

What Kai has said is true. The model for the OPA192 has a CMRR closer to 26dB at 1MHz, so any signal on the supply line in your circuit is actually going to see -6dB of attenuation at the output. If this is an asymmetric ripple on your supplies, then this also contributes a common-mode error. However, the CMRR of the OPA192 is much better at 1MHz (~60dB) than the PSRR so the error this contributes should be negligible in comparison. If you want to understand more about this, you may find the attached presentation from the TI precision labs series useful:

0 Filipe Lage35 over 6 years ago in reply to kai klaas69

Prodigy 80 points

Hi Kai,

Thanks for your response.

This solves my issue. I recalculated the expect output noise and it matches the simulation results.

Thanks

Filipe

0 Filipe Lage35 over 6 years ago in reply to Zak Kaye

Prodigy 80 points

Thanks for the presentation Zak. Very helpful.

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OPA192: Output error due to Power Supply ripple