Transistors configuration.

MrChips

Joined Oct 2, 2009
30,711
Output voltage is the measure of potential difference between two points.
Where you place your reference voltage is irrelevant with respect to AC analysis.

When you use the same reference point for all voltage measurements (as you should) then it becomes relative for all DC analysis.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
The load is LED1 + R2.
That represents your output.
In terms of current and voltage, the output is the current flowing through LED1 + R2.
The voltage is the voltage across LED1 + R2.

Where you connect GND is irrelevant.

Aha so the common collector can also look like this :

1657294001379.png
And like this :

1657294071376.png
And like this? :

1657294100483.png
If yes then this picture (should have info that vout is this LED and R2). Can I say that my Vout on the left is emiter ? Then the config will change ?
1657294129708.png
So I have to determine where is the load and this determines the config ?
 

Jony130

Joined Feb 17, 2009
5,487
Yes, if the output or the load is at the emitter side we have an emitter follower (common collector/wspólny kolektor).
But if the load or the output is at the collector side we have a common emitter (wspólny emiter).
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
Yes, if the output or the load is at the emitter side we have an emitter follower (common collector/wspólny kolektor).
But if the load or the output is at the collector side we have a common emitter (wspólny emiter).
Aaaa cool so if I do something like this :

1657295722814.png

So it is now a common collector. If I'm right here of course.
Then cool and now I think I get it.
 
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Jony130

Joined Feb 17, 2009
5,487
Aaaa cool so if I do something like this :

View attachment 271059

So it is now a common collector. If I'm right here of course.
Then cool and now I think I get it.
But in his case, the emitter is at GND. Thus, Vout = 0V all the time. And clearly in this case the load (LED and the resistor) is at the collector. Thus, CE stage.
 
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MrChips

Joined Oct 2, 2009
30,711
1657296721274.png

Remember, Vout is a voltage measurement between two points A and B = Vout = (VA - VB).
If Vout = A, and GND = B, then Vout = (VA - VB) = 0V, and always 0V.
If Vout = A, and Vcc = B, then Vout = (VA - VB) = (0 - 5V) = -5V. and always -5V.

If Vout = A, and Vc = B, then Vout = (VA - Vc) = 0V -Vc = -Vc
This is still a common emitter circuit because the load is on the collector leg of the transistor.
 

Jony130

Joined Feb 17, 2009
5,487
Also, notice that there is a big difference in the voltage level needed to drive the transistor between CE and CC.

untitled.PNG
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
View attachment 271061

Remember, Vout is a voltage measurement between two points A and B = Vout = (VA - VB).
If Vout = A, and GND = B, then Vout = (VA - VB) = 0V, and always 0V.
If Vout = A, and Vcc = B, then Vout = (VA - VB) = (0 - 5V) = -5V. and always -5V.

If Vout = A, and Vc = B, then Vout = (VA - Vc) = 0V -Vc = -Vc
This is still a common emitter circuit because the load is on the collector leg of the circuit.
Ok but theoretically even if this is 0V this is common collector because Vout is on emiter ?
Because I determine what is the load here/output
 

MrChips

Joined Oct 2, 2009
30,711
One usually designates Vout (or any voltage point) as a node in a circuit.
In fact, any voltage is a difference between two voltages (VA - VB).

Thus in reality, Vout is not just a single point.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
One usually designates Vout (or any voltage point) as a node in a circuit.
In fact, any voltage is a difference between two voltages (VA - VB).

Thus in reality, Vout is not just a single point.
Hmm I don't get it.

So my Vout is wrong ? because it is Vout - GND where Vout node is on emiter while there is no load.
So basically you have mentioned that common collector/emiter/base is where we put our load, so I put it in the emiter :

1657298172376.png

So why is it still common emiter even though I've changed the place of Vout to the emiter.

This one for example is common collector because the load was connected to the emiter :

1657298257740.png

But this is not because ?

1657298172376.png
I think I don't get this part why it is not common collector now.
 

MrChips

Joined Oct 2, 2009
30,711
You can have two loads on the BJT circuit, one on the collector and one on the emitter.
Now you have to decide if this is common emitter or common collector configuration.

If the active load is on the collector (to produce voltage gain) then it is common emitter.
If the active load is on the emitter (to produce current gain) then it is common collector.
 

Thread Starter

Xenon02

Joined Feb 24, 2021
500
You can have two loads on the BJT circuit, one on the collector and one on the emitter.
Now you have to decide if this is common emitter or common collector configuration.

If the active load is on the collector (to produce voltage gain) then it is common emitter.
If the active load is on the emitter (to produce current gain) then it is common collector.
AAaaaaaa

Ok so to summarise it, I'll try to say it with my own words to see if I'm right :D

If there is load in both emitter and collector then I can choose.
If I have only load on collector then it is common emitter
or If there is only load on emitter then it is common collector

Yes ?
 

LvW

Joined Jun 13, 2013
1,752
No. You don't choose.
Which one is the active load?
For my opinion, in such a case (load in the emitter leg as well as in the collector leg) it makes no sense to ask if it is a common collector or common emitter configuration. For which purpose ?
We can say it is a "mixture" or we can say that it is something between - or the best solution:
We should not care and ask: So what? Is it really so important to select a name out of the three basic configurations ?
Remember: Sometimes BOTH outputs are used (are "active") at the same time - phase splitter application.!
 
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