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Purpose of emitter resistor in a common collector amplifier

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I see a lot of common collector amplifiers with the output taken across an emitter resistor. What is the purpose of doing it this way instead of connecting your load directly to the emitter? I should mention that I am referring to an emitter follower biased by the collector output of a previous common emitter stage.

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4 answers

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First, let's nail down what circuit you are asking about. Your description is somewhat vague, so I picked this interpretation:

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Q2 is the emitter follower stage, driven by the previous common emitter stage of Q1. Your question seems to be why R2 exists.

Some load is needed at DC to pull the emitter of Q2 low when its base is being driven low. R2 is therefore needed whenever the load does not provide this pulldown function. There are various reasons for this. Some are:

  1. The load is capacitively coupled.
  2. The load itself is capacitive or has a substantial capacitive component. R2 and this capacitance form a time constant to bring the output low "fast enough" for whatever the application is.
  3. The circuit is designed to work with a wide range of loads.
  4. The circuit must still have a valid output signal even with no load.

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Yup, you nailed the circuit based on my description. This clears things up alot. Thank you. ‭jm567‭ 25 days ago

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If you are AC coupling to a load then you need an emitter resistor to set the DC operating point and quiescent current of the amplifier. An external AC coupled load cannot do this and, it needs to be done to ensure the amplifier works as intended.

If your load is a resistor and directly connected to the emitter then you may be able to avoid having an extra emitter resistor.

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In most cases, the common-collector stage (emitter follower) is used as a buffer stage (unity gain). It is, therefore, an independent stage, whose mode of operation should not be determined by the next stage (which is to be decoupled).

That means: Even without a connected load this stage should offer a low impedance output voltage. Therefore, it must have an emitter resistor RE, which allows the desired DC operating point.

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So, we can simply say that it behaves as an "ideal" (perfect, constant, steady) voltage source? ‭Circuit fantasist‭ 11 days ago

Not "ideal" - but "idealized" and as "good" as possible with such a simple arrangement. ‭LvW‭ 11 days ago

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In addition to the previous considerations, I will add a few more.

Stages with voltage output (such as a voltage follower) are implemented as voltage dividers consisting of two elements in series "stretched" between supply rails. One of them is "pulling up" and the other is "pulling down" the common middle point (circuit output). Thus the circuit output can both source and sink current to/from the load.

If the stage consists of only one active element, it can only source or sink current. In the considered example, the emitter follower implemented by an n-p-n transistor can only source current. This can create more problems besides those already listed.

If the load contains a positive voltage source or a "pull-up" resistor (as TTL gates that try to inject current to the previous stage) the base-emitter junction may become backward biased... the transistor cuts off... and a breakdown can even occur. The same can happen with a capacitive load (charged capacitor).

So a pull-down resistor is necessary... and it has to be low resistive enough to handle any load. But it does not work well when the input voltage varies widely because the current will also significantly vary (a typical example is an emitter follower driving the output stage of a power amplifier). In this cases, the ordinary ohmic resistor is replaced by a "dynamic pull-down resistor" (current sink).

The best solution is to change the pull-down resistance in an opposite direction to the pull-up change. This idea is implemented by replacing the pull-down resistor with another but p-n-p transistor in "complementary emitter followers" (aka "push-pull stages"). They can both source and sink significant current.

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