Common emitter amplifier question (rc || rl)

Think *Thevenin's impedence of an ideal voltage supply*.

In the AC equivalent circuit (where there is no DC - only AC signals) :

1. The +ve supply (Vcc) is ground.
2. The DC-blocking (aka coupling) capacitor between the collector and RL is a short circuit.
3. The transistor is an AC current source

So, the AC equivalent circuit is a current source, RL and RC - in parallel.

A current source with parallel resistance can be transformed to a voltage source with series resistance; therefore, the AC equivalent circuit can also be drawn as a voltage source in series with RL and RC.

In order to compute the output voltage, you will want to look at the equivalent circuit seen by the transistor at the collector- this is clearly rl||rc. Then when it comes time to compute the voltage developed across rl, you look back in and see rc || transistor. Does that make sense?

It's only a convenience for modelling using equivalent circuits. rc doesn't actually exist as a resistance, it's simply a way of modelling the transistor's sensitivity to collector voltage vs collector current when inserted into the appropriate equation..

In comparison Rbb and Ree are real ( bulk resistance of semiconductor material ).

Graham

I read in sci.electronics.design that cheese9988 wrote (in ) about 'Common emitter amplifier question (rc || rl)', on Sat, 29 Jan 2005:

As far as the signal is concerned, the collector DC supply is at ground potential (or should be).

So both Rl and Rc have one end at the collector and one at ground potential. That means they are in parallel.

It's usual to use R for resistors outside the transistor and r for resistors in its internal equivalent circuit. Helps understanding.

They are effectively in parallel because at signal frequencies the all voltage supply points are ground (there is an 'invisible short' across the battery or power supply at signal frequencies so your Vcc and GND are one and the same). This only applies at signal frequencies, of course, so at DC they're still pretty much isolated from eachother.

On Sun, 30 Jan 2005 09:06:35 +0000, John Woodgate attempted to answer

.googlegroups.com>) about 'Common emitter amplifier question (rc ||

signal?

both

ground

stop wasting bandwidth by repeating his question. The question is why is it at Ground. Either give an answer or go and mow your lawn. Maybe you're Kevin Alyward under a different name?

Maybe he is, but no matter what you call yourself, you're an asshole.

Oh sure I'd love to be called that. Yes I sure remember when I untangled you saying that the gain between 2 amplifier stages was equal to the sum of the gains of each individual stage! Do you need the url on that? Now Mr. Brainy can you answer the guys elementary question above?

Ummm I'm not not satisfied with your answers. maybe you can't explain yourself . Why should you bypass something that is there? Why is the impedance of the DC voltage source low? Why is it a short at AC? Try again.

You just were. Want to try for twice?

F*cking liar. You didn't discuss that with *me*. I knew that over 20 yrs ago - if you're talking gain in dB, anyway.

It's already been answered but I have 3 more answers.

1. If the DC supply is bypassed, it presents a low impedance (short) to ground for AC. So connect the Vcc node to ground and you have your ac equiv.
2. Even if it's not bypassed, the source resistance of the DC supply is so low that it looks like a short to ground.
3. If you short the DC source and open the tranny's current source to get the Thevinin or Norton resistance, you get a low resistance in series with RC (Vcc node shorted to ground) in parallel with RL. The output R, Ro, of the tranny is also in parallel with RL and RC.

Any way you look at it. The DC supply is a short to AC. Short Vcc to ground, and it's at ground.

Let's put in the part you snipped just to remind the group that you're a liar:

F*cking liar. You didn't discuss that with *me*. I knew that over 20 yrs ago - if you're talking gain in dB, anyway.

You didn't even try to worm your way out of that.

Uhmm... ask me if I care. But if the OP's still around, maybe he'll benefit.

I prefer the 3rd answer - less hand waving. The first 2 answers are better suited for someone just learning the basics.

Filter cap. Filters out ripple and supplies transient current that might otherwise change the drop across the supply's Thevinin resistance. Keeps the signal off the rail.

Because if it were high, too much voltage would be dropped across its Thevinin resistance, Rth, and it wouldn't make a very good voltage source. The voltage at the source's terminals would be more dependant on the load it's energizing.

In #2, I said it "looks like a short" Relative to RC which is in series with it, it's a short. In #1, the filter cap is the AC short.

It was a lousy answer of the kind Mike's notorious for. He and Kevin are very similar. They only understand half a question and then proceed to waffle on about the *other* half. Personally, I killfiled 'em both several months ago. No regrets.... ;-)

explain

Yeah you're right. Maybe they are twins. I eliminated K.A some time back. This one is going now.