# Gain bandwidth product of unity gain opamp

1. Sep 29, 2012

### issac newton

i have attached a bode plot and the opamp dc second order lpf circuit. the opamp Tl082 has a gain bandwidth product of 4Mhz according to the datasheet. this circuit has a bandwidth of 414.25khz. can we determine the opamp' individual gain-bandwidth product from this plot ??
if yes , how do we find the answer ?

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2. Sep 29, 2012

### yungman

What low pass filter? I don't see no caps. From my short analysis, you have a divided by 11 at the ouput of both opamp. The total gain of the whole circuit is gain of 1 only, but each of the opamp actually have to perform a gain of 11 due to the voltage divider.

For an opamp with GBWP of 4MHz, if it has a gain of 11 due to the divider, the bandwidth is about 370KHz and start rolling off, which you show in the Bode plot. That is the expected result.

I cannot tell the individual GBWP of each opamp as you only have one graph. I assume both have the same BW as they both are using as gain of 11. The phase angle show -100 at about 330KHz which is close to the pole frequency of 45 deg each. You have a little bump at around the pole frequency mainly because the way you connect the amp, that the first opamp has to go through the second opamp to complete the loop. So the phase margin is reduced and cause a little under damp.

That's all I see so far.

Last edited: Sep 29, 2012
3. Sep 29, 2012

### Staff: Mentor

I'm thinking aloud here, but haven't tried it ....

If you were to model each OP-AMP gain by a single pole, viz., A₀/(1 + ω/(GB)) and determine the two-stage amplifier's transfer function, you could equate your measured -6dB corner frequency to that in the transfer function and solve for the GB product. (I see the slope of your graph is -40dB/dec, confirming that it is a 2nd order roll-off.)

4. Sep 29, 2012

### yungman

You can use the GBWP to roll off and treat it as LPF? That's not the most reliable way of doing it!!

5. Sep 29, 2012

### Staff: Mentor

Yes, it saves bulky components (capacitors).

6. Sep 29, 2012

### yungman

Ha, so if you want 40KHz, do a divided by 100!!! But be careful, the output swing is very limited.

7. Sep 29, 2012

### issac newton

thats the transfer function i derived. now how can i find the gain bandwidth product of the opamp theoritically ? how can i determine it from the plot i posted previously and compare my results ?

where alpha = R1/R1+R2

I Substituted A=GB/s not GB/2 sorry for the typo in the pic

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8. Sep 29, 2012

### Staff: Mentor

Try to express the denominator as a double pole, (1+jω/ω₀)² and compare that to the -6db frequency of your plot.

9. Sep 29, 2012

### uart

What yungman said in the first reply would have worked fine if you just needed the approximate GB product, by inspection from the bode plot without going into the full details of finding the TF.

Anyway I just did a quick calculation and that transfer function in your attachment looks correct to me (after of course simplifying it with the assumption $\alpha GB >> 1$).

Write your TF in the following form (where $k = 1/\alpha$),
$$H(s) = \frac{1}{k^2 (s/GB)^2 + k (s/GB) + 1}$$
and notice that when $s = j GB/k$ that the TF simplifies down to $H = 1/j$, which should be pretty easy to locate on your bode plot (unity gain and 90 degrees phase lag).

10. Sep 29, 2012

### yungman

I hope this is only use for the exercise of the brain and for the fun of it. Theory and formulas are very important to learn. I encourage people to go through the formulas.

BUT as a practice engineer, this circuit is a very bad idea.

1)Pole frequency is unpredictable as you cannot count on the exact GBWP. As an IC designer before, component values inside the IC vary +/-20 to 30% between runs. So you cannot count on 4MHz. That's only a typical plot, an approx. value.
2) The GBWP can drift with temperature also.
3) The worst and the kicker, this circuit severely limit the swing of the output. What you have, it can only swing +/-1.3V at it's best with +/15V supplies. If you want lower freq, like 40KHz, you only get 130mV peak to peak. That is not useful as a circuit.

It is very important for student to realize most of the circuits studied in college not necessary useful ( to be very polite). Particular if the professor never have real life working experience. That's the reason why a lot of student that get good grades don't transition into a good engineer. One of the most important thing school fail to teach is common sense engineering. That's the reason why people always said you don't use 90% of what you learn in school. Just keep this in mind, this circuit is only good for discussion, suck big as a real circuit.

Last edited: Sep 29, 2012
11. Oct 1, 2012