Op Amp Calculations
in [Design]
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From: Fred Bloggs on 26 Sep 2005 21:13 John Woodgate wrote: > I read in sci.electronics.design that The Phantom <phantom(a)aol.com> > wrote (in <kfcgj1hadlo6jbiv02pt12jnesa30atfd2(a)4ax.com>) about 'Op Amp > Calculations', on Mon, 26 Sep 2005: > >> (Refer to Jim Thompson's schematic in >> alt.binaries.schematics.electronic) > > > Instead of that, go back up the tread and see my ASCII art re-draw of > the feedback as a simple potential divider across the output, with the > feedback resistor taken from the tap. It's FAR easier to analyse. The T-PI transformation is a natural here because you don't care about the two shunts arms most of the time and the poles/zeroes fall right out: View in a fixed-width font such as Courier. - - - Z - - - ---[R1]--+--[R2]--- | [R3] | === |C | --- R1R2+R1(R3+1/SC)+R2(R3+1/SC) Z= --------------------------- R3+1/SC (R1R2+R1R3+R2R3) ---------------- SC R1 + R2 + 1 Z= (R1 + R2) ------------------------ R3CS + 1 (R12 + R3 )CS + 1 Z= (R1 + R2) ----------------- ;R12=R1||R2 R3CS + 1 +--- Z---+ | | | |\ | Vi>--[Ri]--+-|-\ | | >---+--> Vout +-|+/ | |/ --- (R1 + R2) (R12 + R3 )CS + 1 Vout/Vin=(-) --------- x ----------------- Rin R3CS + 1 Add the input shunt component: ---[R1]--+--[R2]--- | | [R3] Zs | === | |C | --- (R12 + R3 )CS + 1 Zs= (R1 + R2) ----------------- ;R12=R1||R2 R2CS +--- Z---+ | | | |\ | Vi>--[Ri]---+-----+-|-\ | | | >---+--> Vout [Rs] +-|+/ | | |/ === --- |Cs | --- R1 Rs= ( 1 + -- ) (R12 + R3 ) R2 C Cs= ---------- R1 ( 1 + -- ) R2 1 midband fractional gain error= ------------ 1 1 - ------- A(jw)*Rs A(jw)=OA gain
From: Terry Given on 26 Sep 2005 22:35 Fred Bloggs wrote: > > > John Woodgate wrote: > >> I read in sci.electronics.design that The Phantom <phantom(a)aol.com> >> wrote (in <kfcgj1hadlo6jbiv02pt12jnesa30atfd2(a)4ax.com>) about 'Op Amp >> Calculations', on Mon, 26 Sep 2005: >> >>> (Refer to Jim Thompson's schematic in >>> alt.binaries.schematics.electronic) >> >> >> >> Instead of that, go back up the tread and see my ASCII art re-draw of >> the feedback as a simple potential divider across the output, with the >> feedback resistor taken from the tap. It's FAR easier to analyse. > > > The T-PI transformation is a natural here because you don't care about > the two shunts arms most of the time and the poles/zeroes fall right out: > View in a fixed-width font such as Courier. thats how I did it, and why :) Cheers, Terry > > > > - - - Z - - - > > ---[R1]--+--[R2]--- > | > [R3] > | > === > |C > | > --- > > > > R1R2+R1(R3+1/SC)+R2(R3+1/SC) > Z= --------------------------- > R3+1/SC > > > > (R1R2+R1R3+R2R3) > ---------------- SC > R1 + R2 + 1 > Z= (R1 + R2) ------------------------ > R3CS + 1 > > > > > (R12 + R3 )CS + 1 > Z= (R1 + R2) ----------------- ;R12=R1||R2 > R3CS + 1 > > > > +--- Z---+ > | | > | |\ | > Vi>--[Ri]--+-|-\ | > | >---+--> Vout > +-|+/ > | |/ > --- > > > > > (R1 + R2) (R12 + R3 )CS + 1 > Vout/Vin=(-) --------- x ----------------- > Rin R3CS + 1 > > > > > Add the input shunt component: > > > > ---[R1]--+--[R2]--- > | | > [R3] > Zs | > === > | |C > | > --- > > > (R12 + R3 )CS + 1 > Zs= (R1 + R2) ----------------- ;R12=R1||R2 > R2CS > > > > > +--- Z---+ > | | > | |\ | > Vi>--[Ri]---+-----+-|-\ | > | | >---+--> Vout > [Rs] +-|+/ > | | |/ > === --- > |Cs > | > --- > > > R1 > Rs= ( 1 + -- ) (R12 + R3 ) > R2 > > > C > Cs= ---------- > R1 > ( 1 + -- ) > R2 > > > 1 > midband fractional gain error= ------------ > 1 > 1 - ------- > A(jw)*Rs > A(jw)=OA gain >
