hi for NPN transistor normally Collector will be connected with Vcc, resistor then LED. and Emitter will be conencted to Vss. So whenever voltage applied to base LED illuminates.

however when Vcc connected to Emitter and Vss to Collector (the rest remains, LED polarity changed) LED illuminates too when there is a base voltage.

any idea? as far i know NPN conducts from C to E only.


  • \$\begingroup\$ Minor correction to first para : whenever current applied to base, LED illuminates. When voltage > 0.7 applied to base, base-emitter junction takes all the current available from voltage source and transistor burns out. The difference is; for this LED driver app, you need a current limiting resistor, usually in the base circuit. But get used to thinking of bipolar transistors as current amplifiers; it's the resistors around them that convert from voltage to current and vice-versa. \$\endgroup\$
    – user16324
    Commented Dec 24, 2012 at 10:08
  • \$\begingroup\$ Vcc is a bipolar transistor term (the "c" comes from "collector"). Vss is a field-effect transistor term (the "s" comes from "source"). Vee is used for bipolars. \$\endgroup\$
    – Kaz
    Commented Dec 24, 2012 at 22:04

2 Answers 2


A bipolar transistor is basically symmetrical from C to E. In practice, it is optimized for good properties (most important: a high current amplification factor), which requires asymmetry. So a real transistor will have good properties when used as intended, but will still function as a very mediocre transistor when used with its C and E exchanged. In particular it will have a very low Beta (= current amplification factor) when used in this way, think of 10 instead of 200.

  • \$\begingroup\$ So Vb doesn't have to be higher than Ve for the transistor to conduct? \$\endgroup\$
    – m.Alin
    Commented Dec 24, 2012 at 7:55
  • 1
    \$\begingroup\$ @m.Alin - It does, but in this case, the collector is acting as the emitter. Vb is higher then Vc, so you get conductance. \$\endgroup\$ Commented Dec 24, 2012 at 8:15
  • \$\begingroup\$ @m.Alin maybe you read the B in my answer as Base, I meant Beta. Answer edited to explain this. \$\endgroup\$ Commented Dec 24, 2012 at 11:26
  • \$\begingroup\$ I think the key here is that the polarity of the supply reverses with respect to the transistor. So not only is Vb not higher than Ve, but Vbe is reverse-biased (until saturation, at least). Vbc is forward-biased, playing the role of Vbe. \$\endgroup\$
    – Kaz
    Commented Dec 24, 2012 at 22:09

Assuming that you mean the transistor collector is hooked to Vcc THROUGH a resistor and then LED, then yes having the BJT flipped over (using emitter as collector and vise-versa) could work.

There are devices specifically designed to be used this way, commonly called reverse active region operation. Such devices typically have high forward active region gain (\$ h_{fe}\$ of 500 or so), and a higher than normal rated \$V_{be}\$ reverse voltage (say 15V). When operated in the reverse mode would have \$ h_{fe} \$ of around 5, but would also have very low \$ V_{ce} \$ saturation approaching 100 or 200 \$\text{$\mu $V} \$. Here is an example part ... from MicroSemi (2n2432a: http://www.microsemi.com/existing-parts/parts/44905#docs-specs), or from Tohiba (2SC2878: http://www.toshiba.com/taec/components2/Datasheet_Sync/50/6479.pdf). These used to be used more for low drop switching, before MOS technology became so good.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.