Explanation of Reverse Bias Voltage and Breakdown in a diode

Hi guys,

Another newbie question, correct me if I am wrong. If I have a Diode in series with a resistor and I apply +5V as

(a) Forward Bias - will there be ~0.7V across the Diode and the rest across the R.

(b) Reverse Bias - all the voltage across the diode, and none on the resistor (since all reverse current will be blocked). What would happen if the Breakdown Voltage was say 3V, how would the Avalanche Current work, and at breakdown, would Ohms Law be valid? Assume Leakage Current = 0.

This will help us newbies get a grip on Reverse Voltage and Breakdown.

Thanks
AR

diode reverse bias current

well your veiw on A was correct. and on your B since 3V is the breakdown and u are applying 5V it will act like in your forward bias because u have pasted the breakdown voltage therefore causing it to be forward bias.
i hope this helped and if any one knows more to say can also add for i dont really have the time to go in details.

diode reverse bias

a) As diode can conduct infinite(practically large) current above forward bias voltage the voltage across diode could be only forward bias voltage drop of 0.7V. The rest 4.3V(5-0.7) will be dropped across resistor causing a current of 4.3mA(considering R=1K ohm 4.3/R is 4.3mA) in the total circuit 9Within the resistor and diode).
b)Ohms law is applicable for resistor where ever it is used. Here the current flowing through resistor is same as the current in diode.To derive the current flowing in the circuitconsider the following statements.
Until 3V(breakdown voltage) the diode will not conduct current and there is no current flow till 3V. Once the voltage exceeds breakdown voltage across the diode, it will conduct any amount of current, but the drop across diode will be 3V. The remaining 2V(5-3) will drop across resistor causing a current of 2mA(considering R=1K and using ohms law I= V(2)/R(1K)).
Hope this clarifies your queries.

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

diode bias

When Avalanche effect occurs the conduction is through minority carries so the electrons will flow from p to n. Here also the direction of conventional current or hole current will be from resistor to diode. You could visualize the same in the diagram below.