# Diode Modelling

Discussion in 'Homework Help' started by RdAdr, May 8, 2015.

May 19, 2013
214
1
http://en.wikipedia.org/wiki/Diode_modelling

where it says about the Ideal diode in series with voltage source.

What if I apply 1V to the anode of the ideal diode, and 5V to the "minus" of the voltage source? So I have a potential difference of -4V over the Ideal diode in series with voltage source.

Then because the anode of the ideal diode is at 1V and the cathode at 0.7V, then the ideal diode has a potential difference over it of 0.3V. Thus the ideal diode is short circuit.
Thus the Ideal diode in series with voltage source is a short circuit+voltage source. But the potential difference over it is -4V. It should be a short circuit+voltage source only if the potential difference over it is greater than 0.7V.

Or I could say. What if I apply 0V to the anode and -2V to the minus. So I have a +2V over the Ideal diode in series with voltage source. This is greater than 0.7V => it should conduct. But the voltage over the ideal diode is 0-0.7V <0V. Thus the ideal diode is open circuit.

???

Am I saying something wrong? I am confused.

PS: for me Vt = 0.7V

Last edited: May 8, 2015
2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
I think you are confused about how voltages are specified with respect to a reference. In the following, all voltages (except as noted) are measured with respect to the ground node. I create a scenario where V2 takes on only 2 values: 0V and 4V, while V1 is slowly swept from -1V to 6V. I plot V(anode) and V(cathode) with respect to the ground reference vs V1 (x-axis).

Finally, I plot the voltage across the ideal diode, the differential expression V(anode)-V(cathode) to show what is happening to the diode itself as a function of V1.

The test circuit:

The plots:

In all instances, the green trace is with V2=0V, and the red trace is with V2=4V

Do you understand why V(anode) ...the red trace in the middle plot pane.. clamps at ~4.7V?

May 19, 2013
214
1
Yes, because the diode turns ON and it will have a potential difference of 4.7V across it (from the anode to the ground).

Now I get it.
I said:
"What if I apply 1V to the anode of the ideal diode, and 5V to the "minus" of the voltage source? So I have a potential difference of -4V over the Ideal diode in series with voltage source.
Then because the anode of the ideal diode is at 1V and the cathode at 0.7V, "

But the cathode would be at 5.7V. So the ideal diode would have a potential difference of 1-5.7V <0V => ideal diode OFF and not short circuit. Dumb mistake.

Thanks.