forward reverse bias diode
Thanks for your reply
So does that mean the following
(a) in Forward Bias if R = 1KOhm
I = (5V - 0.7V) / 1KOhm
Does that mean that in FB, from curves, I increases exponentially upto the max allowable I from the eqn above.
(b) in Reverse Bias
- BEFORE Breakdown if supply was say 2V instead of 5V,
I = 0
- AFTER Breakdown if supply is back to 5V with Breakdown at 3V
(i) will there be a opposing volt on the diode and the resistor will have to compensate for the rest
in other words
5V (Batt) = - 3V (Diode) + 8V across the Resistor
OR
(ii) will it be 5V (Batt) = + 3V (Diode) + 2V across R.
Sorry for making the issue more complicated.
Added after 26 minutes:
Ah!!! Write Write Constraint - I was writing my post just as you has posted your reply.
Just to clarify, if we look at voltages
in Reverse Bias at 5V
0V -------|>|-------------3V-----------------^^^^^^^--------------5V
Diode Resistor
In terms of current,
in FB, electrons flow from n to p side, so the direction of I is from R to Diode
in RB, what will be the current direction
and is Avalanche effect totally by the minority carriers.
Thanks for the help guys