# Transistor misunderstanding....

Discussion in 'General Electronics Chat' started by cheddy, Dec 1, 2007.

1. ### cheddy Thread Starter Active Member

Oct 19, 2007
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I thought I had a basic understanding of transistors. According to my SPICE simulation I was wrong.

Can somebody please explain to my why the voltage across R4 is 1.4V and not more?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
What you have constructed is a common collector stage (A.K.A. emitter follower).

The value of the voltage at the emitter should be around 0.6 to 0.7 volts less than the voltage applied to the base. That means that your value of 1.4V is in the appropriate range.

Link to AAC's ebook Common collector tutorial. The information at the link provided should help clear thing up for you. If not then feel free to come back with follow-up questions.

hgmjr

3. ### cheddy Thread Starter Active Member

Oct 19, 2007
87
0
I am just missing something from my understanding here. How is the current leaving the emitter the sum of the current from the base and the collector but the voltage across the resistor is only equal to the base voltage minus the voltage drop of the P-N junction between the base and emitter.

My confusion is how come the collector voltage doesn't effect the emitter voltage.

again, there is something I am just not seeing.

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Keep in mind that the primary controlling parameter in a transistor is the base current. The collector voltage in the common-collector is fixed by the voltage applied. The collector current and the emitter current are related to the base current by beta of the transistor.

hgmjr

5. ### cheddy Thread Starter Active Member

Oct 19, 2007
87
0
Explain a little more when you said "The collector voltage in the common-collector is fixed by the voltage applied." please.

6. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Emitter current is 1.4V / 420$\Omega$ = about 2.9mA.

Thus the transistor effective resistance (emitter to collector) is more than (12V / 2.9mA) - 420$\Omega$ = 3718$\Omega$

7. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I could not have put it better myself. Good reply thingmaker3.

hgmjr

8. ### cheddy Thread Starter Active Member

Oct 19, 2007
87
0
So in the end we are just letting the collector current source take over and do the work so the source at the base doesn't have to?

If we took out the transistor completely and just connected the load resistor to the power supply at the base we would have a greater voltage and current than we do in the amplifier.