I see in the specification sheet for a darlington transistor that its collector to emitter breakdown voltage is 100V.
What is its emitter to collector breakdown voltage? The same?
Thank you in advance.
Gerb
Question about transistor breakdown voltage
Oct 23, 2008
18 Replies
It depends on the concept you're trying to understand. If you simply want to have an idea of the maximum voltage that can be applied between collector and emitter before breakdown, assuming that the polarity is in the usual direction, then it could be said that they are the same.
However, to be technically correct and also specify the polarity at the same time, they are not the same. For example:
The voltage applied to the collector of an NPN transistor is normally positive with respect to the emitter. Therefore, the collector-emitter voltage (whether breakdown or normal operating) would be given as X volts, whereas for a PNP transistor, it would be -X volts.
If you give X volts as the emitter-collector voltage, it would mean that the emitter is positive with respect to the collector. This is the reverse of the way an NPN transistor is normally operated. The breakdown voltage would also probably be different because the base-emitter junction and the base-collector junctions are made different in practice, although their basic principles are the same.
Let me add some clarification to my earlier statement.
The darlington transistor is an NPN device used as a low side switch. Emitter is tied to ground. I am interested in understanding what happens when the collector goes negative with respect to ground.
In this instance, wouldn't the emitter to base junction break down first (max rating of 10V)? Current would flow from the emitter, to the base, then from the base to collector (one diode drop). So, total breakdown voltage in the reverse direction (assuning the connection scheme listed above) would be approximately 10.7VDC?
Thank you
Gerb
Let me add some clarification to my earlier statement.
The darlington transistor is an NPN device used as a low side switch. Emitter is tied to ground. I am interested in understanding what happens when the collector goes negative with respect to ground.
In this instance, wouldn't the emitter to base junction break down first (max rating of 10V)? Current would flow from the emitter, to the base, then from the base to collector (one diode drop). So, total breakdown voltage in the reverse direction (assuning the connection scheme listed above) would be approximately 10.7VDC?
Yes, that's what would happen unless the base is prevented from being pulled below ground by some other circuitry. This can be qualitatively demonstrated with a single (non-darlington) transistor and an analog multimeter set at the high ohmmeter range. The 9V battery generally used to drive the high resistance range is higher than the e-b breakdown voltage (usually around
5V). In fact, I often use this to identify the leads of an unknown transistor. The collector-emitter path is essentially open in one direction, but conducts in the other direction, but at a lower level than a single p-n junction.
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JF
** Look at the diagram of the darlington - see a diode drawn from collector to emitter?
All power darlingtons have them.
Means full conduction from 0.6 volts when subjected to reverse polarity.
...... Phil
what's connected to the base?
Bye. Jasen
Phil 1, everyone else 0.
JF
Where did you get the idea that the OP was referring to POWER darlingtons?
Besides, it's not true that all power darlingtons have antiparallel diodes. The General Electric D40K doesn't have one. And, as GE says in one of their manuals, "The modern monolithic Darlington may contain all or none of the auxiliary components shown in Figure 7.2 (referring to the antiparallel diode and built-in base-emitter resistors). It only takes one counterexample to falsify your assertion.
Apparently not.
OK.
Phil 1000, you 1
JF
Given Phil's normal tendency to mercilessly rag on anyone who makes any assumption that he considers unwarranted, he must be held to his own high standard in that respect. Since the OP didn't use the word "power" in any of his posts, Phil gets 0, not 1000, a forfeit for bad behavior.
If anyone else had said what Phil did, he would probably say something like this:
CONSIDER THE CONTEXT, IDIOT!!!! THE OP WOULD HAVE SEEN THE DIODE IF IT WERE IN THE DIAGRAM, AND WOULDN'T BE ASKING THE QUESTION HE DID!!!!!
Only Phil made the assumption that the OP's darlington has a diode; everybody else assumed that it doesn't. Why do you suppose that is? Which is the more reasonable assumption?
Just because I only cited one power darlington without a diode doesn't mean that there aren't more.
But more to the point, it doesn't matter how many power darlingtons have diodes. If the OP's darlington isn't a power darlington, then it's irrelevant how many power darlingtons have diodes. You don't get points for conclusions based on an unwarranted assumption.
Furthermore, if we don't know for sure whether the OP's darlington has a diode or not, and have no good reason to assume that it does, treating the case where it doesn't is the useful thing to do.
"Dr. Posturing Pig "
** 100 volt, NPN low side switch ..... so has all the earmarks.Plus see the " ? " at the end of the line ????
That means something.
** All the popular, readily available ones do.** Really.
Why don't you go jump in front of an express train - f****it.
..... Phil
If there were a diode in the diagram he wouldn't have been asking the question.
Small signal darlingtons are very often used as low side switches to turn on LED's, energize small relays, and they are also often NPN. These things don't imply power.
It means he was asking a question. If a diode were visible in the diagram, he wouldn't need you to ask him if he could see it.
So that means he couldn't be asking about one of the less popular ones?
Really. And more besides.
After you.
100 volt, NPN low side switch. Those qualities are the exact "earmarks" of a BC372 small signal darlington for example, and don't imply "power". The diagram shows there is no diode. If the spec sheet he was looking at had a diagram with a diode, he would have seen it, and wouldn't have been asking the question he asked. That's the context you're ignoring, as you so often accuse others of doing.
I notice he hasn't responded to your question. The answer was so obvious it didn't merit a reply. If there had been a diode in a diagram, he wouldn't have been asking about reverse breakdown.
Vceo, Vceb or Vcev ?
Graham
And why would that happen ?
Never heard of inverse parallel diodes ?
Graham
A fairie.
Perhaps the OP will come forward with the device he was referring to and
end the squabbling?
JF
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