Inverting Amplifier: How to analyze component variation?

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godfreyl said:
No, the reference voltage won't be set to zero. If it was , you'd never be able to do AC analysis of any amplifier circuit because the power supply voltage would be set to zero.
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Well as far as I know, a DC bias point anlysis always proceeds AC analysis. Once you have operating point values (gm etc), you don't need supplies.
 

OK, I see your point. The DC analysis will calculate a value for your feedback resistor, then use that (fixed) value for the AC analysis.
 

Could you just find the laplace formula for a sinewave or exponential or whatever best describes R, and substitute your R value for this in the transfer function? Then make the bode plot?
 
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AlienCircuits said:
Could you just find the laplace formula for a sinewave or exponential or whatever best describes R, and substitute your R value for this in the transfer function? Then make the bode plot?
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Well, I did think that way but there is problem.
Let us say that R2(t)= sin(2pi*f*t).
Then v(t)=i(t)*R2(t), where i(t) is current through R2 and v(t) is voltage across R2. Now multiplication in time domain results in convolution in s-domain.
And that means, I won't be able to get V(s)/I(s) , the way i get for say capacitor or inductor.
 


Your transfer function should be unitless and expressed in vout/vin. no current expression. Also, I think you may be confusing the mechanics of multiplication and convolution with the laplace transform. This action is already built into the transform.


Say Vout/Vin = -R2/R1

then
Let R2(t) = Asin(wt) + Roffset

Then R2(s) = A*(w^2/(s^2 + w^2)) + Roffset/s

so

Vout(s)/Vin(s) = A*(w^2/(s^2 + w^2))/R1 + Roffset/(R1*s)

So you will see
 
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I think Vout/Vin = -Z2(s)/Z1(s) will be valid in s-domain only for R=R, XL=sL and XC=1/sC , not for a rapid varying resistor in question?
 

Is this what you would expect to see? This is magnitude of change in gain with frequency and phase of change in gain with frequency with the above equation used.

I chose a gain of R2/R1 as 0.5
R2 varies by 1% at a frequency of 1kHz.





As I increase the R2 percent variation, the peak gets wider and takes more spectrum. If it is a much smaller variation, the peak is much more local around the R2 frequency. No matter what, they always approach infinity at the R2 frequency and the DC gain is always 0.5 (R2(0)/R1). I am interpreting the infinite peak at the R2 frequency as change in R2 at this frequency divided by change in R1, which would be infinity. At any other frequency, it is the phase and frequency of the input signal interacting with this R2 frequency. Also, if I lower the frequency of R2 change close to DC, the gain is always -6dB (0.5) across the entire spectrum such that the variation has no frequency affect.

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maktoomi said:
I think Vout/Vin = -Z2(s)/Z1(s) will be valid in s-domain only for R=R, XL=sL and XC=1/sC , not for a rapid varying resistor in question?
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You are right. I just am not sure if a rapid varying resistor can be defined in the laplace domain as I did and have it still work out in circuit analysis. But, R != R in your case, so it has to be defined by something else . .
 
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A resistor with a DC voltage across it and modulated by an AC source works as gain element in AC analysis. It can be modelled in SPICE by a behavioral current source, current = voltage across the resistor multiply controlling value.
 


Hi All.

Thanks for your comments.

@AlienCircuits: I could not follow what you tried to do in Matlab. I think I will have to forget about AC in this case and rely on transient analysis result.

Again, Thanks.
 

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