Op Amp Calculations
in [Design]
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From: Roger Lascelles on 21 Sep 2005 23:16 "Roger Lascelles" <despam_rklasl(a)aanet.com.au> wrote in message news:1127349323.52193f9f3c35830fb56e0c14734ee939(a)teranews... > > ASCII art; use Courier font. > > _______R4_______ > | - | > o----R1--+---|\ | > | \_____ out | > | / | | > o------------|/ R2 | > + |----' > R3 > | > o--------------------+------ Gnd This is the formula you are looking for: If R4 = R2 and R3 = K R2 Av = 2 ( 1 + (1/(2K)) ) R2 / R1 derived from the balanced amp case given in Toby, Graeme, Huelsman, "Operational Amplifiers", Burr-Brown. I hope "wiley old engineer" mentioned the drawbacks - this ciruit will *not* give you the max gain possible from an opamp, because it wastes some of the available gain. 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. I think the amplifier can be made quite good by having non-equal values for R2 and R4. Select R4 > 10 R1 to fix the gain wastage problem. Make R2 || R3 much less than R4. Select the ratio of R2 and R3 for the gain you want. The gain is then approx : Av = ( R4 / R1 ) ( R2 + R3 ) / ( R3 ) Roger Lascelles
From: Pooh Bear on 22 Sep 2005 00:23 Roger Lascelles wrote: > "Roger Lascelles" <despam_rklasl(a)aanet.com.au> wrote in message > news:1127349323.52193f9f3c35830fb56e0c14734ee939(a)teranews... > > > > ASCII art; use Courier font. > > > > _______R4_______ > > | - | > > o----R1--+---|\ | > > | \_____ out | > > | / | | > > o------------|/ R2 | > > + |----' > > R3 > > | > > o--------------------+------ Gnd > > This is the formula you are looking for: > > If R4 = R2 and R3 = K R2 > > Av = 2 ( 1 + (1/(2K)) ) R2 / R1 > > derived from the balanced amp case given in Toby, Graeme, Huelsman, > "Operational Amplifiers", Burr-Brown. > > I hope "wiley old engineer" mentioned the drawbacks - this ciruit will *not* > give you the max gain possible from an opamp, because it wastes some of the > available gain. How can it 'waste it' ? Max gain = Avol. End of story. Graham
From: Fred Bartoli on 22 Sep 2005 03:53 "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 :-) 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. Isn't that Mathematica lovely? ;-) -- Thanks, Fred.
From: Roger Lascelles on 22 Sep 2005 04:33 "Pooh Bear" <rabbitsfriendsandrelations(a)hotmail.com> wrote in message news:433231D1.3B0708F5(a)hotmail.com... > > available gain. > > How can it 'waste it' ? Max gain = Avol. End of story. > > Graham Look at this circuit : _______R4_______ | - gnd --R1--+---|\ | \_____ | / out |/ + The amplifier is open loop, with no feedback voltage into R4. The inverting input voltage is divided by the R4, R1 attenuator. The amplifier signal is attenuated, and is now smaller in relation to the amplifier's own noise, offset and drift. Now let the opamp have an open loop gain of Avol - I think that is what you mean by Avol. The maximum gain our overall amplifier can have is : Atotal = Avol (R4 / (R1 + R2 ) ) which is less than Avol . So the input circuit has "wasted" some gain. A high gain opamp makes up for it - but not really, because we actually have "less feedback" in order to get the required level of output. In effect we have an input attenuator followed by a higher gain opamp amplifier. This is why the divided feedback circuit is bad - it can heavily attenuate the input signal. Roger Lascelles It is common to speak of a reserve of gain - the If you open the loop and do a gain analsis, you see that the An opamp is a high gain amplifier with feedback around it. The "extra gain" or loop gain is the differe If you look at the amplifier
From: John Woodgate on 22 Sep 2005 13:42
I read in sci.electronics.design that John Fields <jfields(a)austininstruments.com> wrote (in <run3j19en9a9dpl6dpu17v7siuv7l1os2r(a)4ax.com>) about 'Op Amp Calculations', on Wed, 21 Sep 2005: >I believe the circuit he described looks more like this: > >Vcc------------+ > | > [R1] > | > +----------|+\ > | | >-----+-->Vout >Vin>------[R3]-------+----|-/ | > | | | > [R2] +-[R4]-+-[R5]-+ > | | >GND>-----------+-[C1]--[R6]-+ The bias on the + input is irrelevant in the context of the thread, and C6 is normally so large that it has no effect within the operating bandwidth. -- 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 |