Op Amp Calculations
in [Design]
|
From: Pooh Bear on 27 Sep 2005 00:15 Jim Thompson 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. > > But considering it so low an impedance, that you call it just a > divider, is a degenerate case... and certainly NOT the most useful > one. Just take the source resistance of the divider and add it to the attached feedback R. It resolves very easily. Graham
From: The Phantom on 27 Sep 2005 00:51 On Wed, 21 Sep 2005 12:20:11 -0700, "RST Engineering \(jw\)" <jim(a)rstengineering.com> wrote: >A lot of years ago (38 to be exact) a wise old "rule of thumb" engineer >taught me a slick way of getting maximum gain out of an opamp without >resorting to very high or very low values of resistors. > >As we all know, for an inverting opamp, the gain is given simply by Rf/Ri, >where Rf is the feedback resistor from output to inverting input and Ri is >the resistor between signal and inverting input. The DC level of the output >may be set anywhere you choose by an appropriate bias level on the >noninverting input. For AC amplifiers from a single supply, this is >generally Vcc/2 with capacitive coupling between Ri and the signal. > >However, for very large AC gains, either Rf must be rather large or Ri >rather small. Rf being rather large makes the input voltage/current errors >become significant as regards quiescent DC output point and Ri being rather >small requires large capacitors for coupling and loading errors from the >signal source. > >So, sez old wily rule-of-thumb, just break Rf into two reasonable sized >equal value resistors equal to Rf/2 and run them in series from output back >to (-) input. And, from the midpoint tap on these two resistors run a >series RC circuit to ground. Bingo, the AC gain improves greatly. > >And guess what, it works. How do I calculate the R in the series RC circuit >I asks old wily. The answer comes back "Tweak it until you get the gain you >want." (Assume that C can be made appropriately large to get the >low-frequency gain you want.) > >I haven't used that trick in an awfully long time, but I've got an >application that needs it. And, if I want to use Diddle's constant in a >simulation program I can fool around (ahem, heuristically experiment) to get >the gain I need. > >However, I can't convince myself that I can mathematically come up with the >resistor value. I have googled the problem and come up short. Anybody got >a pointer to a URL that goes through the math of how this configuration >works? And what I'm doing to my phase margin? > >Jim > I remember Robert Pease in his column in Electronic Design magazine dealt with this topic: http://www.elecdesign.com/Articles/Index.cfm?ArticleID=3929 Since he shows a circuit with no input resistor, but rather just an input current (this would be equivalent to an infinite input resistance), his comments on noise gain would have to be modified to apply to the circuit under consideration here.
From: The Phantom on 27 Sep 2005 02:10 On Tue, 27 Sep 2005 16:09:44 +1200, Terry Given <my_name(a)ieee.org> wrote: >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. That's why in my discourse on the use of the Y-Delta transformation, I began by saying: " In what follows, I'll assume the capacitor is replaced with a short and only first order effects will be considered." But for those who care, using Woodgate's reference designators, the signal gain with infinite Aol is: (R3*R4 + R2(R3 + R4))/(R1*R3) and if the amplifier gain is finite, the signal gain is: Aol*(R3*R4 + R2*(R3 + R4)) -------------------------------------------- R1*(R2 + R3 + Aol*R3)+ R3*R4 + R2*(R3 + R4) As one might expect, when trying to get a gain of 1000 out of the closed loop circuit, an open loop gain of 100 times that, or 100,000 would seem reasonable to reduce errors to the 1% level, and in fact that's what I get when I compute the error in the infinite gain expression compared to the more exact finite gain expression with an Aol of 100,000. I would agree, we need to be "far, far below UGBW" to get a closed loop gain of 1000 with a fairly small error from neglecting the shunt arm on the minus input. > >Hey, its only a little bit of baby maths. > >Cheers >Terry
From: Fred Bloggs on 27 Sep 2005 08:57 Terry Given wrote: > > I recently built about 50,000 of this circuit, with a feedback cap too > (mathcad rather than mathematica, and a pencil to start with for the > analysis), and 15 inputs thru 100k resistors. the effect of the 14 > "grounded" resistors shifted the center frequency by about 10% - Aol was > about 50. power consumption (and cost) constraints meant I couldnt use a > faster opamp, so instead I stopped assuming and started calculating :) > What was the transfer function you were shooting for, and which amp?
From: John Woodgate on 27 Sep 2005 01:34
I read in sci.electronics.design that Jim Thompson <thegreatone(a)example.com> wrote (in <2sngj1pjheo5h7aifs3tklqatrf5ohdesn(a)4ax.com>) about 'Op Amp Calculations', on Mon, 26 Sep 2005: >In the "interesting" cases the "divider" impedance is within the same >order of magnitude as the first feedback resistor. In which case the 'bottom arm' is two resistors in parallel, and the 'first feedback resistor' is one resistor in series with two in parallel. Not rocket science, jut the normal 'loaded potential divider'. -- Regards, John Woodgate, OOO - Own Opinions Only. If everything has been designed, a god designed evolution by natural selection. http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk |