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The datasheet for a BD679 transistor lists amongst the absolute maximum ratings that the "Emitter Base Voltage" has a maximum of 5v.

This figure confuses me - my mental model of a (BJT) transistor has the path from the base to the emitter equivalent to that of a diode and the potential difference is irrelevant - it is the current that controls the gate.

I have searched for this term and among the results get ones like this which appear to be talking about a different property of the transistor.

The notation ('Emitter Base Voltage' as opposed to 'Base Emitter Voltage') makes me think this might be referring to the maximum 'negative voltage' that could be placed across the Base-Emitter, instead of the maximum in normal operation. Is this correct?

If not, what is this figure, and what causes this junction to have such a low maximum compared with the rest of the device?

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4 Answers 4

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"Emitter Base Voltage" is the maximum voltage that may be applied when the base-emitter diode is in reverse; not conducting. This is generally much lower than a small signal diode in reverse can handle.

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  • \$\begingroup\$ Thank you, I am glad this is the case as it made the most sense! \$\endgroup\$
    – sebf
    Commented Jul 3, 2012 at 20:56
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You suspicion is correct - it's emitter-base, which means it's the (positive) voltage from emitter to base, not base to emitter (which would be given as -5V)
So basically it means you can't let the base drop >5V below the emitter (or emitter 5V above the base ;-) )

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  • \$\begingroup\$ Does the polarity of voltage matter in this case ? I wanted undertand what happens when negative voltage is applied to base or emitter \$\endgroup\$
    – Embedded Enthusiast
    Commented Mar 12 at 15:32
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Your guess about the order of the terminals is right: it is the maximum reverse voltage that the emitter-base diode can block. Some people claim this makes a good zener diode , others use the zender mode as noise source.

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  • \$\begingroup\$ Thanks for those links they are very interesting (I would have thought violating the absolute maximums would have killed the transistor - or maybe it did and the junction remained a working zener diode anyway?) \$\endgroup\$
    – sebf
    Commented Jul 3, 2012 at 20:55
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    \$\begingroup\$ For an ideal zener a voltage above the threshold results in an infinite current - in that sense vioaltion of that voltage limit will be fatal. But a current-limited 'violation' might be non-destructive. But the transistor is not designed for this mode of operation, so the manufacturer does not bother to spec these limits. \$\endgroup\$
    – Wouter van Ooijen
    Commented Jul 3, 2012 at 21:21
  • \$\begingroup\$ A dedicated Zener diode is typically used with high current in a power source. This application is only for small signal effects at the edge of breakdown, like the noise generator, if it's avalanche mode. Edit: see also Wikipedia on the difference between avalanche and Zener breakdown. \$\endgroup\$
    – Potatoswatter
    Commented Jul 4, 2012 at 8:20
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You say "path from the base to the emitter equivalent to that of a diode and the potential difference is irrelevant - it is the current that controls the gate." This is completely wrong. The BJT, like all types of transistors" is a voltage-controlled current source: it is the base - emitter voltage and not the base current that controls the collector current. The base current is an unavoidable error caused by forward biasing the base - emitter junction, but it is not the basis of the BJT's operation.

For further information, see the following:

http://d1.amobbs.com/bbs_upload782111/files_29/ourdev_554203.pdf

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