Op Amp Calculations
in [Design]
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From: Jim Thompson on 24 Sep 2005 12:43 On Sat, 24 Sep 2005 12:57:46 +1000, "Roger Lascelles" <despam_rklasl(a)aanet.com.au> wrote: >"Jim Thompson" <thegreatone(a)example.com> wrote in message >news:qak8j1tkkfnhejnitafai3qrdbq1p3sr80(a)4ax.com... > >> See... >> >> Newsgroups: alt.binaries.schematics.electronic >> Subject: Op Amp Calculations (from S.E.D) - MultiPathFeedback.pdf >> Message-ID: <d6k8j1l0hrju6f8mhuhkkg9fv17kmc04mn(a)4ax.com> > >I don't seem to have access to alt.binaries.schematics.electronic - perhaps >someone can tell me how. > >I do take your point Jim, that the T circuit is not "bad". Engineering is >about making "how much" judgements. I don't think I wrote off the circuit, >in fact I suggested using unequal resistances each side of the tap to >minimise the penalty. > >My point is - use the ratio of resistances at the opamp minus node to >ESTIMATE what you are doing to noise, offset, drift and bandwidth. Applies >for any opamp circuit, not just the T. Most of the time you have extra >performance to burn. > >The T feedback can be evil. Easy as pie to peel 6 or 10db off your >equipment spec and make it worse than the competition. > >The original post disturbed me because it advertised the "T" as a free lunch >and seemingly equivalent to a feedback resistor. > >Roger Lascelles > Roger, Go to my website and send an E-mail with your E-mail address, and I'll send you the PDF. ...Jim Thompson -- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food.
From: Terry Given on 26 Sep 2005 01:47 John Woodgate wrote: > I read in sci.electronics.design that Roger Lascelles > <despam_rklasl(a)aanet.com.au> wrote (in > <1127358972.f51b6a91d2505dd8d1089bfe370aa720(a)teranews>) about 'Op Amp > Calculations', on Thu, 22 Sep 2005: > >> Looking at the circuit open loop, for a high gain amp, you want R4 >> >> R1, so most of the input signal hits the opamp minus input, but you >> end up with the opposite. > > > The inverting op-amp circuit works by having NO signal on the - input. > That's what 'virtual earth' means. Hi John, obviously your "no signal" is quite a bit larger than my "no signal" thats a convenient approximation, but it is most surely an approximation. to prove it, place a 10 Ohm resistor from the -ve input to 0V, and watch the circuit perform differently. I blame Jiri Dostal. Cheers Terry
From: John Woodgate on 26 Sep 2005 06:29 I read in sci.electronics.design that Terry Given <my_name(a)ieee.org> wrote (in <1127713562.620391(a)ftpsrv1>) about 'Op Amp Calculations', on Mon, 26 Sep 2005: >John Woodgate wrote: >> I read in sci.electronics.design that Roger Lascelles >><despam_rklasl(a)aanet.com.au> wrote (in >><1127358972.f51b6a91d2505dd8d1089bfe370aa720(a)teranews>) about 'Op Amp >>Calculations', on Thu, 22 Sep 2005: >> >>> Looking at the circuit open loop, for a high gain amp, you want R4 >>>>> R1, so most of the input signal hits the opamp minus input, but >>>you end up with the opposite. >> The inverting op-amp circuit works by having NO signal on the - >>input. That's what 'virtual earth' means. > >Hi John, > >obviously your "no signal" is quite a bit larger than my "no signal" I don't know what that means. > >thats a convenient approximation, but it is most surely an approximation. Yes, it assumes that the op-amp has infinite open-loop gain. Not a bad approximation at frequencies very much smaller than the unity-gain bandwidth. > >to prove it, place a 10 Ohm resistor from the -ve input to 0V, and >watch the circuit perform differently. > You can 'prove' all sorts of things by changing a circuit. If you measure the signal voltage on the inverting input, you find, unless you are using the op-amp at an unwisely-high frequency, that the voltage there is TINY compared with the input voltage. Not 'most of the input signal'. -- 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
From: The Phantom on 26 Sep 2005 13:42 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. 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: The Phantom on 26 Sep 2005 14:27
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. > > 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. I made a transcription error here. The expression for noise gain of the multi path network should be 1 + R6'/R1 + R6'/R4', where R6' is the series arm in the equivalent delta network, and R4' is the 10090.7 ohm 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. This sentence should read "Since R4' is about 10 times R1, we should expect..." >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%, This should be "around 10%". >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% This also should be "about 10%". >higher noise gain. >The total impedance in the feedback loop is much lower than the two-resistor circuit. |