When setting up a transistor as a switch is there any difference between putting the load on the collector or emitter?

As far as I can see the only difference is in calculating the Vbe i.e. calculating what voltage is required to turn the transistor to saturation because of the voltage drop across the load

  • \$\begingroup\$ Actually in the emitter follower configuration (load on the emitter), the transistor doesn't saturate. This can be a advantage if the turn-off speed is important. \$\endgroup\$ Mar 28, 2013 at 14:06

2 Answers 2


The most important difference is that for the common collector (that's the one with the load on the emitter side) you'll need a higher drive voltage. While for the common emitter 0.7 V is already enough, for the common collector the voltage must be 0.7 V + the voltage across the load.

Suppose your load is a 12 V relay, and you also supply 12 V to the collector. If you want to control that by a 5 V microcontroller then that 5 V is the maximum you can supply to the base. The emitter will be 0.7 V lower, that's 4.3 V, which is too low to activate the relay. The voltage can't go higher, because then there wouldn't be no base current anymore. So if the load voltage is higher than the control voltage you can't use common collector.

Also different is how you calculate base current. Suppose you apply 5 V on the base, the load on the emitter's side is 100 Ω and the transistor's \$h_{FE}\$ is 150. Maybe you would expect the current to be 4.3 V/100 Ω = 43 mA. That won't be the case. A base current of \$I_B\$ will cause 150 \$\times\$ \$I_B\$ through the 100 Ω resistor, not \$I_B\$. Therefore the created voltage \$V_E\$ = 150 \$\times\$ \$I_B\$ \$\times\$ 100 Ω. So the resistance seen by the base current is \$ {R_E}^{'} = \frac{V_E}{I_B} = \frac{150 \times I_B \times 100 \Omega}{I_B} = 150 \times 100 \Omega = 15 k\Omega\$.
So that 100 Ω resistor will cause a base current of only \$\frac{5V -0.7V}{15 k\Omega}\$ = 290 µA.

That's why you often won't need a base resistor in common collector configuration. You will need one though if the load consists of LEDs for instance, because contrary to the resistor these will cause a more or less constant voltage drop.

  • \$\begingroup\$ Once again @stevenvh - succinct, loaded with example scenarios and a good dash of experience. \$\endgroup\$ Aug 20, 2012 at 18:17
  • \$\begingroup\$ Minor clarification of the last paragraph: In the common collector configuration (also called emitter follower), the resistor goes in series with the emitter, not the base. The output of the emitter looks like a voltage source, so you need a resistor in series with the LED to make the current predictable, just like when driving a LED from any other voltage source. \$\endgroup\$ Mar 28, 2013 at 14:04
  • \$\begingroup\$ so is there a way to supply a higher voltage to the emitter? \$\endgroup\$
    – samsamm777
    Sep 13, 2020 at 19:34

You're basically correct. The main difference is in how you calculate or generate the voltage/current from the base to emitter.

Normally the emitter would be connected to power or ground rail, a steady voltage, to make things easier but there is no reason why it couldn't be connected somewhere else.

A similar thing applies to MOSFETs as well.


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