# Transistor, AC amplification

Discussion in 'General Electronics Discussion' started by newuser, Dec 16, 2012.

1. ### newuser

9
0
Dec 16, 2012
Hi,

looking at this

circuit i dont get how setting the base-emitter voltage to half of vcc (lets say vcc=12v, so the base-emitter gets 6v) enables the transistor to also amplify the negative ac waves.

as far as i know a transistor starts working at about 0.7v, now if you put 2/3/4/5/6 volt at its input i dont get why you would do that - shouldnt make a difference.
isnt Ib the current thats interresting. i mean doesnt Ib open/close the collector-emittor line more/less?

#### Attached Files:

• ###### CommonEmitter.gif
File size:
6.4 KB
Views:
309
Last edited by a moderator: Dec 16, 2012
2. ### Harald KappModeratorModerator

12,465
2,987
Nov 17, 2011
Let's start with R1 and R2. These resistors form a voltage divider. Without any load the voltage at the midpoint were Vb=12k/(12k+150k)*24V = 1.78V. The schematic says 1.67V because some current from R1 is used as base current for the transistor, thus increasing the voltage drop across R1.
So let's assume 1.67V is correct.
The 0.7V you're talkinmgh about is the base-emitter voltage Vbe. However, in this circuit there is a resistor Re from emitter to GND. The emitter current creates a voltage drop across this resistor. The schematic states that the emitter voltage is 1.17V. That makes Vbe=1.67V-1.17V=0.6V That's reasonable since Vbe=0.6...0.8V for a typical transistor (note that this includes your 0.7V).

Re is used to stabilize the circuit. Assume there is no Re: Vbe is temperature dependent with ~-2 mv/degree C. That means that the base current follows temperature and so does the collector current. The whole circuit would be rather temperature sensitive.
Enter Re. If the Vbe should decreas, Ibe will rise. An increase in Ibe means an increase in Ie and thus an increase in the voltage drop across Re. Thus Vbe is decreased, Ibe is reduced etc. The circuit becomes much more stable.

This is called negatve feedback and is often used to stabilize any system (mechanical, electrical, biological etc.).

Sonce the negative feedback will redice the AC gain, too, you will often find this circuit modified by a capacitor in parallel to Re. The capacitor is a short circuit for AC and thus reduces teh negative feedback.

Harald

3. ### newuser

9
0
Dec 16, 2012
In the first equation you probably switched 12k and 24v, is that right?

4. ### BobK

7,682
1,689
Jan 5, 2010
The equation is not parenthesized incorrectly, it should be:

(12K / (12K + 150K)) * 24V

This is just the equation for 2 series resistors across a voltage source, i.e. a voltage divider. The voltage at the juncion is the ratio of the lower resistor to the sum of the two resistors times the voltage source.

I believe the question you were asking is how can this amplify AC when the output is always positive? Well, AC is really more generic than you think, it is any varying voltage, whether positive, negative or both. When it is always positive we would call it AC with a DC offset (1/2 V+ or 12V in this case).

In any case, the amplifier should have a capacitor between the collector and the output. This will make the output pure AC, going above and below the ground level. How? Remember that the voltage at the collector is varying. Lets say at some point it is 15V. The capacitor charges up to have 15V on one side and 0V on the other. Then when the output changes to 9V, what is the other side of the capacitor? It will still be 15V lower then the side connected so it will go to -6V. So even though the collector is always at a positive voltage. So the other side of the output capacitor can swing both positive and negative.

Bob

5. ### Harald KappModeratorModerator

12,465
2,987
Nov 17, 2011
I wrote:
which is completely correct, considering standard mathematical rules.

The second pair of parenthesis introduced here is correct, but not necessary. A second pair of parenthesis would have been necessary had I intended to write
But I didn't mean to.

6. ### BobK

7,682
1,689
Jan 5, 2010
Harald,

Sorry, you are right. The text wrapped in the middle of the expression, which made it hard to read.

Bob

12,465
2,987
Nov 17, 2011
No problem

8. ### newuser

9
0
Dec 16, 2012
looks like i have a lot to learn, please forgive me if i ask some random questions that dont really look like they could help with the first.

if Re is to make the circuit more temperature stable, wouldnt the same happen without Re?
I mean, the volate drop that gets bigger across Re when the transistor heats up is also bigger on R1 where Ib goes through right?