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Diode reverse bias

 

I'm trying to figure out at what voltage (reverse) would a 1n4148
shut off current flowing through it. e.g. 5 volt supply being
opposed by a 3.3V reverse voltage.

On the datasheet, I see Vfr - (Forward voltage recovery) at 2.5V.
Would that mean that it needs 2.5 or greater volt difference
between anode and cathode before current can flow?
[email protected]
 

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In a message dated 4/8/2004 7:10:47 PM Central Standard Time, ghidera2000@... writes:
On the datasheet, I see Vfr - (Forward voltage recovery) at 2.5V.
Would that mean that it needs? 2.5 or greater volt difference
between anode and cathode before current can flow?
No.? About 0.6 volt will cause SOME current to flow through a 1N4148 in the "forward" direction.? You may find that SOME will begin to conduct with less voltage across them.? I have NO clue what Vfr is, as I are igernt.
Keith
 

--- In Electronics_101@..., "ghidera2000"
<ghidera2000@y...> wrote:
I'm trying to figure out at what voltage (reverse) would a 1n4148
shut off current flowing through it. e.g. 5 volt supply being
opposed by a 3.3V reverse voltage.

On the datasheet, I see Vfr - (Forward voltage recovery) at 2.5V.
Would that mean that it needs 2.5 or greater volt difference
between anode and cathode before current can flow?
Zero volts and any reverse voltage up to the maximum for a 1N4148 wil
stop current flowing though it (except for the leakage current). A
small positive voltage will produce a small current flow. For
example, at room temperature you will get 1uA of forward current at
300mV of forward voltage. At 0.675V it will be around 1mA.

Keith.
 

Look at the fig. 8 of this datasheet:

The Vfr means the maximum voltage peak that's produced when the diode is switched on, due to the accumulated loads (I think so; corrections are welcome)

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