Hi,

I wanted to know how you can derive the soft limiter maximum and minimum points where it changes in slope as part of the voltage at Va and Vb.

In the attached graph, the derivation would be:
Va = Vref*(R4/R3) - Vd1*(1+R4/R2)
Vb = Vref*(R6/R5) + Vd2*(1+R6/R5)

where Vd1 and Vd2 is the voltage drop due to the diode.

Thanks.

 

What attached graph?

Please also include your best efforts at an answer. It is, after all, YOUR homework.

 

Well,

I understand where the slope of the graph comes from since when one of the diodes is in forward bias mode, we have R4 and R2 in parallel with the R3 as the input resistor to a inverting amp.

When in reverse bias, we get a open circuit of the diode and we get a simple inverting amp.

What I dont understand is how the node voltages at A and B are derived.

Thanks.

(Also updated with attachment)

 

You need to take the analysis in pieces. In this case you have four possible regions:

Neither D1 nor D2 is conducting.
Only D1 is conducting.
Only D2 is conducting.
Both D1 and D2 are conducting.

My guess is that the last region is not accessible for this circuit.

In each region, redraw the circuit with non-conducting diodes removed and conducting diodes replaced by shorts (or batteries having the nominal forward-biased diode drop).

Determine the output plots for each circuit and plot them. Where they cross are the transition points.

To calculate the transition points directly, you are looking for the input condition in the circuit having both diodes removed in which the voltage, between the points where one of the diodes actually is, is equal to the forward diode drop voltage.