Common emitter amplifier question (rc || rl)

[cheese9988]

Hi, I have been trying to figure this out and its bugging me. On a
common emmitter amplifier, you have rl (load resistance) and rc
(collector resistor). How are these two in parallel with an ac signal?
It looks more to me like rl is in parellel with the transistor and both
being in series with rc. Can anyone explain this?

 

[JeffM]

How are these two in parallel with an ac signal?

It looks more to me like rl is in parellel with the transistor
and both being in series with rc.
cheese9988

Think *Thevenin's impedence of an ideal voltage supply*.
http://www.google.com/search?&q=Thevenin

 

[Andrew Holme]

Hi, I have been trying to figure this out and its bugging me. On a
common emmitter amplifier, you have rl (load resistance) and rc
(collector resistor). How are these two in parallel with an ac signal?
It looks more to me like rl is in parellel with the transistor and
both being in series with rc. Can anyone explain this?

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.

 

[Fred Bloggs]

Hi, I have been trying to figure this out and its bugging me. On a
common emmitter amplifier, you have rl (load resistance) and rc
(collector resistor). How are these two in parallel with an ac signal?
It looks more to me like rl is in parellel with the transistor and both
being in series with rc. Can anyone explain this?

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?

 

[Pooh Bear]

Hi, I have been trying to figure this out and its bugging me. On a
common emmitter amplifier, you have rl (load resistance) and rc
(collector resistor). How are these two in parallel with an ac signal?
It looks more to me like rl is in parellel with the transistor and both
being in series with rc. Can anyone explain this?

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

 

[John Woodgate]

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

Hi, I have been trying to figure this out and its bugging me. On a
common emmitter amplifier, you have rl (load resistance) and rc
(collector resistor). How are these two in parallel with an ac signal?
It looks more to me like rl is in parellel with the transistor and both
being in series with rc. Can anyone explain this?

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.

 

[Paul Burridge]

Hi, I have been trying to figure this out and its bugging me. On a
common emmitter amplifier, you have rl (load resistance) and rc
(collector resistor). How are these two in parallel with an ac signal?
It looks more to me like rl is in parellel with the transistor and both
being in series with rc. Can anyone explain this?

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.

 

[lemonjuice]

I read in sci.electronics.design that cheese9988
.googlegroups.com>) 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).

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?

 

[Active8]

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


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.

 

[lemonjuice]

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?

 

[Active8]

Oh sure I'd love to be called that.

You just were. Want to try for twice?

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?

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.

Now Mr. Brainy can you answer the guys elementary question above?

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.

 

[lemonjuice]

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.

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.

 

[Active8]

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

*******

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?

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.

Ummm I'm not not satisfied with your answers.

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.

maybe you can't explain
yourself . Why should you bypass something that is there?

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.

Why is the
impedance of the DC voltage source low?

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.

Why is it a short at AC?

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.

 

[Paul Burridge]

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.

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.... ;-)

 

[lemonjuice]

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.... ;-)

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