Npn transistor with a diode

  • #1
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Homework Statement



The absolute max voltage rating for the base-emitter voltage is the lowest of all values. This low value means the B-E junction must be protected from a high reverse voltage. A diode is used for an npn transistor. Explain its function and conditions under which protection is given. Draw the equivalent circuit for pnp transistors

**see attached diagram**

Homework Equations





The Attempt at a Solution



I am not really sure about this one, but i know that because of the diode no current will flow from B to E through the diode due to the bias of the diode. However, it seems that current will still be able to flow from E to B through the diode, with a .6V voltage drop. Not really sure if i'm on the right track here or not... any help would be great!!!
 

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Answers and Replies

  • #2
gneill
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"...the B-E junction must be protected from a high reverse voltage", and you said, "...current will still be able to flow from E to B through the diode, with a .6V voltage drop".

0.6V seems like a pretty tame reverse voltage! So, how's the diode functioning?
 
  • #3
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Well it looks like since it is reverse bias when current is flowing from B, so it has to go through the transistor. When current flows from E back to B it is going through the diode which is forward biased, which will create a .6V drop in the voltage... which i think they said is already pretty low to begin with...

Don't know if this is what you were looking for :/
 
  • #4
gneill
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Suppose that there was a large negative voltage being applied at the base, perhaps from some large amplitude AC source on the negative portion of its cycle. If the diode were not there, what would be the maximum reverse bias (Base-Emitter voltage) that the transistor would see? How about with the diode in place?
 
  • #5
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Well without the diode, wouldn't there have to be a bigger voltage drop in the base-emitter junction? With the diode, wouldn't it drop some of that voltage so the junction doesn't have to drop it all?
 
  • #6
gneill
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Well without the diode, wouldn't there have to be a bigger voltage drop in the base-emitter junction? With the diode, wouldn't it drop some of that voltage so the junction doesn't have to drop it all?

Yes, but not just some. What is the maximum reverse voltage that can occur across the base-emitter junction with the diode there? Remember, the base-emitter junction itself is reverse biased at this point, so it shouldn't be passing any current (it should "look" like an open circuit).
 
  • #7
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Well theoretically it should drop all of it :D, so the max should be zero with the diode

I realized that after I typed it....

So for the PNP would I just reverse the bias of the diode?
 
  • #8
gneill
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Yes!

The diode is acting as a voltage clamp for the B-E junction. It's preventing large negative excursions by shorting them out. Real diodes conduct at about 0.6V, so it's a 0.6V clamp.
 
  • #9
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Thanks so much, that makes sense!!

However, am I right about the PNP being the other direction because then the current would be flowing in the opposite direction, so then the reverse current would be limited by a diode that is biased from B to E?
 
  • #10
gneill
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Thanks so much, that makes sense!!

However, am I right about the PNP being the other direction because then the current would be flowing in the opposite direction, so then the reverse current would be limited by a diode that is biased from B to E?

Yes. For a PNP configuration, the diode would be oriented in the opposite direction.

But it's the voltage that's being limited, not the current per say. The reverse current through the junction should be negligible because it's reverse biased (Well, we want it to be negligible! If the junction fails due to too high a reverse voltage, it will conduct and probably destroy the transistor).
 
  • #11
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Good catch :D

Thanks so much for the help!!! I really appreciate it!
 

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