If PSRR is defined in uV/ V, then ideally it should be zero. That means even if there is any change in the supply voltage, there should not be any change in the input offset voltage, and effectively in the output of the op-amp. But when it is defined in dB, then it is -20 dB log (ΔVio/ ΔVs). In datasheets of the op-amp, typically it is defined in dB. And in dB, higher value is more preferable. In the video, please watch it 5:40 onwards.
For notes on Op-amp PSRR, check this link:
bit.ly/40KEEdH
For more videos related to op-amp, check this playlist:
bit.ly/AAE_OperationalAmplifier
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Nice and clear explanation. Thank you.
Why can't you make video on electromagnetics....if u made videos on this subject, that will be more helpful for students like me
How is 0 the ideal state? if it suppresses the ripple more, psrr increases according to your equation.
If PSRR is defined in uV/ V, then ideally it should be zero. That means even if there is any change in the supply voltage, there should not be any change in the input offset voltage, and effectively in the output of the op-amp. But when it is defined in dB, then it is -20 dB log (ΔVio/ ΔVs). In datasheets of the op-amp, typically it is defined in dB. And in dB, higher value is more preferable. In the video, please watch it 5:40 onwards.
How did you calculate the ∆vs to be 0.2v
Thanks
Ripple is +- 100mV. So total change in the supply is 200mV. ( Peak to peak Ripple)
@@ALLABOUTELECTRONICS ok
Thanks
Isn't PSRR for an ideal op amp infinite??
Awsome! Thank you my friend.
Very very nice 👍👍
Nice wrk
it is off+input * psr gain not only off*psr gain