As was discussed in another thread a while ago, shouldn't this be:

Rout=( (R1 || R2 || Rsource)/(beta+1) + Hib ) || Re

A very small difference as we both know, but in the interests of accuracy....

Also, it seems a little strange to me to have a common base h parameter in this expression; I would probably have put it:

Rout=( (R1 || R2 || Rsource)/(beta+1) + re ) || Re

\(r_e\) is what I was referring to with 0.017Ω

 

Also, it seems a little strange to me to have a common base h parameter in this expression; I would probably have put it:

Rout=( (R1 || R2 || Rsource)/(beta+1) + re ) || Re

That's fine too. I'm not sure of the general consensus on this, but I was taught with this notation. I just checked my old school book, and they don't have any qualms about the mixing notation. The formula for the common collector reads as follows:

\( Z_o={{h_{ie}+(ri||Rb)}\over{h_{fe}+1}}=h_{ib}+{{(ri||Rb)}\over{h_{fe}+1}}\)

I think the author's attitude is that once an expression is defined it can be used in any context. But, it doesn't matter to me.

 

That's fine too. I'm not sure of the general consensus on this, but I was taught with this notation. I just checked my old school book, and they don't have any qualms about the mixing notation.

Hmmm. I've never seen that kind of mixing. Just out of curiosity, what is the title and author of the book?

The formula for the common collector reads as follows:

\( Z_o={{h_{ie}+(ri||Rb)}\over{h_{fe}+1}}=h_{ib}+{{h_{ie}+(ri||Rb)}\over{h_{fe}+1}}\)

I think the author's attitude is that once an expression is defined it can be used in any context. But, it doesn't matter to me.

Surely you mean:

\( Z_o={{h_{ie}+(ri||Rb)}\over{h_{fe}+1}}=h_{ib}+{{(ri||Rb)}\over{h_{fe}+1}}\)

 

Hmmm. I've never seen that kind of mixing. Just out of curiosity, what is the title and author of the book?

I made the correction back in my previous post. I was sloppy with the cut and paste.

The book is "Electronic Circuits, Discrete and Integrated" by Donald L. Schilling and Charles Belove, Published 1979

EDIT: By this way, I really like this book, but I find there are mixed opinions about it. Some people love it, and some don't.

 

That is what I suspected. At least that's what I thought dynamic range implies myself. The maximum possible Voltage swing would be at 0V right? So i can swing to -30 V (cutoff) or 30 V(saturated).
Hrm, I'm currently playing with some values, I determined Re to be 17Ω which lead to R1 and R2 being 165 Ωand 172 Ω respectively. This would give me 0.79Ω output impedance at the Quiescent point. This is also neglecting Hie, since the current is so large. (Hie i calculated to be 0.017 Ω).

I still really am not sure if this is correct, since I simply chose some values to make it work out to 0.8Ω output impedance. No way I'm getting 5 Watts this way. By the way, the quiescent current Ie is 1.77 Amps.

Also, thanks for all your help so far, particularly steve

I went back to look at your circuit. It's almost solid black in my browser; I had to adjust the brightness and contrast.

Was I surprised at what I saw!!!

You can't possibly get 5 watts using a 2N2222!

The maximum allowable (continuous) collector current is .8 amp; you need a quiescent current of about 1.5 amps.

The maximum allowable collector-emitter voltage is 30 volts. You've got ±30 volt supplies. Under cutoff conditions, you could have 60 volts across the transistor.

Finally, the maximum total device dissipation is .8 watts. Have you calculated what the transistor dissipation would be when delivering 5 watts to the load?

With an 8Ω load and a β of 100 (and even if a 2N2222 could handle 1.5 amps, β would not be 100 at that current), the input impedance at the base can't be more than 800Ω. Your signal source is going to suffer a substantial loss if the source resistance is 5k.

The 2N2222 is a small-signal transistor; you need a power transistor. You could use a darlington such as the TIP120. It has a 60 volt rating, can dissipate 65 watts (on a heatsink) and has a guaranteed β of 1000 at a collector current of 3 amps.

The output capacitor, C2, will have to be fairly large value and able to carry nearly 1 amp to drive the load to 5 watts.

 

I went back to look at your circuit. It's almost solid black in my browser; I had to adjust the brightness and contrast.

Was I surprised at what I saw!!!

You can't possibly get 5 watts using a 2N2222!

The maximum allowable (continuous) collector current is .8 amp; you need a quiescent current of about 1.5 amps.

The maximum allowable collector-emitter voltage is 30 volts. You've got ±30 volt supplies. Under cutoff conditions, you could have 60 volts across the transistor.

Finally, the maximum total device dissipation is .8 watts. Have you calculated what the transistor dissipation would be when delivering 5 watts to the load?

With an 8Ω load and a β of 100 (and even if a 2N2222 could handle 1.5 amps, β would not be 100 at that current), the input impedance at the base can't be more than 800Ω. Your signal source is going to suffer a substantial loss if the source resistance is 5k.

The 2N2222 is a small-signal transistor; you need a power transistor. You could use a darlington such as the TIP120. It has a 60 volt rating, can dissipate 65 watts (on a heatsink) and has a guaranteed β of 1000 at a collector current of 3 amps.

The output capacitor, C2, will have to be fairly large value and able to carry nearly 1 amp to drive the load to 5 watts.

Wow, ok Still failing on my part. The Design should be an unknown generic Transistor. Not the 2N2222 in paticular. Again I used it when i did a SPICE simulation.
Sorry about that.

 

Also, the assignment continues in the next part by asking me to design an inverting amplifier with Gain of -4. this will send it's output to the power amplifier just designed.

The part after that, part 4, asks us to replace the single transistor in the inverting amp with a darlington pair. So i guess you guys really want the whole assignment questions.

 

So, I'm not sure if we came to a conclusion on whether it is ok to up the bias voltage. if doing so allows me to get a max of 5 Watts across the load. The dynamic range would be changed, but we can get 5 Watts.

Is there any other way to satisfy the 5 Watt criterion without knowing the input source signal?