Oh, I see. Sorry for the misunderstanding, I wasn't implying somebody made a sign error. It was just *me* that made the sign error when first writing the problem and got obviously wrong results. Error corrected in a snap.
By saying:
I was just emphasing how easy it is to correct such errors vs the paper/pencil method.
Yes it is.
Well, you can't just make A0 and C1 go to the limit and expect the DC response. You also have to either make WT go to the limit, or make s=0.
Then all goes fine.
But then you can see that it's easy, even for someone "sufficiently well versed in complex arithmetic as used nowadays" (to take your words), to forget something in the process. Hence my preference for writing A(s). It's just one thing less to remember.
From some of your previous posts I believe you have mathematica, so you'll find the code herewith.
--
| 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.
You're absolutely correct. It is really nice, having spent hours on a really complicated problem, discovering a stupid mistake that would entail more hours of computation after having discovered the mistake. to be able to just key in a small change, hit enter, and !bang!, the final result, a result derived from a large amount of algebraic drudgery!
Except that you didn't use that particular formulation. You used a non-standard (in my experience) expression, namely;
A(s) = WT/((WT/A0 + s)) and this caused your results to contain both A0 and WT. There's no need for this since WT incudes A0.
(I'm going to use Wp for the pole frequency in the single pole model for an op amp in what follows.)
A standard expression for op amp gain, (which is exactly equivalent to yours, but not involving *both* A0 and WT explicitly) such as found on page 19 of the classic Schaumann and Van Valkenburg, is:
A(s) = WT/(s + Wp)
another alternative would be:
A(s) = (A0 Wp)/(s + Wp)
Solving the circuit with either of these expressions gives results that explicitly display the effect of Wp, and only one variable that involves the DC gain of the op amp.
(I personally like to use the expression:
A(s) = A0/(1 + sT), where T is the time constant of the pole.)
Yes, you can, if you use the second version of the op amp gain I gave above. If you use the first, standard, expression for A(s), then you have to make WT go to infinity (but not A0 since it's no longer in the result expressions) and then C1.
Only if you have used an expression for A(s) which involves *both* A0 *and* WT, which is not necessary.
In your Mathematica program, you have these four lines (change mode temporarily from using s to using p :-) ):
The fourth limit operates on the result of the second limit, which expression already lacks the variable p, so of course the result of this limit is the same as the result of the second limit and therefore accidentally gives the correct DC gain; but, in general, just letting p -> 0 won't give the DC gain if capacitors are involved, as I explain.
It is the admittance of the capacitor(s) we want to become infinite to get DC gain; that is, the product p*C1 must go to infinity. I notice that if you let p -> 0 first (with C1 still finite), and then let A0 go to infinity, the result is -(R4 + R5)/R3, because the product p*C1 has gone to zero. Letting both p -> 0 and C1 -> infinity would only work if p and C1 vary in such a way that the product p*C1
-> infinity. Thus, letting p -> 0 won't give the DC gain if C1 remains finite. But this is not a problem, since letting C1 ->
infinity does the job.
When I originally took your full expression involving A, B, C, and D and took the limit as A0 -> infinity and C1 -> infinity, I got the same result you got after the second limit (above) and I couldn't understand why WT was still in the result; that's why I initially thought your result was incorrect. Other than that unnecessary complication, your result is, of course, correct.
The basic three equations are easy to write down by inspection as you did, and Mathematica did just what you told it to, and yes, Mathematica is indeed lovely.
You're dwelling on open loop anomalies- tie that collector to the rightmost transistor collector if it bothers you. There is no condition under which the TLV274 will put sustained maximum current into the base circuit- not even close- the output will rail at uA drive.
And it STILL will go up in flames during power-up. Sheeesh! Such DESIGN ;-)
[snip]
...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.
This should be Limit[%%, p -> 0] of course. I proofread all this several times, and still this slipped by. :-(
I just noticed that this is not true; I think I must have been making small changes to the program (fooling around to explore some things) and so what I was looking at when I said this was not your original program.
It gives the correct result because products such as p*C1 have been eliminated by the second limit.
I think this is true if any products of p and C1 are present in the expression before letting p -> 0. If you let p -> 0 after the first limit of the four above, then you get the result -(R4 + R5)/R3 (WT drops out, too), which I believe happens as explained below.
-- | 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 | |
formatting link
| 1962 | I love to cook with wine. Sometimes I even put it in the food.
shouldnt there be something to limit current though? +5V...Vce...Vbe with nary a resistor in sight. ultimately the base current could be as high as the opamp output current (assuming negligible contribution from the summing junction)....
That, (which can be corrected by moving Q2's collector) plus Q1/Q2 operate at vanishingly low IC so that the input impedance has nothing to do with that of a CFB, even at moderate audio frequencies.
Take IC3=1mA, HFE=100 for all the Qs. That makes IC1=25nA and gmQ1=1uA/V and a total transconductance one third of it or gm = 0.33uA/V.
The TLV GBW product is 3MHz, so at 10kHz the input impedance is 3M/300=10K and inductive (a 0.16H inductance). With a 3pF input it'll resonate at a low 230kHz and give a 2nd order roll-off above this frequency.
The 3rd stage miller capacitance will also roll-off in that frequency range.
Not quite, essentially all the incremental base drive voltage is developed across hie of the first transistor. Your reasoning is wrong and your estimate is high.
Yeah- right, i,in= A(jw)*gm*vin or Zin= re,Q1/A(jw), for the TLV274 with A(jw)=6.28E6/(jw), this becomes Zin= re,Q1 * jw /6.28E6 so that at 1KHz you have re,Q1E-3. Taking re,Q1=beta^2*0.026/IcQ3,Q this makes Zin about 2.6K at 1KHz. But you have a fairly huge gain of vout=A(jw)*gm*beta^2*2*Rc*vin, and since gm=1/(re,Q3*beta^2), Av=A(jw)*2*Rc/re,Q3, so that Yf becomes Av/Rf, and the current divide ratio is Av/(Av+A(jw)*gm*Rf) or 1/(1+Rf/(2*Rc*beta^2))- this is looking more like a CMF all the time- a broadband split neglecting a few poles here and there.
That is very easily taken care of- but it's not my job to educate you...all you need to know is that the CE rolloff is not an issue.
Nah- "hardly usable" is what your boss says about you. SPICE does not agree with your conclusions- what does your IQ non-enhancer, Mathematica, tell you now? The one drawback to the triple Darlington is the same peaking effect due to error amplifier impedance growth with frequency, it is possible to replace Q1 with a PNP emitter follower feedback to OA input, interchange OA (-) and (+), but then this creates another headache by introducing a dominant pole in the loop.
"hie"... what BS! Fred must be of the "Kevin School" of transistor operation ;-)
No limit on IB of 1st transistor except output capability of OpAmp
IC of 1st transistor IS limited somewhat by FB resistors, but substantial IB is introduced into 2nd transistor
Absolutely nothing to limit IC of 2nd transistor except current crowding
My best guess is that all three transistors will flame or be seriously damaged when Murphy chooses ;-)
(And the bandwidth will be nothing like claimed.)
What was the original intent (OP)? I didn't get in on this thread at the beginning.
...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.
I read in sci.electronics.design that Jim Thompson wrote (in ) about 'Op Amp Calculations', on Wed, 5 Oct 2005:
Offer him a half-share in 'Snooooort!'
--
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
That's still insufficient. Good engineering practice would limit base drive from the OpAmp AND also the collector current of the 2nd transistor.
...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.
I did that once on snow. Nothing like sliding sideways, down a hill yet, and dodging traffic, to put some "age" on you ;-)
...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.
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