Dear All

Can you please solve this problem, attached in this Thread. I tried using super position theorem, but there is an diode between the circuit, so that I couldn’t apply super position theorem, Can you please help me how to approach this problem. Thanks

 

What are you suppose to solve for??

 

Redraw the circuit for two mesh loops and replace the diode with its model (a .65 V independent voltage source with the positive at node one and the negative at node six).

 

Redraw the circuit for two mesh loops and replace the diode with its model (a .65 V independent voltage source with the positive at node one and the negative at node six).

And analyze your results carefully...particularly with regard to your diode bias.

 

But how to model the diode as 0.65V for this condition, there is no much current flowing through the diode, the diode may not completely forward biased, The current through the diode approximately 27uA, the diode drop may not be 0.65V in this condition. Thanks

 

But how to model the diode as 0.65V for this condition, there is no much current flowing through the diode, the diode may not completely forward biased, The current through the diode approximately 27uA, the diode drop may not be 0.65V in this condition. Thanks

Which I suspect is the purpose of such an exercise. So, how do we model a diode which is not forward biased? Also, from a design perspective, if we wanted to have that diode forward biased, what could we change in the circuit to accomplish that goal?

 

To determine if the diode is really forward biased or not, evaluate the voltage across the diode connections with the diode removed from the circuit. Does the voltage turn out to be a reverse bias?

Watch out for the sign of the 15V supply by the way. Isn't it -15V? Did you assume that when you got a diode current value of 27μA?

http://www.onsemi.com/pub_link/Collateral/BAS21SLT1-D.PDF

 

As it concerns the reversed biased diode, I would just replace it with an open circuit. Leakage currents can be safely ignored.

For the forward biased diode, there are several levels of simplification.
The simplest one is to replace it with a short, with no voltage drop.
A bit more complex is to add a 0.7V DC source in series to it, with its polarity impeding the flow if the current towards the Ground.
Finally, if you have the diode model or some measurements on terminal voltage - diode current, you could add the internal resistance of the diode, in series as well.

 

In this particular case the ohmic resistance voltage drop should not be terribly big!

 

As it concerns the reversed biased diode, I would just replace it with an open circuit. Leakage currents can be safely ignored.

For the forward biased diode, there are several levels of simplification.
The simplest one is to replace it with a short, with no voltage drop.
A bit more complex is to add a 0.7V DC source in series to it, with its polarity impeding the flow if the current towards the Ground.
Finally, if you have the diode model or some measurements on terminal voltage - diode current, you could add the internal resistance of the diode, in series as well.

This is the response I was hoping to elicit from the OP But, yeah, this is how I have always solved these.

 

This is the response I was hoping to elicit from the OP But, yeah, this is how I have always solved these.

Oops, sorry for the spoiler. I just thought that it is more important to have a direction to work for, rather than stumble with trial and error. Doing the math and solving the equations is just as instructive in learning the solution of the problem.