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I'm reading a datasheet for a small NPN BJT transistor (UTC S8050).

In the datasheet there are two entries regarding Cut-Off Currents.

Here's a table screenshot

enter image description here

I've tried Googling and YouTubing about this parameters and ended up confused.

Let's take for example collector cut-off. If, say, when there's 20 volts across collector and base there's no current flow, but when the voltage increases to 30 (according to the datasheet) some cut-off happens. Cut-off of what current?

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3 Answers 3

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When there is 30V across the collector and base (which is the breakdown voltage, so there had better be no more than 30V in any circuit!), and no current following out of the emitter, (for example, if someone's connected a 30V battery and meter across the collector and base), at most 1uA flows between the collector and base.

Ideally there should be none at all (the transistor is off), but at the maximum voltage for the CE junction, a little leaks through, but no more than this even at this extreme voltage.

A similar analysis holds for the emitter-base junction with the other figure.

I assume this is considered a useful measurement because the reader is expected to assume there will be the same or less leakage at lower voltages.

Think of it like this, if you have a water tap it will have a maximum pressure it can handle. That will be in its "datasheet". At exactly that maximum pressure, even if the tap is off, a little may leak through the tap. That's what this figure represents.

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    \$\begingroup\$ Now it all makes sense. Thank you a lot!!! \$\endgroup\$ Commented Nov 1, 2019 at 16:05
  • \$\begingroup\$ Is there a typo in the answer? The text "but at the maximum voltage for the CE junction" should be "but at the maximum voltage for the CB junction" ?? \$\endgroup\$
    – Dat
    Commented Sep 3, 2021 at 17:47
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The symbol gives information about the parameter:

"Icbo" means the current (I) from collector (C) to base (b) with the third electrode (emitter) open (O).

"Iebo" means current (I) from emitter (e) to base (b) with the third electrode (collector) open (O).

You may also see parameters such as Ices - which is with the base shorted to the emitter, or Iceo - which is with the base open.

Because any leakage between the collector and base will be multiplied by the gain of the transistor Iceo would be expected to be higher than Ices where the leakage current is diverted to the emitter by the external short.

The test conditions columns states the details of how the test is applied.

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  • \$\begingroup\$ thank you, makes much more sense now \$\endgroup\$ Commented Nov 2, 2019 at 16:26
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when there's 20 volts across collector and base there's no current flow, but when the voltage increases to 30

No. You are reading too much into the numbers. You are treating the numbers as a curve or trend, which they are not. Nowhere does it say no current flows at 20V.

some cut-off happens

Also, this sentence doesn't make sense. "Some cutoff happens" doesn't mean anything. It's like saying "some fast happens" for a car. Maybe you meant to say "some cutoff current flows"?

That is simply one measurement taken under the specified conditions.

Forget the fact it even says cutoff for a second. It does not change the meaning or interpretation of the measurement at all. The cutoff label just tells you this is intended to give you an idea of leakage currents when the BJT is off (in cutoff), but doesn't change what the measurement is saying.

i.e. Disconnect the emitter so Ie = 0, apply 30V between C and B and you get 1uA flowing through C and B. In other words, wire the BJT as just a reverse-biased diode using the collector and base junction and measure the leakage at 30V. Nothing more and nothing less. Maybe that's what you were missing? The emitter was disconnected. That's the only way they can 100% make sure Ie = 0. So the leakage current measured must be between B and C.

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  • \$\begingroup\$ Thank you a lot! \$\endgroup\$ Commented Nov 1, 2019 at 16:07
  • \$\begingroup\$ VCB = 30V mean VC - VB =30V right? Or VB - VC = 30V ? \$\endgroup\$
    – Dat
    Commented Aug 2, 2021 at 4:19
  • \$\begingroup\$ @Dat By convention you subtract them in the order they are listed (the one that comes second is the reference) so Vab = Va - Vb \$\endgroup\$
    – DKNguyen
    Commented Aug 2, 2021 at 5:45

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