Understanding BJT Emitter Resistance: Explained by Horowitz & Hill

  • Thread starter Thread starter garibaldi
  • Start date Start date
  • Tags Tags
    Bjt Resistance
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
5 replies · 15K views
garibaldi
Messages
1
Reaction score
0
Hey everyone,

Im currently reading Horowitz & Hill and am a little confused on the topic of the intrinsic emitter resistance (re). I understand that this resistance is essentially the dynamic resistance of the base-emitter diode.

I assumed that when analyzing a circuit this resistance should always be placed in series with the emitter. However H&H seem to place this resistance from the emitter to ground in the emitter followers they demonstrate.

In the case of the emitter follower circuit this makes sense because the low re means that the circuit will have a low output resistance as expected. I just don't understand why they placed it to ground.

The circuits in question are attached.
 

Attachments

  • re circuits.JPG
    re circuits.JPG
    10.7 KB · Views: 776
Engineering news on Phys.org
What makes you think that the resistors from emitter to ground shown in the images are re, the intrinsic emitter resistance? They aren't labeled re.
 
I think you mean r'e or pronounced 'R prime E' or 'R E prime'. Summarized, R'e is the AC impedance of the base emitter junction. Normally the output imedance of an emitter follower is Re (actual resistor from emitter to ground) in parallel with [(beta*the base circuit impedance) + r'e].
 
You say:

"Normally the output imedance of an emitter follower is Re (actual resistor from emitter to ground) in parallel with [(beta*the base circuit impedance) + r'e]."

Shouldn't the "(beta*the base circuit impedance)" expression be something more like "(the base circuit impedance/beta)"? Actually, I think the "beta" should be "beta+1".
 
The Electrician said:
You say:

"Normally the output imedance of an emitter follower is Re (actual resistor from emitter to ground) in parallel with [(beta*the base circuit impedance) + r'e]."

Shouldn't the "(beta*the base circuit impedance)" expression be something more like "(the base circuit impedance/beta)"? Actually, I think the "beta" should be "beta+1".

Yes, my bad.
 
It is usually a good idea to have a pull-down resistor on the emitter output to keep the output potential from floating up to the base potential when the external load is not connected. It is absolutely necessary if the external load does not have a dc path to ground.