Op Amp Calculations
in [Design]
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From: Fred Bloggs on 28 Sep 2005 06:26 Jim Thompson wrote: > On Wed, 28 Sep 2005 08:44:46 +1200, Terry Given <my_name(a)ieee.org> > wrote: > > [snip] > >>Dostals method also allowed me to >>directly calculate the phase margin. > > [snip] > >>Cheers >>Terry > > > What IS Dostal's method? The Loop Gain & Phase analyser on my website > is based on R.D. Middlebrook's laboratory technique, and is VERY > accurate, since the loop is never actually broken. > > ...Jim Thompson Middlebrook is still going strong too: http://ardem.com/index.asp
From: The Phantom on 28 Sep 2005 08:38 On Thu, 22 Sep 2005 09:53:22 +0200, "Fred Bartoli" <fred._canxxxel_this_bartoli(a)RemoveThatAlso_free.fr_AndThisToo> wrote: > >"The Phantom" <phantom(a)aol.com> a ?crit dans le message de >news:7sr3j157pdfap23guks1go3p9g7qjrvn5k(a)4ax.com... >> On Wed, 21 Sep 2005 17:36:57 -0500, John Fields ><jfields(a)austininstruments.com> wrote: >> >> >On Wed, 21 Sep 2005 21:03:48 +0100, John Woodgate >> ><jmw(a)jmwa.demon.contraspam.yuk> wrote: >> > >> >>I read in sci.electronics.design that "RST Engineering (jw)" >> >><jim(a)rstengineering.com> wrote (in <11j3cjlkfbcqj6e(a)corp.supernews.com>) >> >>about 'Op Amp Calculations', on Wed, 21 Sep 2005: >> >> >> >>>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? >> >> >> >>I believe Win has sort-of proprietary rights on this circuit. But it's >> >>easy if you re-draw it. Why do these things come up in s.e.d. instead of >> >>a.b.s.e? >> >> >> >>ASCII art; use Courier font. >> >> >> >> _______R4_______ >> >> | - | >> >>o----R1--+---|\ | >> >> | \_____ out | >> >> | / | | >> >>o------------|/ R2 | >> >> + |----' >> >> R3 >> >> | >> >>o--------------------+------ Gnd >> > >> >--- >> > >> >I believe the circuit he described looks more like this: >> > >> >Vcc------------+ >> > | >> > [R1] >> > | >> > +----------|+\ >> > | | >-----+-->Vout >> >Vin>------[R3]-------+----|-/ | >> > | | | >> > [R2] +-[R4]-+-[R5]-+ >> > | | >> >GND>-----------+-[C1]--[R6]-+ A >> > >> > >> >Where, for DC, >> > >> > >> > R4 + R5 Vcc R2 >> > Vout = -Vin --------- + --------- >> > R3 R1 + R2 >> > >> >For AC, however, (assuming Vin and Vout are perfect voltage sources) >> >we wind up with Vout trying to force the - input of the opamp to >> >stay at whatever voltage the + input is set to by R1 and R2. >> >> For AC, assuming that C1 is very large so that it can be treated as an >AC short, and >> the op amp gain is infinite, the gain is: >> >> -((r5*r6 + r4*(r5 + r6))/(r3*r6)) >> >> If A is the op amp gain and C1 is included in the calculation, the gain >is : >> >> -A*(r4 + r5) - A*c1*(r5*r6 + r4*(r5 + r6))*s >> ----------------------------------------------------------------------- >-- >> r3 + A*r3 + r4 + r5 + c1*(r5*r6 + r4*(r5 + r6) + r3*(r5 + r6 + >A*r6))*s >> > >Well, you've forgotten the GBW in all that :-) No, I didn't forget at all; it's implicit in A. >If we set WT=2.pi.GBW then we have > > -A - B p >----------------- with > 1 + C p + D p^2 > > A0(R4+R5) >A = ---------------- > R3(1+A0)+R4+R5 > > A0.C1(R4 R5 + (R4 + R5) R6) >B = ---------------------------- > R3(1+A0)+R4+R5 > > (A0(R3 + R4 + R5 ) + C1.WT((R3 + R4)R5 + R6(R3 + A0.R3 + R4 + R5)) >C = -------------------------------------------------------------------- > (R3 (1 + A0) + R4 + R5 ) WT > > > A0.C1 ((R3 + R4) R5 + (R3 + R4 + R5) R6) >D = ------------------------------------------ > (R3 + A0 R3 + R4 + R5) WT > > >Less than 2 min to work this out from scratch, incl. sign error correction. What sign error is that? >Isn't that Mathematica lovely? ;-) Unfortunately, the result doesn't seem to be correct. Put the expressions for A, B, C (the numerator of the "C" expression seems to be missing a closing parenthesis), and D into the first expression, namely: -A - B p ----------------- 1 + C p + D p^2 Then when you have it written out in all it's glory, find the limit as A0 --> infinity, and then the limit as C1 --> infinity. You *should* get: -((r5*r6 + r4*(r5 + r6))/(r3*r6)) But, alas, you don't.
