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What kind of opamp is that? 
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#55
Mar512, 01:09 AM

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P: 5,365

The circuit is also definitely an inverting amplifier, gain = 1.4 v/v. Using the lower end of R_{1} 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 opamp so as to allow operation from a single polarity power supply. 


#56
Mar512, 03:26 AM

P: 669

QFever....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 opamp as a huge openloop 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 = (0.650.42)(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=5vε, Vout=12V and G=10000. Then you get G(5V  (5Vε))+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). 


#57
Mar812, 06:33 AM

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P: 5,365

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! 


#58
Mar812, 06:57 AM

P: 669

Thanks for checking.



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