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Design >> Analog Design >> Actual gain Vs. Bodeplot gain ?
https://designers-guide.org/forum/YaBB.pl?num=1384009397

Message started by AAOAA on Nov 9^th, 2013, 7:03am

Title: Actual gain Vs. Bodeplot gain ?
Post by AAOAA on Nov 9^th, 2013, 7:03am

Dear all,

I have the gain @100Khz in bode plot almost=0dB, while the actual
gain= 20 log(Vout/Vin)=20 log(3/0.005)= dB55.56 (from transient)

Could someone please explain why I have this weird bode plot gain reading?

Regards

Title: Re: Actual gain Vs. Bodeplot gain ?
Post by AAOAA on Nov 9^th, 2013, 7:05am

Input transient

Title: Re: Actual gain Vs. Bodeplot gain ?
Post by AAOAA on Nov 9^th, 2013, 7:05am

Bode plot

Title: Re: Actual gain Vs. Bodeplot gain ?
Post by AAOAA on Nov 9^th, 2013, 7:06am

Schematic

Title: Re: Actual gain Vs. Bodeplot gain ?
Post by aaron_do on Nov 10^th, 2013, 5:51pm

Hi,

it might be because your design is non-linear. You have driven your circuit in an unconventional way. I assume you are trying to model some kind of sensor. Your transfer function is proportional to

(100Vin + 100k)/(100Vin + 200k), whose derivative is,

10M/(100Vin + 200k)²

assuming my maths is right. From the equation, the nonlinearity shouldn't really show up, but that's the only thing I noticed that might be causing issues.

Why don't you just model your sensor in a more conventional manner (unless you are trying to see the nonlinearity)? For instance a Norton or Thevenin equivalent circuit? Also, R1 and R4 don't seem to serve any purpose.

Aaron

Title: Re: Actual gain Vs. Bodeplot gain ?
Post by tm123 on Nov 11^th, 2013, 7:02am

I think you are incorrectly calculating the gain from the transient plot. Can you explain why you calculate the transient gain as 20*log(3/0.005)? Just trying to read some approximate values off of the plot I calculate the transient gain as ~1.5dB which matches the Bode plot pretty well. Remember the input to your circuit is the voltage source V2 not the voltage at the 'impedance' node. At least that is how the Bode plot is generated. If you are calculating the transient gain as V(out)/V(impedance) then that is not the same as the gain found in the Bode plot which is V(out)/V(CNT).

Hope this helps.

Tim

Title: Re: Actual gain Vs. Bodeplot gain ?
Post by AAOAA on Nov 11^th, 2013, 8:37am

Dear aaron_do and tm123,

Both your answers fixed the problem, by putting an equivalent source to the opamp it gave the right gain now.

Thanks,
A.A