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I have two electronics textbooks and for some reason, both of them just gloss over configurations that are not "Common Emitter". Regarding the emitter follower, I understand that there's no voltage gain. It was very easy to see why. But then the author just says, it has a very high current gain. Even though he didn't bother to show why, I went on to calculate the ratio and indeed there was a large current gain. The problem is my mind is tuned to "Common Emitter" circuits where the base current is multiplied by beta. This current is large and so the output voltage is large. Hence the voltage gain. That is quite understood. But when it comes to the emitter follower, where is the current "hiding" because I read that it can supply loads that require large current. I know I'm confusing things. That's why I am asking for an explanation.
Thanks.

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    \$\begingroup\$ A common emitter is not an emitter follower. A common emitter has a voltage gain, an emitter follower is a buffer (i.e. voltage gain is approximately 1). Which circuit are you talking about? \$\endgroup\$ – Null Sep 22 '14 at 3:55
  • \$\begingroup\$ I'm sorry. I am talking about an emitter follower. \$\endgroup\$ – medwatt Sep 22 '14 at 4:28
  • \$\begingroup\$ A transistor's base current \$I_{B}\$ is always multiplied by \$\beta\$ to give \$I_{C}\$. This is a high current gain since \$\beta >> 1\$ usually. But the voltage gain for the emitter follower is approximately unity despite the current gain. Is the difference between high current gain and unity voltage gain what is tripping you up? \$\endgroup\$ – Null Sep 22 '14 at 4:42
  • \$\begingroup\$ What I'm trying to ask in short is how can the circuit supply large currents. What difference does it make if a circuit supplies a large current instead of a large voltage ? What I mean why should a load care as long as the power is in the either of them. \$\endgroup\$ – medwatt Sep 22 '14 at 4:58
  • \$\begingroup\$ Transistor does not supply current or power, that comes from an external source. Emitter follower configuration emitter voltage tends to follow the base voltage (minus nominal 0.7 Vbe) and collector current is as much as needed to sustain this bias condition. \$\endgroup\$ – MarkU Sep 22 '14 at 5:09
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Let me try a short - and more descriptively - explanation without formulas (which you already know): The input signal causes a signal output current Ic. Now it is important how this current is translated to voltage:

1.) Common emitter: There is a collector resistor which produces a corresponding output voltage Vc, which does NOT react back to the input. As a consequence, we can have a rather good voltage gain.

2.) Common Collector: Now, there is a emitter resistor which also "translates" the output current (forget that Ie is little larger than Ic) into a voltage Ve. However, this voltage strongly reacts back to the input because it is a part of the current-controlling quantity Vbe=Vb-Ve. More exact: It follows the input voltage at the base Vb - and, thus, does not allow any voltage amplification. This is the result of negative feedback.

(May I add the following - although I am aware that not all forum members are happy? The explanation under 2.) clearly shows that the BJT is a voltage-controlled device and that the base current Ib is not the controlling quantity. The working principle of an emitter follower cannot be explained using the current-control model.)

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Here is a basic emitter follower:

schematic

simulate this circuit – Schematic created using CircuitLab

The input supplies the base current \$I_{B}\$ through its source resistance \$R_{S}\$ but the \$\beta\$ multiplied \$I_{C}\$ is supplied by the \$V_{CC}\$ power supply. That's how there is a large increase in the current. \$I_{B}\$ and \$I_{C}\$ combine to form \$I_{E}\$, which flows into the emitter resistor and the load. Although \$I_{E} = (\beta + 1)I_{B}\$, there is not necessarily an increase in power.

If the load has a high input resistance \$R_{L}\$ and \$R_{L} >> R_{E}\$ then almost all of \$I_{E}\$ flows through \$R_{E}\$ and very little to \$R_{L}\$. Since the emitter follower has unity voltage gain the power \$P = VI\$ delivered to the load decreases -- the extra current (and power) is dissipated in \$R_{E}\$. Usually an emitter follower is simply buffering a voltage signal from one circuit to the next, so the current dissipated in \$R_{E}\$ isn't a problem. An emitter follower is not a power amplifier so a power gain is not necessarily expected.

In the case where \$R_{E} >> R_{L}\$ then most of the current flows into the load rather than \$R_{E}\$ and there can be a power increase. But this extra power delivered to the load again comes from \$V_{CC}\$, not from the emitter follower transistor itself.

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  • \$\begingroup\$ Thanks for taking your time writing this. I really know all of this stuff but I just have a feeling that I don't fully understand it. Thanks !! BTW, do you know of a good book that is very well rounded in terms of solid explanations and mathematical depth. Neither of the books I have use calculus, so things tend to get messy quickly. \$\endgroup\$ – medwatt Sep 22 '14 at 5:45
  • \$\begingroup\$ @medwatt, I like "Art of electronics" as a text. But there is not a whole lot of mathematics. I sometimes find that too much math just makes things harder to understand. The best thing is to build some circuits and see what happens. \$\endgroup\$ – George Herold Sep 22 '14 at 13:02
  • \$\begingroup\$ @medwatt "Analysis and Design of Analog Integrated Circuits" by Gray, Hurst, et. al is a classic that I, my college classmates, and many of my coworkers have used. It is fairly rigorous in that it tends to use more accurate hybrid-\$\pi\$ models (i.e. that include \$r_{o}\$) and it does use calculus where needed. It might be worth a look for you. \$\endgroup\$ – Null Sep 22 '14 at 13:44

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