Op Amp Calculations
in [Design]
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From: RST Engineering (jw) on 21 Sep 2005 15:20 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
From: John Woodgate on 21 Sep 2005 16:03 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 You can see then that R2 and R3 just form a potential divider across the output, allowing R4 and R2 to be much smaller for a given gain: 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). The circuit is highly redundant, which means that you get to choose highly inappropriate values for all the resistors and then wonder why it doesn't work, even though the gain calculation works out.(;-) You can probably avoid that by making R1 something in the 1 k to 10 k range, but don't take bets. I remember one of my fellow students making a 'see-saw' [1](what we now call an inverting amplifier) with 1 kohm and 1 Mohm around a 6J5 triode and wondering why the gain wasn't 1000. [1] I think that's a 'teeter-totter' in US English, but I wonder if the circuit was called that, or 'see-saw'. -- 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: Noway2 on 21 Sep 2005 17:05 Adding an RC circuit to the feedback path would cause the feedback impedance to become frequency dependant, which would produce different gains depending on the frequency. I am having a bit of trouble understanding how this configuration would increase the gain (Zf / Zi) as opposed to reducing it (apparently I am not envisioning the circuit correctly). Wouldn't placing the RC combination in the feedback path cause the effective Rf (parralllel combination) to be lower, in which case the gain would go down? I have seen similar ideas used in common emitter transistor amplifier circuits, where large AC gain is desired around a DC bias point. In those cases, the capacitor shunts the AC thereby decreasing the emitter resistance and increasing the gain (Rc / Re). I don't see how this works for the op-amp case, though.
From: jmeyer on 21 Sep 2005 17:21 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
From: John Fields on 21 Sep 2005 18:36
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. 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... -- John Fields Professional Circuit Designer |