So I've seen alot of explanations on two stage voltage amplifiers and how how their gains can be multiplied together. My question is when you are using a two stage common emitter amplifier for current amplification, does the current gain from each stage multiply together or add together?

I am thinking they add together because if their base, emitter, and collector are tied together, their currents must add together to form a linear relation with voltage and power. Could someone verify this?

  • 3
    \$\begingroup\$ Can you link to a diagram of your circuit? \$\endgroup\$ – Oli Glaser Jul 18 '12 at 1:30

If you mean something like below, then the currents add:

parallel transistors

As Alfred mentions, this is not known as a "two stage CE amplifier" though.
With BJTs, the emitter resistors are needed to prevent thermal runaway. This is caused by the fact that as a BJT gets hotter it passes more current. The emitter resistor provides some negative feedback (higher current = more voltage across resistor = lower Vbe to stop runaway) but wastes power.
FETs don't have this problem and can be connected directly, so are better for parallel operation.

  • \$\begingroup\$ Yes this is what I am looking for. If this is what is being done, I understand that it will prevent thermal runaway, will the current gain be the same as if its one BJT? \$\endgroup\$ – user1207381 Jul 18 '12 at 15:37
  • \$\begingroup\$ The overall current sunk from +V will triple in the example above, but will require 3 times the base current, so the \$ h_{FE} \$ (current gain) will be the same, yes. \$\endgroup\$ – Oli Glaser Jul 18 '12 at 16:25

I am thinking they add together because if their base, emitter, and collector are tied together

I take the above to mean that the transistors are connected such that B1 is connected to B2, E1 to E2, and C1 to C2.

Such a connection is not even remotely considered a "two-stage CE amplifier". In a cascade amplifier, the output from the previous stage is the input to the next stage. The overall voltage and current gains are the products of the individual stage's (loaded) voltage and current gains.

The output of a CE amplifier is from the collector while the input is to the base. So, in a two-stage CE cascade, the bases aren't connected to each other and the collectors aren't connected to each other (though the emitters may be, at least for signals).

  • \$\begingroup\$ Agrees with Alfred.. The other two designs by Oli and SH are not CE cascade. CE Cascade in fact is not a current amplifier but now configured as a voltage amplifier with a constant bias. Cascading voltage amplifiers adds the gain in dB or multiplies it in linear scale. for Gain [dB] = 20 log10 [G1 * G2] . I wont add this as answer out of respect for Alfred who identified the correct answer. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 18 '12 at 21:10

What you describe, with bases connected, collectors connected, and emitters connected is not a two stage amplifier. In theory it would indeed add the currents. But the world isn't perfect, and base-emitter voltages of the two transistors may differ slightly. Then the transistor with the smallest base-emitter voltage will draw most of the base current, until its voltage levels with the other one. But one transistor will draw more current than the other one.

A two stage amplifier could be a Darlington:

enter image description here

Here the base current of the first transistor gets amplified, passing a higher current to the base of the second, which amplifies that again. So the total HFE (current amplification) of a Darlington is the product of the two separate transistors' HFEs. While a general purpose transistor may have an HFE in the order of 100, a Darlington may have an HFE of more than 10 000.


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