When BJT's collector and base are shorted, then what happens for the circuit or may I say the change of BJT's performance in the circuit as there Vbc is zero.


simulate this circuit – Schematic created using CircuitLab

Is it work like normal BJT where there is Ic and Ib current ?

Or, It just work like a diode as there is only one biasing voltage difference between one pn junction and zero voltage difference ( so no foreword or rivers bias ) for other junction?

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    \$\begingroup\$ If doing this in real life, make sure VCC is never above 5.5-6V, unless your BJT has special maximum voltage ratings for the base pin. \$\endgroup\$ – KyranF Apr 30 '14 at 3:31
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    \$\begingroup\$ @KyranF Why would Vcc > 6V be a problem in this circuit? \$\endgroup\$ – apalopohapa Apr 30 '14 at 10:17
  • \$\begingroup\$ @apalopohapa well it can actually be whatever you want, just put a Zener Diode with 5.1V rating or similar to GND as a protection measure. The base of transistors for every part i've ever seen has an absolute max rating of 6V, recommended max of 5-5.5V. If you fry the base, it may fail from base->emitter as a short, and then your "diode" effect is lost, thus possibly causing other damage to other parts of your circuit (otherwise, why else have a diode there??). So in this schematic the voltage of that node may not always be at VCC, but if it does go that high, make VCC < 6V, or use Zener! \$\endgroup\$ – KyranF Apr 30 '14 at 11:01
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    \$\begingroup\$ @KyranF Voltage of the base with respect to what? the emitter? And how would this voltage build up, given the two 100kohm resistors in the circuit, if Vcc was for example 10 V? \$\endgroup\$ – apalopohapa Apr 30 '14 at 17:51
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    \$\begingroup\$ @KyranF No, the transistor has no idea where ground is defined. Zero potential (ground) is arbitrarily defined, all that matters are potential differences. I suggest you read up on this. Vcc in the circuit could be 20V and nothing bad would happen. \$\endgroup\$ – apalopohapa Apr 30 '14 at 23:13

Connecting the base to the collector makes it act as a diode, and will drop ~0.7V-1V

Also, for more information about this topic please check out this answer

  • \$\begingroup\$ Haha, i actually clicked your circuit, the "simulate this circuit" link, and replaced VCC with a 5V voltage source. The simulation reported node voltages and currents that support my answer. Voltage drop across the simulated BJT was ~550mV (because of low current) and the current through the circuit was ~22 microamps. We expect 22 microamps because of 5V/ ~200K Ohms. \$\endgroup\$ – KyranF Apr 30 '14 at 3:52
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    \$\begingroup\$ Where did you get this strange idea about having a limit on Vcc? Are you thinking of the reverse Vbe rating of the transistor? That wouldn't apply as long as the junction is forward-biased. \$\endgroup\$ – Dave Tweed May 1 '14 at 1:34
  • \$\begingroup\$ @DaveTweed indeed I am wrong and will edit my answer to remove that warning. I was confused by the voltage the base pin would see is related to ground, but Apalopohapa pointed out this is incorrect in the comments of the original question. When I apply voltages to the base of transistors I always do it with reference to ground which in my case it could perhaps be an issue. \$\endgroup\$ – KyranF May 1 '14 at 5:03

There is a really good video on this circuit, how it works and its applications:


Here are the lab book notes:



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