Op Amp Calculations
in [Design]
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From: The Phantom on 21 Sep 2005 20:08 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]-+ > > >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 > >As the frequency of Vin increases, the reactance of C1 will >decrease, diverting some of the signal from the output of the opamp >to ground, with the result that the output voltage will have to rise >in order to keep the voltage on the - input of the opamp the same as >the voltage on the + input. > > >Looking at R5R6C1 as a lowpass filter with single pole, we have: > > > Vout > | > [R5] > | > +----Vfb > | > [R6] > | > [C1] > | > GND > >And,suddenly, it's too close to dinner (Thai beef salad and a nice >white for me and Thai beef salad and a nice red for she) and too >complicated to get into right now. > >Ma?ana, talvez...
From: Roger Lascelles on 21 Sep 2005 20:35 <jmeyer(a)nowhere.net> wrote in message > From a Burr Brown ap note, > http://focus.ti.com/lit/an/sboa061/sboa061.pdf, .... > > . A very high resistance feedback resistor is MUCH better > than a low resistance in a T network. See Figure 5. Although > transimpedance gain (eOUT/iSIGNAL) is equivalent, the T network > will sacrifice performance. The low feedback resistance > will generate higher current noise (iN) and the voltage > divider formed by R1/R2 multiply input offset voltage, drift, > and amplifier voltage noise by the ratio of 1+ > R1/R2. In most electrometer amplifiers, these input specifications > are not very good to start with. Multiplying an already > high offset and drift (sometimes as high as 3mV and 50mV/ > ?C) by use of a T network becomes impractical. By using a > far better amplifier, such as the OPA128, moderate T network > ratios can be accommodated and the resulting multiplied > errors will be far smaller. Although a single very-high > resistance will give better performance, the T network can > overcome such problems as gain adjustment and difficulty in > finding a large value resistor. > > Jim "The other one." Meyer ASCII art; use Courier font. _______R4_______ | - | o----R1--+---|\ | | \_____ out | | / | | o------------|/ R2 | + |----' R3 | o--------------------+------ Gnd The input signal at the opamp minus input is reduced by a voltage divider formed by R1 and R4 + (R2 || R3). The *least* reduction of the input signal occurs when you use a straight feedback resistor, not the R2, R3, R4 combination. Since the input signal is reduced but the overall circuit has the same gain, the opamp must be making up the deficit from its reserve of loop gain. In other words, more noise, offset, drift, less bandwidth. People who want variable gain balanced amps use a pot or single gain setting resistor. Not too bad for a small range of gain increase, but you often see R2 < R1. An example : overall gain of 1: 1. Omit R3, R2 = 0, R4 = R1 . Input signal is reduced to 0.5 . This is the optimum case with singe feedback resistance. 2. R2 = R4 = 0.25 R1 and R3= 0.125 R1. Input signal is reduced to 0.25 . Now you need 6db more gain from the opamp. Roger Lascelles
From: Winfield Hill on 21 Sep 2005 20:39 The Phantom wrote... > >> And,suddenly, it's too close to dinner (Thai beef salad and a nice >> white for me and Thai beef salad and a nice red for she) and too >> complicated to get into right now. Hmm, a bottle of red for you, a bottle of white for her, plus a little Thai beef salad on the side... -- Thanks, - Win
From: Pooh Bear on 21 Sep 2005 21:47 jmeyer(a)nowhere.net wrote: > On Wed, 21 Sep 2005 12:20:11 -0700, "RST Engineering \(jw\)" > <jim(a)rstengineering.com> wroth: > > >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. > > > > From a Burr Brown ap note, > http://focus.ti.com/lit/an/sboa061/sboa061.pdf, .... > > ? A very high resistance feedback resistor is MUCH better > than a low resistance in a T network. See Figure 5. Although > transimpedance gain (eOUT/iSIGNAL) is equivalent, the T network > will sacrifice performance. The low feedback resistance > will generate higher current noise (iN) and the voltage > divider formed by R1/R2 multiply input offset voltage, drift, > and amplifier voltage noise by the ratio of 1+ > R1/R2. In most electrometer amplifiers, these input specifications > are not very good to start with. Multiplying an already > high offset and drift (sometimes as high as 3mV and 50mV/ > ?C) by use of a T network becomes impractical. By using a > far better amplifier, such as the OPA128, moderate T network > ratios can be accommodated and the resulting multiplied > errors will be far smaller. Although a single very-high > resistance will give better performance, the T network can > overcome such problems as gain adjustment and difficulty in > finding a large value resistor. > > Jim "The other one." Meyer If the R to ground is simply ac coupled ther's no degradation of DC performance though. I used a similar trick once with a non-inverting configuration on the input to produce a 'bootstrapped' Very high input Z. Graham
From: The Phantom on 21 Sep 2005 23:08
On 21 Sep 2005 17:39:31 -0700, Winfield Hill <Winfield_member(a)newsguy.com> wrote: >The Phantom wrote... >> >>> And,suddenly, it's too close to dinner (Thai beef salad and a nice >>> white for me and Thai beef salad and a nice red for she) and too >>> complicated to get into right now. > > Hmm, a bottle of red for you, a bottle of white for her, plus a > little Thai beef salad on the side... Actually, it was John Fields who wrote this. |