External AC Emitter Resistance Formula

1. Jun 12, 2014

?n0t_A_nUmb3R?

I have worked on an example earlier today. It's the section on BJT Amplifiers.
I have been studying the sub-section on Common-Collector Amplifier. I've come across a rather peculiar formula, which has not been mentioned anywhere in my prescribed textbook besides a worked example.
Here is the formula:
Re = RE || RL

I am pretty much confused about the application of this formula.
Can the formula be used in a Common-Emitter circuit?

Last edited: Jun 12, 2014
2. Jun 12, 2014

willem2

The emitter resistance and the load resistance are actually parallel in a common collector amplifier, so it makes sense to compute the equivalent resistance, for example when you want to find the power dissipation in the transistor, or the input impedance.

In a common emitter circuit the load resistance isn't parallel to the emitter resistance

3. Jun 12, 2014

?n0t_A_nUmb3R?

Hi There,

I've just flipped back to the section on common emitter and saw that there was no load resistor with the bypass capacitor. It was my error.

I did manage to find the Re formula. It was in small writing with not much detail.

Another thing I would like to mention. In a common collector amp, won't the bypass capacitor effectively short the emitter to ground?

4. Jun 13, 2014

Staff: Mentor

No, because in the common collector amplifier there is no bypass capacitor in the emitter, the emitter is where you take the output! You don't want to short it to ground!

I think you meant in the common emitter amplifier? If so ... at AC, yes, when the bypass capacitor does its job. But at DC, no.

That's why it is named the common emitter, precisely because the emitter is grounded (at the frequencies of interest).

Last edited: Jun 13, 2014
5. Jun 13, 2014

?n0t_A_nUmb3R?

I am aware that at Common Emitter, the bypass capacitor shorts the emitter to ground.

But why is the emitter not shorted in the Common Collector?
Does the Load Resistor play a fundamental role?

I'm just asking out of curiosity.

6. Jun 13, 2014

Staff: Mentor

7. Jun 13, 2014

?n0t_A_nUmb3R?

Last edited: Jun 13, 2014
8. Jun 13, 2014

Staff: Mentor

There are any number of circuits there. Can you screen-capture just the one you are pointing to and attach it to your post here as a JPEG?

While paying for expensive cellular data, I hesitate at downloading a whole chapter when all that is of interest is one figure.

Last edited: Jun 13, 2014
9. Jun 15, 2014

?n0t_A_nUmb3R?

10. Jun 15, 2014

Staff: Mentor

11. Jun 20, 2014

?n0t_A_nUmb3R?

I wasn't planning on doing anything. I just wanted to know what role the Load Resistor plays.
I am aware that if the Load Resistor is removed, the bypass capacitor shorts the emitter.

12. Jun 21, 2014

Staff: Mentor

Ah, I think I see the problem ......

There are a few misunderstandings. Let's see if we can clear them up.

When you use this type of circuit, you don't solder a resistor RL onto the circuit board. That load resistor RL shown in the schematic represents something that the emitter follower is to send a signal to, and it may be a small loudspeaker, or a small light-bulb, or a power amplifier's input, or anything else that you wish to drive with an AC signal from a low impedance. For the purpose of design calculations, RL is assigned a value equal to that of the input impedance of the following stage or device.

The capacitor you see connected to RL is not an emitter bypass capacitor, and it never will be. That is not its purpose. It is a coupling capacitor; it's there to deliver the signal from the emitter to the load.

If the load were to be removed, this would mean we'd show RL as an open circuit, so the coupling capacitor would end in mid-air. It would be connecting the emitter nowhere, and definitely would not be shorting the emitter to ground or anywhere else.

Does that help?

Now, a question for you. Why is a coupling capacitor inserted between the emitter and the load? Why don't we connect RL directly to the emitter, and save the cost of the capacitor?

Last edited: Jun 21, 2014
13. Jun 23, 2014

?n0t_A_nUmb3R?

It makes perfect sense now. It was not explicitly stated that the capacitor is a coupling capacitor. I assumed that it was a bypass capacitor.
But I was aware that the load resistor could represent an external device.

To answer your question, we need a coupling capacitor because current tends to take the path which has less resistance, in this case from emitter to ground (assuming that the emitter capacitor is removed).
If we added the emitter capacitor, the capacitor would be able to produce the signal required to power the load.

14. Jun 23, 2014

Staff: Mentor

0/10, too vague to earn any marks. :uhh:

Need to think about it some more.