In a CB Amplifier, the Base is common to both Input and Output. Why is Input applied at the Collector and Output at the Emitter?

Could it have been the opposite?

Similarly, for CC and CE amplifiers why are the chosen Input and Output terminals selected?

  • \$\begingroup\$ Possible duplicate of NPN transistor CE or EC? This question is about why a transistor still works when wired backwards, and answers which explain how a transistor is logically symmetric, but not made to work as well when reversed. Sorry about the close vote, but I will give you an upvote. :) \$\endgroup\$
    – Kaz
    May 6, 2013 at 16:36
  • \$\begingroup\$ @Kaz I kept looking and found another link. To confirm I understood the content correctly I'm summarizing. Using the BJT Amplifier and interchanging I/P & O/P, the amplifier will work but with a reduced Beta. The BJT is biased so EB is forward biased and CB is reversed biased i.e. BJT is working in Forward-active mode and not the Reverse-active mode. This is cause Emitter is more doped than the collector \$\endgroup\$
    – user23564
    May 6, 2013 at 17:39
  • \$\begingroup\$ Do you have a circuit diagram to support your question? \$\endgroup\$
    – jippie
    May 6, 2013 at 20:14
  • \$\begingroup\$ !Wiki CB Amplifier. Now interchange the Vin and Vout. \$\endgroup\$
    – user23564
    May 7, 2013 at 3:07
  • \$\begingroup\$ CB amplifiers can have lots of different topologies inside. Assuming any port is directly connected to a collector or emitter makes no sense and is quite likely to be invalid. Also, when using a CB amplifier you have to be careful not to exceed the maximum allowed power into the antenna, which I vaguely remember is something like 5 W. \$\endgroup\$ Apr 2, 2014 at 19:44

1 Answer 1


Radio Frequency amplifiers quite often use common base circuits and I believe the main reason is because "miller capacitance" does not have the same detrimental effect.

On a normal common-emitter configuration, the input is fed to the base and the output is at the collector but, internally the capacitance between collector and base acts as negative feedback and can reduce gain.

To combat this, if the base is held at a fixed potential to ground (as per in its normally biased state), and this is supplemented by decent capacitance to ground, the miller capacitance is effectively shunted to ground. The down-side is that feeding an input to the emitter requires a stronger drive voltage because the emitter input impedance is lower.

It is lower because the emitter fed input has to also "handle" collector currents as well as the little bit of emitter-to-base current normally needed for amplification.

Consider also the differential amplifier: -

enter image description here

This is another circuit that uses the emitter as an input. On this occasion the input to the emitter comes from the emitter (outputting) of the other BJT. In fact, both emitters are simultaneously inputs and outputs.

  • \$\begingroup\$ And the second stage in a cascode is effectively a common base, for the same reason : to disconnect the miller capacitance. A cascode is effectively a semiconductor implementation of a pentode... \$\endgroup\$
    – user16324
    May 6, 2013 at 16:52
  • \$\begingroup\$ @BrianDrummond good call. I'd forgotten about that (which is bizarre because I've recently had to re-design one at work!!!) \$\endgroup\$
    – Andy aka
    May 6, 2013 at 16:58
  • \$\begingroup\$ In addition to the small-signal improvement in bandwidth, a cascode also improves linearity in a large signal context. I remember reading this article in Audio Magazine back in the late 70s: passdiy.com/project/amplifiers/cascode-amplifier-design \$\endgroup\$ May 6, 2013 at 21:35
  • 1
    \$\begingroup\$ @AlfredCentauri that is one of those articles that should be of biblical significance to BJT designers \$\endgroup\$
    – Andy aka
    May 6, 2013 at 22:15
  • \$\begingroup\$ You are right. Basically, the tail source keeps the current stiff, so yes, each input transistor acts as a driver for current, where the opposite transistor looks like a common base. (Which is literally true when we pin one base to the ground.) It's like they are cascoding each other. \$\endgroup\$
    – Kaz
    May 6, 2013 at 23:52

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