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Op Amp circuit

How can I obtain the transfer function graph of this circuit, assuming the zener diodes voltage are 4.3V and silicon diodes 0.7V.

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  • \$\begingroup\$ I could be wrong, but it was my understanding that the concept of a transfer function only really applies to linear circuits, no? This doesn't look linear at all; both of those zeners will be conducting. \$\endgroup\$ – Hearth May 16 '18 at 2:02
  • \$\begingroup\$ @Felthry In most cases, yes, transfer functions assume linearity. However, in rare cases, you can use a describing function to estimate a time-invariant system... However that is not applicable here. \$\endgroup\$ – KingDuken May 16 '18 at 3:13
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    \$\begingroup\$ Do you mean a graph of output voltage vs. input voltage? \$\endgroup\$ – Chu May 16 '18 at 7:24
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While the answer given by Humpawumpa is as far as I can tell correct, the question seemed to be "how do I obtain ...". So here's a little bit more information.

You usually start with an easy point on the graph, which is typically the origin. I will call the top node 1 and the bottom node 2.

\$V_i = 0V \Rightarrow V_o = 0V\$

It is also important to look at nodes 1 and 2. Since the diodes are reversely biased, the Zener diodes are biased to their Zener voltage via their respective resistors.

\$V_1 \approx -V_Z = -4.3V\$

\$V_2 \approx +V_Z = 4.3V\$

The way they are biased will make these two voltages follow the output voltage. So they will move in opposite direction from the input. The regular diodes are both reversely biased and can be left out for determining the graph. What is left is a regular inverting amplifier. We find that

\$V_o = -\frac{10k\Omega}{10k\Omega}V_i = -V_i\$

\$V_1 \approx -4.3V - V_i\$

\$V_2 \approx 4.3V - V_i\$

The situation will change once \$V_1\$ or \$V_2\$ cross the negative input voltage (\$V_-\$) of the opamp (remember, due to negative feedback, the negative input pin is virtually shorted to ground). When this happens, a low impedant path is created through the regular diode, continuing along the zener diode to the output. All excess current flowing from input to \$V_-\$ is deviated via the diode, preventing the output to decrease (or increase for negative inputs).

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The general transfer function is that of an inverted amplifier so $$V_{\text{out}} =-\frac{R_2}{R_1}V_{\text{in}}$$ but the diodes will limit your maximum output to \$\pm(4.3\text{ V}+0.7\text{ V})\$. Like already mentioned in the comments, that is not a linear behavior and therefore cannot be represented in a transfer function - or at least I don't know how.

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