# BJT Emitter Resistance

1. Mar 12, 2009

### garibaldi

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 dont understand why they placed it to ground.

The circuits in question are attached.

#### Attached Files:

• ###### re circuits.JPG
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2. Mar 12, 2009

### The Electrician

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.

3. Mar 12, 2009

### Averagesupernova

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].

4. Mar 12, 2009

### The Electrician

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".

5. Mar 13, 2009