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What is the need for adding a current source (\$I_\text{E}\$) at the output of an emitter follower topology at the emitter leg of the transistor?

enter image description here

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    \$\begingroup\$ Where did the picture come from? \$\endgroup\$
    – Andy aka
    Jan 20, 2017 at 18:11
  • \$\begingroup\$ Wikipedia, en.wikipedia.org/wiki/Common_collector#/media/… \$\endgroup\$
    – G.Hajj
    Jan 20, 2017 at 18:23
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    \$\begingroup\$ The current source reduces 2HD caused by the transistor being better able to source current via the collector than sink it via RE. \$\endgroup\$
    – user207421
    Jan 20, 2017 at 21:19

3 Answers 3

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When biasing a transistor, you want to Q point to be stable. Since \$\beta\$ varies a lot, you want to design in such a way that changes in \$\beta\$ do not disturb the bias point.

By placing a current source there, \$I_E\$ will not depend on \$\beta\$ (and \$I_C\approx I_E\$) and that way you can keep the transistor in the active region regardless of changes in \$\beta\$.

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what is the need for adding a current source (IE) at the output of an emitter follower topology at the emitter leg of the transistor?

  1. increasing the gain closer to 1x;
  2. setting the bias point for the BJT;
  3. maintaining better linearity towards the ground;
  4. lower production cost in a chip - a CCS is much easier to make than a resistor on a chip.
  5. ...

Basically, very limited benefits, so they are not as widely used.

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The wiki page (on common collector topology) that the picture in the question comes from says this: -

Figure 4: NPN voltage follower with current source biasing suitable for integrated circuits.

In other words, inside a linear integrated circuit, current source biasing is usually preferred over the use of a resistor. In some instances this is beneficial for maintaining stability against temperature changes however, in this instance, the use of a bias current at the emitter is less advantageous. The wiki article also says this: -

sometimes an active current source is used instead of RE (Fig. 4) to improve linearity and/or efficiency1.

At the end of the quote it links to the design of a 20 watt class A amplifier but it's all rather tenuous given the type of output stage it has and the amount of negative feedback it uses: -

enter image description here

So, my personal opinion is that the use of a current source in the emitter is justified inside a chip (because they are easier to fabricate) but on an amplifier like this it might be pushing the boat out a little too far in favour of the audiofools I mean philes.

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  • \$\begingroup\$ The current source makes the class A Amplifier more efficient than it would be with a resistor .The circuit as shown would be less than 25% efficient .If the current source was somehow modulated with the Audio signal then your theory max efficiency would be 50% . \$\endgroup\$
    – Autistic
    Jan 20, 2017 at 20:28
  • \$\begingroup\$ I agree but the wiki article is tenuous in its justifications. \$\endgroup\$
    – Andy aka
    Jan 20, 2017 at 21:08
  • \$\begingroup\$ that circuit has three "ccs": 1) the lower leg of the output stage; 2) the bootstrap circuit; 3) that 6.8k resistor. \$\endgroup\$
    – dannyf
    Jan 20, 2017 at 22:10

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