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What kind of op-amp is that?

by Femme_physics
Tags: kind, opamp
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NascentOxygen
#55
Mar5-12, 01:09 AM
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Quote Quote by rude man View Post
The circuit is definitely a non-inverting amplifier, gain = +2.4V/V. If you were to change the input from +5V to say + 5.5V you would get 2.4*5.5 = +2.64V output.
True, but that's only half the story...

The circuit is also definitely an inverting amplifier, gain = -1.4 v/v. Using the lower end of R1 for input, if you change the input voltage from 0v (ground) to -0.5v, the output rises from +12v to +12.7v. Viewed this way, the zener functions to bias the op-amp so as to allow operation from a single polarity power supply.
M Quack
#56
Mar5-12, 03:26 AM
P: 669
Q-Fever....OK, this is an attempt to see if the working point at V+=V-=Vout=0V is stable.

Note that my knowledge about electronics is very limited and that this may be a completely wrong way of looking at the problem.

Let's assume the following:

The op-amp as a huge open-loop gain G such that Vout=G(V+ - V-) + Vs/2. G will be something like 10^4 or 10^6. Vs/2 is the midpoint between the supply and ground.

Looking at the characterstic curve of the Zener diode: Around 0V, the Zener diode behaves like a resistor with a fairly large impedance Rz. I guess Rz>RI/4 (you will see later why). Let's say Rz=6.5 kΩ just to have a number.

Now assume V+=ε, a small voltage and V-=0 (it works the same way with V=-ε or V+=ε1 and V-=ε2).

You then get Vout = G ε + Vs/2, which leads to

V-=RI/(Rb+RI)*(Gε+Vs/2)=0.42 (Gε+Vs/2)
V+=Rz/(Ra+Rz)*(Gε+Vs/2)=0.65(Gε+Vs/2)

V+ - V- = (0.65-0.42)(Gε +Vs/2) which is larger than the starting point (which was ε), so the situation is unstable and Vout will drift up until the Zener no longer behaves as a
resistor. For a very small value of ε the constant offset Vs/2 on the output voltage dominates. It does not even matter if ε is positive or negative

For this to be the case RI cannot be too large, RI < Rb/Ra Rz.

For those who wonder if the assumption Vout=G(V+ - V-) + Vs/2 is compatible with the "normal" operation of the circuit:

Assume V+=5V and V-=5v-ε, Vout=12V and G=10000.
Then you get G(5V - (5V-ε))+Vs/2 = Gε + 12.5V = 12V, i.e. ε=-0.5V/G=-50μV which can be neglected.

At the operating point with Vz=5V, the Zener diode has a very steep characteristic curve, i.e. a very small impedance. If Vout changes, the change in V+ will be much smaller than that in V-, so that the negative feedback dominates and Vout is stabilized.

So, NascentOxygen, am I thinking along the right lines here? (I've just made it up, don't be too polite if it's all BS).
NascentOxygen
#57
Mar8-12, 06:33 AM
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Quote Quote by M Quack View Post
The op-amp as a huge open-loop gain G such that Vout=G(V+ - V-) + Vs/2.
I guess that's right. Op-amps usually are powered by double supplies, so your Vs/2 term is unusual, but I can't see it could be anything else.
looking at the characterstic curve of the Zener diode: Around 0V, the Zener diode behaves like a resistor with a fairly large impedance Rz. I guess Rz>RI/4 (you will see later why).
That's right.

situation is unstable and Vout will drift up until the Zener no longer behaves as a
resistor.
So there is no stable state with Vout=0. In practice, you usually won't get the op-amp delivering an output equal to either supply rail, in any case, so you wouldn't be able to obtain Vout=0.

You analysis is spot on. Below the zener breakdown, the amplifier has more positive feedback than negative. This could send Vout towards 0 but nothing happens that would keep it there, and any noise that sends Vout towards +Vs will succeed, owing to a nett positive feedback.

Well done!
M Quack
#58
Mar8-12, 06:57 AM
P: 669
Thanks for checking.


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