Why Vceo is alway less than Vcbo in BJT

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KCH3
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Why Vceo is alway less than Vcbo in BJT

Post by KCH3 » Thu Nov 05, 2009 1:11 am

In all the BJT spec's why Vceo, (the reverse breakdown voltage between collector and emitter when base is open) is always less than Vcbo (the reverse breakdown voltage between collector and base while emitter is open) even though it is across two junctions instead of only one junction of collector and base as in Vcbo. Can anyone explain the mechanism involve in it in plain term. I can't find anywhere about it and confused.

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MrAl
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Re: Why Vceo is alway less than Vcbo in BJT

Post by MrAl » Thu Nov 05, 2009 2:38 am

Hi,

For one thing, the collector to base voltage is often higher than the collector to emitter voltage
in the application. This would require a transistor who's ratings were bigger for collector to base
than collector to emitter.
LEDs vs Bulbs, LEDs are winning.

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Joseph
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Re: Why Vceo is alway less than Vcbo in BJT

Post by Joseph » Thu Nov 05, 2009 6:50 am

I think one reason is that the current leakage from collector to base tends to turn the transistor on.

KCH3
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Re: Why Vceo is alway less than Vcbo in BJT

Post by KCH3 » Thu Nov 05, 2009 9:25 pm

Hi, I'm talking about the off characteristics (reverse breakdown between two terminals while the third terminal is left opened), it might be related to the doping level of semiconductor or something else. If anyone have knowledge please explain.

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Re: Why Vceo is alway less than Vcbo in BJT

Post by dyarker » Fri Nov 06, 2009 5:11 am

I'm pretty sure Joseph got it.

With reverse voltage across base/collector the base and collector layers are a reverse biased diode.

With reverse voltage across emitter/collector with base LEAD open, the junction between base layer and emitter layer is NOT reverse biased, but the junction between the base layer and collector layer is reversed biased. However, any electrons going from the emitter through the base layer toward the reversed base/collecter junction would turn the transistor on.

So it takes less voltage for Vceo to put enough electrons (or shortage of holes) in the base layer to force turn on, than it does for Vcbo to cause an avalanche.

I'd check that for you, but my semiconductor book from college is almost a thousand miles away.

Cheers,
Dale Y

KCH3
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Re: Why Vceo is alway less than Vcbo in BJT

Post by KCH3 » Fri Nov 06, 2009 7:07 am

Hi dyarker, your explanation make sense to me. But at this moment I can not take in the whole picture and comprehend it all yet. I'm going to analyze it more closely base on your assumption in pencil and paper method (for all the diffusion current, drift current, polarities etc. in detail), because it's a little hard for me to see through and understand every thing involve very quickly. Thank you all for your replies.

KCH3
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Re: Why Vceo is alway less than Vcbo in BJT

Post by KCH3 » Fri Nov 06, 2009 9:36 pm

Ok, this is what I'd come up with. When Vcbo is applied, it's a simple reverse biased diode and as the base and collector are lightly doped compare to emitter, the depletion region will be spread farther apart and leakage current will be small.
But when Vceo is applied, the emitter junction is not reverse biased (this will negate the thinking of reverse voltage is applied across two junctions), and because emitter is heavily doped, plenty of majority carriers from emitter will flow into the base, and because the base is very thinly constructed, all those (now minority carriers in the base) carriers will be swept away into the collector by the bias voltage and the BJT is easily turn on and reach the break down condition in lower voltage of Vceo.

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Re: Why Vceo is alway less than Vcbo in BJT

Post by MrAl » Sat Nov 07, 2009 4:32 am

Hi there,


Here is a good explanation of the phenomenon:

http://books.google.com/books?id=NVC75H ... bo&f=false

I didnt want to copy and paste because of possible copyright infringement.

I have designed many many transistor circuits without having to think about this, which might shed some
light on just how important this is. Usually you look at the leakage current with temperature and try to
provide a base resistor (at least) that will guarantee that the transistor will stay in the 'off' mode
(completely) so that the transistor does not partly turn on and destroy itself or simply hampers correct
circuit operation, and also the max Vce comes into great importance of course so the device can handle
the required operating voltage in the first place.
The criterion for the base resistor is that the expected max leakage current (at elevated temperature)
does not cause enough forward voltage across the base emitter to turn the device on or even
partly on. A good rule of thumb would be 0.3v but others have used 0.4v too, being that many
transistors start to turn on at about 0.6v. It really depends a lot on the particular transistor though.
The criterion for the max operating Vce is something like 3/4 of max Vce but that would partly depend
on the application too.
LEDs vs Bulbs, LEDs are winning.

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