From: The Phantom on 26 Sep 2005 23:51 On Mon, 26 Sep 2005 19:24:53 +0100, John Woodgate <jmw(a)jmwa.demon.contraspam.yuk> wrote: >I read in sci.electronics.design that The Phantom <phantom(a)aol.com> >wrote (in <kfcgj1hadlo6jbiv02pt12jnesa30atfd2(a)4ax.com>) about 'Op Amp >Calculations', on Mon, 26 Sep 2005: >> (Refer to Jim Thompson's schematic in >>alt.binaries.schematics.electronic) > >Instead of that, go back up the tread and see my ASCII art re-draw of >the feedback as a simple potential divider across the output, with the >feedback resistor taken from the tap. It's FAR easier to analyse. I guess you and I will have to disagree about what "...FAR easier..." means. You first give an expression which is only approximate and then to make it exact, you say: "replace R4 by R4 + (R2R3/(R2 + R3)". By the time this is done, the amount of algebra is not significantly (if any) less than a Y-Delta method: Knowing that the shunt arms of the equivalent delta have no effect on the signal gain, we need only compute the series arm and use the old Rf/Ri gain formula for the two-resistor case, with Rf replaced by the expression for the delta series arm. Thus, the series arm is given by (R3*R4 + R2*R3 + R2*R4)/R3. Since the input resistor is R1, we have the exact gain expression of Rf/Ri = (R3*R4 + R2*R3 + R2*R4)/(R1*R3). I can't see that, as you said, "Gain = (R4/R1) x {(R2 + R3)}/R3, if R4 is very much larger than R3. If it isn't, replace R4 by R4 + (R2R3/(R2 + R3)." is "...FAR easier..." than recognizing that the equivalent Rf is (R3*R4 + R2*R3 + R2*R4)/R3, which when divided by R1 gives the signal gain.
From: Terry Given on 27 Sep 2005 00:09 The Phantom wrote: > On Mon, 26 Sep 2005 19:24:53 +0100, John Woodgate <jmw(a)jmwa.demon.contraspam.yuk> wrote: > > >>I read in sci.electronics.design that The Phantom <phantom(a)aol.com> >>wrote (in <kfcgj1hadlo6jbiv02pt12jnesa30atfd2(a)4ax.com>) about 'Op Amp >>Calculations', on Mon, 26 Sep 2005: >> >>> (Refer to Jim Thompson's schematic in >>>alt.binaries.schematics.electronic) >> >>Instead of that, go back up the tread and see my ASCII art re-draw of >>the feedback as a simple potential divider across the output, with the >>feedback resistor taken from the tap. It's FAR easier to analyse. > > > I guess you and I will have to disagree about what "...FAR easier..." means. You first > give an expression which is only approximate and then to make it exact, you say: > "replace R4 by R4 + (R2R3/(R2 + R3)". By the time this is done, the amount of algebra is > not significantly (if any) less than a Y-Delta method: > > Knowing that the shunt arms of the equivalent delta have no effect on the signal gain, > we need only compute the series arm and use the old Rf/Ri gain formula for the > two-resistor case, with Rf replaced by the expression for the delta series arm. > > Thus, the series arm is given by (R3*R4 + R2*R3 + R2*R4)/R3. Since the input resistor > is R1, we have the exact gain expression of Rf/Ri = (R3*R4 + R2*R3 + R2*R4)/(R1*R3). > > I can't see that, as you said, "Gain = (R4/R1) x {(R2 + R3)}/R3, if R4 is very much > larger than R3. If it isn't, replace R4 by R4 + (R2R3/(R2 + R3)." > > is "...FAR easier..." than recognizing that the equivalent Rf is (R3*R4 + R2*R3 + > R2*R4)/R3, which when divided by R1 gives the signal gain. except that if the frequency of interest is not far, far below UGBW, the shunt arm connected to the -ve input cannot be ignored. Hey, its only a little bit of baby maths. Cheers Terry
From: Pooh Bear on 27 Sep 2005 00:13
John Woodgate wrote: > I read in sci.electronics.design that The Phantom <phantom(a)aol.com> > wrote (in <kfcgj1hadlo6jbiv02pt12jnesa30atfd2(a)4ax.com>) about 'Op Amp > Calculations', on Mon, 26 Sep 2005: > > (Refer to Jim Thompson's schematic in > >alt.binaries.schematics.electronic) > > Instead of that, go back up the tread and see my ASCII art re-draw of > the feedback as a simple potential divider across the output, with the > feedback resistor taken from the tap. It's FAR easier to analyse. I can't imagine analysing it any other way ! Graham |