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BJT Collector to Emitter voltage

  1. Apr 24, 2015 #1
    I have been checking the BJTs on a car audio amplifier board with my DMM's diode test feature, and I need a little help understanding something I have been noticing:

    The base to emitter and base to collector voltage drops are present, and show an open circuit when the leads are reversed as they should. But, when I connect the pos. lead to the collector, and the neg. lead to the emitter, my DMM reads a voltage drop of anywhere from 0.5v to 1.0v (with no voltage being applied to the base). Reversing the leads, it shows an open circuit.

    Is this how these are supposed to operate? I thought that in the absence of a voltage being applied to the base, there should be no conductance from the collector to the emitter, so I am confused as to why I am seeing the voltage drop from collector to emitter. This amplifier failed after the input stage was fried by too strong of an input RCA voltage (had about 9.0v coming in; amp is only built to handle up to 6.0v). Thanks in advance for any input.
     
  2. jcsd
  3. Apr 24, 2015 #2

    Baluncore

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    Science Advisor

    Welcome to PF.
    For simple tests, a transistor can be modelled as a couple of diodes. Think of an NPN transistor as being two diodes with common anodes at the base, cathodes are the emitter and collector.
    Without a circuit or some idea of your meter we cannot tell much more at this stage.
     
  4. Apr 24, 2015 #3

    meBigGuy

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    Gold Member

    When the base is open, there will be leakage current from the collector base junction which will cause the transistor to conduct for collector to emitter. An open circuit isn't always 0 volts.

    If you ground the base (through a resistor or directly) then the transistor will definitely turn off.

    You can probably ground it with a pretty large resistor. I expect even connecting another voltmeter to the base and emitter will turn off the device (or, if it doesn't, it will indicate there is a base voltage).
     
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