From: The Phantom on 28 Sep 2005 18:11 On Mon, 26 Sep 2005 10:42:52 -0700, The Phantom <phantom(a)aol.com> wrote: >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 > > (Refer to Jim Thompson's schematic in alt.binaries.schematics.electronic) > > In what follows, I'll assume the capacitor is replaced with a short and only first order >effects will be considered. > > An easy way to see what is happening in the "T" (multi path) network feedback >arrangement is to transform the "T" to a "Pi" network. >http://encyclopedia.thefreedictionary.com/Y-delta+transform > > First, consider what happens in an inverting amplifier of the sort in Jim's schematic >when you connect a resistor from the "-" input of the amp to ground. The "signal" gain >remains unchanged, but the "noise gain" of the circuit changes. > > Now apply the well known "Y-Delta" transformation. Jim shows a "T" with (left to right; >input to output side) a series arm of 9000 ohms, a shunt arm of 1000 ohms, and a series >arm of 99200 ohms. This is equivalent to a pi network (left to right) with a shunt arm of >10090.7 ohms, a series arm of 1.001 megohms, and a shunt arm of 111222 ohms. The 10090.7 >ohm shunt arm on the left goes from the "-" input to ground and so has no effect on the >signal gain. The 1.001 megohm series arm provides essentially the same feedback as the >1.000 megohm resistor in the two resistor amplifier circuit. The 111222 ohm shunt arm >from output to ground is simply an additional (small) load on the amplifier and has no >effect on circuit performance. Why hasn't someone pointed out that by exchanging the left and right sides of the T network, we could have a shunt arm on the minus input of 111222 ohms, instead of 10090.7 ohms, a factor of 10 better. The 10090.7 ohm arm will then be a load on the amplifier output, still without effect. The series arm is unchanged of course, so the net effect may be an improvement. A possible undesirable consequence might be that the input capacitance of the op amp might be greater than the stray capacitance at the junction of the 3 T-network resistors, so with the reversed arrangement the op amp capacitance is fed with 99200 ohms, rather than 9000 ohms. One would have to check this possibility. Why didn't *I* point this out before now? > > The noise gain in the two-resistor circuit is 1 + R3/R1; in the multi path feedback >circuit, the noise gain is 1 + R3/R1 + R4'/R1, where R4' is the 10090.7 shunt arm in the >equivalent delta network. Since R4' is about .01 times the value of R3, we should expect >no easily observable low-frequency difference in the performance of the two circuits, as >Jim's simulation shows. The noise gain is also the gain applied to the op-amp offset >voltage, and the capacitor in the multi path circuit prevents the increase in this gain at >DC. But in this particular multi path circuit, the increase in noise gain is only around >1%, so we could omit the capacitor. > > This method (using the Y-Delta transformation) of analyzing the circuit indicates to me >that Jim's assertion: > >"...the 1:1 case has no useful value toward improving (lowering) >the total impedance in the feedback loop." > > is incorrect. Consider the following: > > In Jim's multi path circuit, make R4 9803.92 ohms, make R6 98.0392 ohms, and make R5 >9803.92 ohms. Now the Pi equivalent network has the two shunt resistors equal to 10000 >ohms, and a series arm equal to 1.000 megohms, which gives a (low-frequency) circuit >performance identical to the two-resistor circuit, except for about 1% higher noise gain. >The total impedance in the feedback loop is much lower than the two-resistor circuit.
