# Maximum Power Transfer for Coupled Emitter Follower

1. Apr 15, 2013

### engineer_ja

1. The problem statement, all variables and given/known data

For the attached emitter follower circuit, calculate the maximum ac power (Sinusoidal signal) delivered to an attached load.

I have already calculated Rout of the circuit as 19.2Ω, and the voltage at the emitter at operating point (no input signal) as 10.7V (I know these to be correct).

2. Relevant equations
Ohms Law
Power equations

3. The attempt at a solution

Max Power transfer implies a load resistor f Rl=Ro=19.2Ω
The maximum amplitude sine wave that could be applied at emitter before railing is 20V-10.7V=9.3V
This is an rms value of6.58V
A sine wave of this rms voltage across the load would give (6.58V)^2/19.2Ω = 2.25W of Power.

The answer however says "emitter voltage at operating point = 10.7V. this is equivalent to an rms of 7.57V" - I dont understand why we us this voltage.
it then says "only half this appears across the load resistor (3.79V)" - I don't understand why. Isn't the ac voltage across the load equal to the ac voltage at the emitter?
this then gives the correct answer as "(3.79V)^2/19.2 = 0.75W"

Any help in explaining this would be greatly appreciated.

2. Apr 15, 2013

### engineer_ja

Attachment

Image was .tiff so didn't attach, here is a jpeg.

#### Attached Files:

• ###### EF.jpg
File size:
10 KB
Views:
63
3. Apr 15, 2013

### rude man

How did you compute R_out? Not with small-signal analysis I hope.

Anyway, the question makes little sense. The max load in reality would be set by max transistor power dissipation. A 19.2 ohm load is absurd. Even with a beta of 100 tyat's a 1.9K load reflected back to the base,;the tranbsistor would be grossly loaded down. The bias operating point would be grossly upset.
Etc.

4. Apr 15, 2013

### rude man

Presumably the other half apears across the decoupling capacitor.

5. Apr 15, 2013

### Staff: Mentor

It's a bit puzzling. If you have simulation software, I think you should simulate this and run it for various values of load and input. Record the waveshapes to see how well it holds up.

6. Apr 16, 2013

### engineer_ja

Hi, this isn't real life, its a textbook question.

The question started with identifying the class of amplifier and then finding Ib for hFE=50 and VBE=0.7 for no input supplied. this comes out as 105 muA.
Then using small signal parameters calculate small signal gain and voltage. This again gave m G=0.981 and Ro=19.2 Ohm. these are correct by the mark scheme.

It then asks what the max ac power that can be transferred is. Following 'theory', not reality, this suggests matching the impedances and putting 19.2 Ohm as the load resistor (again correct by the markscheme). My problem is then how the answer calculates max power. I don't how they get their theoretical maximum sine wave as 10.7V, or why they say only half the voltage appears across the load (surely the load is modelled in parallel with output so the full ac voltage appears across it?).

7. Apr 16, 2013

### Staff: Mentor

I understand your line of thought, and your method is the way I would have proceeded.

If you model the output as a Norton source feeding a parallel pair of resistances, then, yes, you can say each resistance gets the full voltage though only half the current. Equivalently, you could see the output as a Thévenin source feeding a pair of series resistances, with each getting the full current though only half the voltage. It is two different ways of expressing the same thing.

8. Apr 23, 2013

### engineer_ja

Right, I'll have to work through that but it seems to make sense.

Thanks for the help!