Hello, I just have a quick question that has been bugging me for a while. We just covered diode circuit models and I would like to have to have some clarification.

The technique that I was taught, (for multiple diode circuits), is to turn all the diodes either OFF to begin and see which diodes pass or fail the assumptions, and turn ON the diode with the largest forward bias (if there is one). I have no problem with this for the ideal model analysis, but I'm having a little bit of trouble with the CVD model.

The CVD model replaces a pn junction diode with an ideal diode in series with a voltage source (usually 0.7 V) connected to its negative terminal. When starting a circuit analysis, I turn all the diodes off so that their currents are zero, but I'm not sure if I am to leave the 0.7 V source in the circuit or not.

For example if I have this circuit, and it says to find the operation of the diode (On/Off, Vd, Id), using the Constant Voltage Drop method:

When I begin, and turn the diode off for analysis, do I put the 0.7 V source in, like this:

or do I leave it out:

Just because for the diode to turn On, shouldn't it have a voltage greater than 0.7 V?

Thanks,
JP

 

I haven't used that technique, but I guess you can leave it or take it out, but for all of the diodes of the circuit.
That way the diode with the larger forward bias will still be the same device, independently of whether it has the 0.7V drop or not.

 

Thank you, I figured it out.

JP