From: Fred Bloggs on 29 Sep 2005 09:33 Terry Given wrote: > Fred Bloggs wrote: > >> >> >> 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? >> > > a summing band-pass (ish) filter. 40 x TLV274. > > I didnt want to AC-couple the inputs (that would have cost me 240 > capacitors) so I used the bridged-T feedback network with an RC shunt to > give a DC gain of about 1/16 - any DC is basically common-mode, and the > next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper than an > RLC. But 100k/14 = 7k in parallel with the -ve shunt arm, enough to move > Fc 10% or so. > > SPICE clearly showed it, so I went back and re-did my opamp analysis > using Dostals approach (originally I did it using the Woodgate > approximation), and voila - out popped the same answer. Mr HP3577 also > agreed with spice and mathcad. Dostals method also allowed me to > directly calculate the phase margin. Since then, I have analysed all > opamp circuits thusly - but I use the Woodgate approach with pencil & > paper as a bullshit detector :) > > Cheers > Terry You can achieve a wild increase in effective GBW by going to current mode feedback. The peaking and rapid rolloff due to that low impedance -ve shunt is eliminated from any frequency bands usable with the voltage feedback circuit. The output gain of 2x deals with the CMR input range of the TLV274- requires about a volt of headroom to V+ - facilitation odds and ends not shown... View in a fixed-width font such as Courier. .. .. .. >--[Ri]-+-------+------[R1]--+--[R2]------------+ .. | | | | .. o | | [R3] +--[R]---+-->Vout .. | | +5V | | | .. | | | C | | |\ | .. o | | +-----||-+ +--|-\ | .. | | | | | >--+ .. | | +-----[Rc]-----+------|+/ .. o | | | | | |/ .. | | | | | .. >--[Ri]-+ | +-----------+ | [R] .. | | | | | | .. >--[Ri]-+ +------|>|---+ | | | .. | | | | | | | .. >--[Ri]-+ +--------------------------+ .. | | | | | | .. | | | | | | .. | | |\| c | | .. | +-|+\ |/ | | OA TLV274 .. | | >-+---| c | .. 2.5V>-+-----|-/ | |\ |/ | .. |/| | e+ c .. | | |\ |/ .. | [Rb] e+ .. | | |\ .. | | e .. | | | .. GND---------+--+-----------+------------ .. .. .. .. .. .. .. .. .. .. .. ..
From: Terry Given on 29 Sep 2005 18:10
Fred Bloggs wrote: > > > Terry Given wrote: > >> Fred Bloggs wrote: >> >>> >>> >>> 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? >>> >> >> a summing band-pass (ish) filter. 40 x TLV274. >> >> I didnt want to AC-couple the inputs (that would have cost me 240 >> capacitors) so I used the bridged-T feedback network with an RC shunt >> to give a DC gain of about 1/16 - any DC is basically common-mode, and >> the next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper >> than an RLC. But 100k/14 = 7k in parallel with the -ve shunt arm, >> enough to move Fc 10% or so. >> >> SPICE clearly showed it, so I went back and re-did my opamp analysis >> using Dostals approach (originally I did it using the Woodgate >> approximation), and voila - out popped the same answer. Mr HP3577 also >> agreed with spice and mathcad. Dostals method also allowed me to >> directly calculate the phase margin. Since then, I have analysed all >> opamp circuits thusly - but I use the Woodgate approach with pencil & >> paper as a bullshit detector :) >> >> Cheers >> Terry > > > You can achieve a wild increase in effective GBW by going to current > mode feedback. The peaking and rapid rolloff due to that low impedance > -ve shunt is eliminated from any frequency bands usable with the voltage > feedback circuit. The output gain of 2x deals with the CMR input range > of the TLV274- requires about a volt of headroom to V+ - facilitation > odds and ends not shown... > View in a fixed-width font such as Courier. > > . > . > . >--[Ri]-+-------+------[R1]--+--[R2]------------+ > . | | | | > . o | | [R3] +--[R]---+-->Vout > . | | +5V | | | > . | | | C | | |\ | > . o | | +-----||-+ +--|-\ | > . | | | | | >--+ > . | | +-----[Rc]-----+------|+/ > . o | | | | | |/ > . | | | | | > . >--[Ri]-+ | +-----------+ | [R] > . | | | | | | > . >--[Ri]-+ +------|>|---+ | | | > . | | | | | | | > . >--[Ri]-+ +--------------------------+ > . | | | | | | > . | | | | | | > . | | |\| c | | > . | +-|+\ |/ | | OA TLV274 > . | | >-+---| c | > . 2.5V>-+-----|-/ | |\ |/ | > . |/| | e+ c > . | | |\ |/ > . | [Rb] e+ > . | | |\ > . | | e > . | | | > . GND---------+--+-----------+------------ I'll study that a bit later. unfortunately it also achieves a wild increase in parts count and cost - this circuit is replicated many, many times :) Plus of course the original works just fine, and is in production. Cheers Terry |