Op Amp Calculations

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What IS Dostal's method? The Loop Gain & Phase analyser on my website is based on R.D. Middlebrook's laboratory technique, and is VERY accurate, since the loop is never actually broken.

...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 | |

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| 1962 | I love to cook with wine. Sometimes I even put it in the food.

Reply to
Jim Thompson
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On what page(s) in Dostal is this approach described?

Reply to
The Phantom

thats how I did it, and why :)

Cheers, Terry

Reply to
Terry Given

There is a 2001 paper describing extensions of Middlebrook's method. It's somewhat obscure since it was published in one of the IEEE magazines rather than a regular journal. I've posted it over on ABSE with the subject "Small Signal Stability".

Reply to
The Phantom

You first

say:

algebra is

signal gain,

input resistor

R2*R4)/(R1*R3).

very much

R2*R3 +

except that if the frequency of interest is not far, far below UGBW, the shunt arm connected to the -ve input cannot be ignored.

Hey, its only a little bit of baby maths.

Cheers Terry

Reply to
Terry Given

I read in sci.electronics.design that The Phantom wrote (in ) about 'Op Amp Calculations', on Mon, 26 Sep 2005:

I would really be reluctant to try to get a gain of 1000 out of one op-amp at today's prices. I suppose that sometimes there is no choice, but I think that's very rare. 31.62 is your friend.

--
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
Reply to
John Woodgate

No, I didn't forget at all; it's implicit in A.

What sign error is that?

Unfortunately, the result doesn't seem to be correct. Put the expressions for A, B, C (the numerator of the "C" expression seems to be missing a closing parenthesis), and D into the first expression, namely:

-A - B p

----------------- 1 + C p + D p^2

Then when you have it written out in all it's glory, find the limit as A0 --> infinity, and then the limit as C1 --> infinity. You

*should* get: -((r5*r6 + r4*(r5 + r6))/(r3*r6))

But, alas, you don't.

Reply to
The Phantom

I read in sci.electronics.design that Terry Given wrote (in ) about 'Op Amp Calculations', on Wed, 28 Sep 2005:

How do you get any signal at the -ve input with the input grounded?

--
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
Reply to
John Woodgate

a summing band-pass (ish) filter. 40 x TLV274.

I didnt want to AC-couple the inputs (that would have cost me 240 capacitors) so I used the bridged-T feedback network with an RC shunt to give a DC gain of about 1/16 - any DC is basically common-mode, and the next stage was AC coupled. 3 Rs and 2 Cs was a whole lot cheaper than an RLC. But 100k/14 = 7k in parallel with the -ve shunt arm, enough to move Fc 10% or so.

SPICE clearly showed it, so I went back and re-did my opamp analysis using Dostals approach (originally I did it using the Woodgate approximation), and voila - out popped the same answer. Mr HP3577 also agreed with spice and mathcad. Dostals method also allowed me to directly calculate the phase margin. Since then, I have analysed all opamp circuits thusly - but I use the Woodgate approach with pencil & paper as a bullshit detector :)

Cheers Terry

Reply to
Terry Given

Hi Jim,

its the same one used by Jerald Graeme.

Hideal(s) = -a(s)/B(s)

a(s) = feedforward factor = signal at opamp -ve input when output grounded

B(s) = feedback factor = signal at opamp -ve input when input grounded

Hactual(s) = Hideal(s) ------------- 1+Aol(s)/B(s)

I got Dostals book about 12 years ago, but until last year never used this method, as the -ve input = 0V method worked well enough. Its only when I started really pushing an opamp that it became necessary, mostly because I wasnt happy with simply twiddling component values in SPICE.

I havent yet managed to get your SPICE trick to work in Simetrix :(

but I did read RDMs paper.

Cheers Terry

Reply to
Terry Given

Middlebrook is still going strong too:

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Reply to
Fred Bloggs

first order

schematic

"signal" gain

(left to right;

a series

shunt arm of

The 10090.7

effect on the

feedback as the

shunt arm

has no

Why hasn't someone pointed out that by exchanging the left and right sides of the T network, we could have a shunt arm on the minus input of 111222 ohms, instead of

10090.7 ohms, a factor of 10 better. The 10090.7 ohm arm will then be a load on the amplifier output, still without effect. The series arm is unchanged of course, so the net effect may be an improvement. A possible undesirable consequence might be that the input capacitance of the op amp might be greater than the stray capacitance at the junction of the 3 T-network resistors, so with the reversed arrangement the op amp capacitance is fed with 99200 ohms, rather than 9000 ohms. One would have to check this possibility.

