Trouble with OP amplifier + Zenner circuit

[malmoit]

Hi there folks!

This time I have some trouble solving the out-signal as a function of the in-signal, with regard to the circuit (see attached files). The circuit includes an OP amplifier and a Zenner-diode --- which both are assumed to be IDEAL! ---

My suggested approach for solving the problem starts off like:


KVL:

(1) -u,out + R2*i + u = 0

(2) -u,out + R2*i + R1*i u,ref = 0 ==> i = (u,out - u,ref) / (R1 + R2)

equation (2) in (1) gives

(3) u = (u,out*R1 + u,ref*R2) / (R1 + R2)

Now, when u,in < u then u,out = 0 V

If we denote this with u = u1, and use KVL to obtain u,ref = 7 V, and use the values in equation (3), we should get

u1 = (0*100*E3 + 7*200*E3) / (100*E3 + 200*E3) = 8 V

----- which is wrong answer according to my book!! -----


Corrispondingly when u,in > u = u2 then u,out = 10 V and

u2 = (10*100*E3 + 7*200*E3) / (100*E3 + 200*E3) = 4,667 V

----- which is wrong answer according to my book!! -----



THE RIGHT ANSWERS TO THE PROBLEM ARE:

u,in < u1 = 6,67 V gives ==> u,out = 10 V

u,in > u2 = 3,33 V gives ==> u,out = 0 V


Well, reading the prerequisites of the problem in my text-book I note that "Rs" is given. However, I haven't used Rs in my calculations above, so I'm sure I'm doing something wrong! But why do I need to know Rs...?

Would be very grateful if someone could help a newbie learning more about OP amplifiers, and tell me how the right solution to the problem should look like.

 

[Roff]

I have to confess I don't have the patience to review your calculations, but I would proceed as follows:
We know that the op amp will switch when Uin is equal to the voltage at the junction of R1 and R2 (U in your drawing).

U=(Uout-Uz)(R1/(R1+R2))+Uz

Simply substitute 0V and 10V for Uout in the equation.

And your suspicions about Rs are correct. It is a red herring.

 

[MrAl]

With Uin=0, any small voltage on the non inverting terminal
forces the output high, so we'll start there.

With Uin=0, Uout=10v (assuming the output goes that high).
With Vout=10, the voltage on the non invert terminal is
(Vout-Uz)*R1/(R1+R2)+Uz
which for the values shown would come out to:
Vp1=5*1/3+5=6.666666v.

As Uin increases, eventually it reaches this level of Vp1,
and as it does the output starts to ramp down quickly.
This means the voltage at the non invert terminal is also
ramping down. This means the voltage at Uin is now
always greater than the non invert terminal and so the
output continues to ramp down until it hits 0v (approx).
Now that it's at 0v, Vp changes to
Vp2=Uz*(R2/(R1+R2))
which for the values shown comes out to:
Vp2=5*2/3=3.333333v.
This is still less than Uin so the output stays low.

Now if Uin increases, the output stays low, but if
Uin decreases, eventually it equals Vp2 and the output
starts to ramp up. As it does, this makes Vp2 ramp up
also, so that Vp is always greater than Uin. This
keeps the output ramping up, until it hits the max, 10v.

Thus, the output is going to be in one of four states:
1. Low (0v)
2. High (10v)
3. Ramping down quickly
4. Ramping up quickly

but in many cases you can assume only states 1 and 2,
so the output will either be 0v or it will be 10v.