I'm looking through my notes and came across this example. I can't seem to grasp it.

The question is:
Design a limiter with Av = -10 in the +- 5v output range and Av = -1 outside that range.

The attached picture shows the circuit that was drawn with the question.

Thus far I have made the following assumptions:
- The diodes will always be in opposite bias. One will be an open circuit while the other will be a short circuit.
- The op-amp is used to get the Av = -10 gain thus \(\frac{R_{f}}{R_{1}} = -10\)

I realize that R2 thru R5 are used to scale down the gain back to -1 when the output voltage falls outside the given range.

So how does one go about finding the R2-R5 values and some explanation as to how this circuit operates would be great !

Thanks!

 

Stare at this sim. Note the slopes of the three regions of the transfer curve. When neither diode is conducting, the gain is just -R1/R3, or -5. When D1 conducts, R4 is parallel with R1, or 40K*10K/(40K+10J) = 8K, so the overall gain is -8K/8K or -1. When D2 conducts, then R5 is in parallel with R1, or (40K+4.32K)/(40K+4.32K), so the overall gain is - 3.9K/8K, or a bit less than -0.5

It is a little more complicated than that if you are also trying to set the clipping levels at the same time...

 

That all makes sense. Thanks ! Now how do the values of R2 and R6 come into play ?

I'm guessing they set the boundaries for the three regions of the transfer curve. I was thinking they create a simple voltage divider.
Va = R2 / (R2+R4) * Vcc
Thus allowing the diodes to conduct when the voltage coming in is larger than the voltage from the voltage divider.

Although calculating it out, the numbers don't seem to fit, where am I going wrong ? Am I missing some some component of Va due to it being also connected to Vo and thus there is some sort of feedback ?