Why didn't *I* point this out before now?

feedback

arm in the

should expect

circuits, as

offset

this gain at

only around

indicates to me

make R5

to 10000

circuit

noise gain.

circuit.

Reply to
The Phantom

somewhat

regular journal.

aargh, my new news server (news.slingshot.co.nz) hasnt seen *any* new posts on abse for 5 days....

can you please email it to me...

cheers terry

Reply to
Terry Given

in the simple case,

a(s) = Vin(s)*Zf(s)/[Zin(s) + Zf(s)] B(s) = Vo(s)*Zin(s)/[Zin(s) + Zf(s)]

H(s) = -Zf(s)/Zin(s)

Cheers Terry

Reply to
Terry Given

You can achieve a wild increase in effective GBW by going to current mode feedback. The peaking and rapid rolloff due to that low impedance

-ve shunt is eliminated from any frequency bands usable with the voltage feedback circuit. The output gain of 2x deals with the CMR input range of the TLV274- requires about a volt of headroom to V+ - facilitation odds and ends not shown... View in a fixed-width font such as Courier.

. . . >--[Ri]-+-------+------[R1]--+--[R2]------------+ . | | | | . o | | [R3] +--[R]---+-->Vout . | | +5V | | | . | | | C | | |\\ | . o | | +-----||-+ +--|-\\ | . | | | | | >--+ . | | +-----[Rc]-----+------|+/ . o | | | | | |/ . | | | | | . >--[Ri]-+ | +-----------+ | [R] . | | | | | | . >--[Ri]-+ +------|>|---+ | | | . | | | | | | | . >--[Ri]-+ +--------------------------+ . | | | | | | . | | | | | | . | | |\\| c | | . | +-|+\\ |/ | | OA TLV274 . | | >-+---| c | . 2.5V>-+-----|-/ | |\\ |/ | . |/| | e+ c . | | |\\ |/ . | [Rb] e+ . | | |\\ . | | e . | | | . GND---------+--+-----------+------------ . . . . . . . . . . . .

Reply to
Fred Bloggs

I'll study that a bit later. unfortunately it also achieves a wild increase in parts count and cost - this circuit is replicated many, many times :)

Plus of course the original works just fine, and is in production.

Cheers Terry

Reply to
Terry Given

I can't comment on your application since only you know what that is. You may find this strange, but the idea is to overcome the limitations of low inverting input shunt impedance and not to improve your product.

Reply to
Fred Bloggs

I think that anyone sufficiently well versed in complex arithmetic as used nowadays to write transfer functions will know that if they want the DC gain, they can just use the TF I gave with A constant, and if they want AC results, then of course they will know that A must be a function of frequency..

Did you notice that three different people contributed to the posting that you originally replied to? John Woodgate did the first ASCII schematic, John Fields did the second and provided an equation, viz:

R4 + R5 Vcc R2 Vout = -Vin --------- + --------- R3 R1 + R2

All I (Phantom) posted was a couple of transfer functions. I didn't use John's equations, and I don't think there was a sign error in what I posted. Most of your reply was directed to me, so I thought you were suggesting I had made a sign error.

It would make it easier for others to comment without ambiguity and misunderstanding on your posting if you would cut and paste in the equation you think is in error, perhaps even indicating what you think the correct equation should be.

You set WT=2.pi.GBW, but you haven't indicated what kind of op amp model you're using. Was it the standard one pole model:

A(s) = WT/(s + Wa)

or something else? I can't comment further until you tell me your model, but I still think there is something out of kilter with your expression:

-A - B p ----------------- 1 + C p + D p^2

after A, B, C and D are substituted.

Reply to
The Phantom

Sorry for the late reply.

"The Phantom" a écrit dans le message de news: snipped-for-privacy@4ax.com...

gain

-----------------------------------------------------------------------

Well, yes, I somewhat figured that. But not mentionning A(s) when you explicitly use s somewhere else might lead the not so careful to some error, i.e. forgetting its phase which is -90d over almost the bandwidth.

correction.

That was in the starting equations (wrong opamp gain sign).

But then I do :-) I've doubled checked the results the brute force way, taking the A,B,C,D directly from here back into mathematica and all the limits are OK.

--
Thanks,
Fred.
Reply to
Fred Bartoli

It is a common collector pre-drive for the third transistor which is CE. If extreme gain accuracy is not needed it can be pulled, the CE should be a high beta type at low Ic.

Reply to
Fred Bloggs

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