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Op Amp Calculations

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

Reply to
RST Engineering (jw)
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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.

Reply to
Noway2

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

Reply to
The Phantom

---

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.

Mañana, talvez...

-- John Fields Professional Circuit Designer

Reply to
John Fields

Hmm, a bottle of red for you, a bottle of white for her, plus a little Thai beef salad on the side...

--
 Thanks,
    - Win
Reply to
Winfield Hill

Actually, it was John Fields who wrote this.

Reply to
The Phantom

I read in sci.electronics.design that "RST Engineering (jw)" wrote (in ) about 'Op Amp Calculations', on Wed, 21 Sep 2005:

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
Reply to
John Woodgate

On Wed, 21 Sep 2005 12:20:11 -0700, "RST Engineering \\(jw\\)" wroth:

From a Burr Brown ap note,

formatting link
....

? 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

Reply to
jmeyer

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

Reply to
Pooh Bear

How can it 'waste it' ? Max gain = Avol. End of story.

Graham

Reply to
Pooh Bear

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

AC short, and

is :

--

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.
Reply to
Fred Bartoli

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.

  1. 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

Reply to
Roger Lascelles 

--
Geez, Win, you usually stay away from commentary like that. 

Plus, no smiley.  Got a bug up yer ass or something?
Reply to
John Fields

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

Reply to
Roger Lascelles

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

Reply to
Roger Lascelles

I read in sci.electronics.design that John Fields wrote (in ) about 'Op Amp Calculations', on Wed, 21 Sep 2005:

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
Reply to
John Woodgate

I read in sci.electronics.design that Roger Lascelles wrote (in ) about 'Op Amp Calculations', on Thu, 22 Sep 2005:

The inverting op-amp circuit works by having NO signal on the - input. That's what 'virtual earth' means.

--
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
[snip]

Roger,

I take issue with your comment on "T" feedback circuits. If you choose the more common situation where you looking to reduce the feedback impedance, say by a factor of 10X, the difference is barely noticeable.

I simulated two configurations both with GBW=10MHz and GDC=100K. The difference in high-end roll-off response is less than 1dB.

If you add a capacitor in series with the R to ground, in the "T" configuration, you can knock down Vos sensitivity by nearly a factor of 10X.

I will post the results on A.B.S.E shortly.

...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.
Reply to
Jim Thompson

See...

Newsgroups: alt.binaries.schematics.electronic Subject: Op Amp Calculations (from S.E.D) - MultiPathFeedback.pdf Message-ID:

...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.
Reply to
Jim Thompson

selection.

Hello John - also Graham (Pooh Bear). You people are making me defend my not very academic grasp of opamps!

I can see people don't like me talking about the voltage divider at the input, but it is there when you open the feedback loop and therefore affects noise, offset and drift. I first saw it when calculating signal to noise ratios for audio amplifiers using opamp stages.

For example :

_______R2______ | - | o----R1--+---|\\ | | \\_________|__ out | / Vn --|/ +

R1 = R2 for gain = 1. Vn is noise, offset & drift referred to input and has gain of (1 + R2/R1) = 2. So signal to noise is half as good as the opamp can do with the signal going straight in. Now, short the output to ground and look - the signal IS halved by the R1 R2 divider in exactly that 2 to 1 ratio.

The ratio holds for whatever R1 and R2 you choose. R1 and R2 are dividing down the input signal before it hits the opamp input terminal. I know this continues to be correct when I close the loop, because I continue to see the same signal to noise ratio at the amplifier output - although gain changes.

Makes sense really - simple feedback does not change the S/N performance the network had before you added feedback.

Certainly you don't see much voltage at the virtual earth point when you close the feedback loop, and I see that I needed to explain what I meant by 'input attenuation".

So when I see R2 < R1, I immediately think "input attenuation", poor noise, offset and drift. Just like the T feedback circuit we were discussing. If you can get R1 >> R2 you can really tap into available opamp performance.

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

The other issue is that the T feedback circuit runs with less loop gain for the same overall gain than a simple two resistor amplifier. This means less bandwidth and accuracy. If anyone is interested, I can post the maths, but the 'input attenuation" line of thought gives it to you for free - the input signal is attenuated, but you get the same overall gain, so you must have sacrificed loop gain.

Roger Lascelles

Reply to
Roger Lascelles

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