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While I was designing a constant current source using a 2N3904 I noticed a behavior on the bjt that prompted me to look closer into it.

The issue is that although the datasheet specifies a minimum BVceo = 40V. Any voltage above 8VDC on the collector with base either floating or grounded (as shown on schematic) creates an sudden increase in collector current (suggesting breakdown) and the voltage across C-E is decreased as to limit power across C-E. I breadboarded the circuit below to investigate this.

My current mirror is to be used on the output stage of an audio headphone amplifier with Vcc = 12V. Any help understanding this would be greatly appreciated.

schematic

simulate this circuit – Schematic created using CircuitLab enter image description here

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    \$\begingroup\$ You sure you don't have the collector and emitter switched? \$\endgroup\$ – John D Mar 19 '14 at 0:02
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    \$\begingroup\$ 8V-9V is roughly what would be expected for the reverse breakdown voltage of the E-B junction. If you do have E and C switched (as @JohnD suggested) you may have noticed your mirror worked, but had low performance because reverse beta is usually relatively low. \$\endgroup\$ – Spehro Pefhany Mar 19 '14 at 0:05
  • \$\begingroup\$ See this: youtube.com/watch?v=rpGOKGrcpAk Here he uses this breakdown voltaje to make an oscillator and shows the curve in Curve Tracer instrument. Indeed he shows it is not only a Breakdown voltage, this also have a little "negative resistance" zone. \$\endgroup\$ – user296760 Mar 4 at 1:51
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thanks much for pointing this out. Indeed it seems i was actually looking at the E-B junction reverse breakdown instead. I had my bjt not properly hooked up.

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