12
\$\begingroup\$

Due to the base current, in a common emitter circuit the emitter current is a bit higher than the collector current:

\$ I_e = \dfrac{1+\beta}{\beta} I_c \$

I was wondering how this is for a phototransistor, like in an optocoupler. I would expect emitter and collector current to be equal because there is no base current. But a photodiode can create a current with zero volt across it, so it seems the photons can create electrons. So is \$I_e = I_c\$ for phototransistors?

\$\endgroup\$
6
\$\begingroup\$

It must be; there is nowhere else for the charge to go.

Photons don't create electrons; they create electron-hole pairs. The electron is accelerated to the collector, the hole to the emitter (for NPN phototransistors; opposite for PNP ones).

In a photodiode, some of the electrons and holes end up on opposite sides of the barrier and can't easily recombine until (a) they build up a voltage of about 0.5V to attract them back across or (b) you provide an easier external current path. This is turning into quite a big industry...

\$\endgroup\$
3
  • \$\begingroup\$ What is turning into quite a big industry? Shoveling electrons? :-) \$\endgroup\$ – Anindo Ghosh Feb 8 '13 at 10:32
  • 2
    \$\begingroup\$ shoveling electrons through photodiodes! \$\endgroup\$ – user_1818839 Feb 8 '13 at 10:38
  • \$\begingroup\$ I don't get it. Photodiodes can most certainly generate a current, even at zero volts if reverse biased appropriately, and thus provide a base current in a phototransistor. \$\endgroup\$ – Scott Seidman Feb 13 '13 at 14:19
4
\$\begingroup\$

Of course \$ I_e = I_c \$ for a phototransistor.
Anything else would violate Kirchhoff's current law.

\$\endgroup\$
2
\$\begingroup\$

Both of the other answers are right, but I suspect you're getting a little confused. Yes, there is energy (photons) going into the base, and yes those photons create a current in the base which is acted upon by the Hfe of the transistor to control the C/E current. That base current will be reflected in an increased C/E current flow because it has to flow through there. But that increase will be tiny in comparison to the main current flow. There isn't current flow from your circuit into the base, the photo-transistor is still a two terminal device with respect to your circuit.

\$\endgroup\$
0
\$\begingroup\$

it seems the photons can create electrons

Photons don't create new electrons. Photons make the already existing electrons move. In this particular case, when a photon collide with an electron, photon increases kinetic energy of the electron. Then, the electron moves from base to emitter. This physical phenomenon creates a pseudo base current which turns the transistor on.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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