Soft limiter circuit with diodes and DC power supplies

Thread Starter

xhulion66

Joined Feb 6, 2020
3
I have already worked on it but I can't seem to make it precisely as it should be.
So the problem asks to create a circuit according to the transfer characteristic of the soft limiter shown in the figure below (blah).
I managed to create a circuit (blah2) but it doesn't work because I need Vo=1/2Vi for -30<Vi<-2 and Vo=Vi for -2<Vi<2 and Vo=1/2 Vi for 2<Vi<30 ....but I only get 1/2 Vi +- 1/2 2V
So if anyone can help I would be very grateful...
 

Attachments

MrAl

Joined Jun 17, 2014
7,457
I have already worked on it but I can't seem to make it precisely as it should be.
So the problem asks to create a circuit according to the transfer characteristic of the soft limiter shown in the figure below (blah).
I managed to create a circuit (blah2) but it doesn't work because I need Vo=1/2Vi for -30<Vi<-2 and Vo=Vi for -2<Vi<2 and Vo=1/2 Vi for 2<Vi<30 ....but I only get 1/2 Vi +- 1/2 2V
So if anyone can help I would be very grateful...
Hi,

In the case of an ideal diode, you need to calculate the value of the two voltage sources in with the output. This means your two resistors can not both be 1k because you have the added 2v. In fact, there is the possibility it wont work using that method because the 2v is always present and so it makes the output transfer function non linear (non linear in the strictest sense).

In the case of the non ideal diode (diode has its own voltage drop) you also need to consider the voltage drop of the diode as being added to the output when the circuit begins to clip. Again the exact circuit you are using so far may not be able to meet the criterion so you may need to rethink just a little bit knowing that you have some constant voltages to include in the calculation.
 

Thread Starter

xhulion66

Joined Feb 6, 2020
3
Hi,

In the case of an ideal diode, you need to calculate the value of the two voltage sources in with the output. This means your two resistors can not both be 1k because you have the added 2v. In fact, there is the possibility it wont work using that method because the 2v is always present and so it makes the output transfer function non linear (non linear in the strictest sense).

In the case of the non ideal diode (diode has its own voltage drop) you also need to consider the voltage drop of the diode as being added to the output when the circuit begins to clip. Again the exact circuit you are using so far may not be able to meet the criterion so you may need to rethink just a little bit knowing that you have some constant voltages to include in the calculation.
Thank you for your response...it seems that the required output was different than I though...and you are right about everything but now its solved and thanks anyway :)
 
Top