Power gain, output power and efficiency of an amplifier

Discussion in 'Homework Help' started by epsilonjon, Apr 2, 2012.

1. epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Hi,

i'm just reading a chapter on basic power amplifiers, and it starts with the class A amplfier. I'm a bit confused by how they're defining output power and efficiency and was wondering whether someone could clarify it for me?

They start by writing the power gain as $A_p=A_vA_i$ where $A_v=V_c/V_b$, $A_i=I_c/I_s$ and $I_c$ is the ac current through the parallel combination $R_c=R_C||R_L$. Then later in the chapter they define the power gain as "the ratio of output power (power delivered to the load) to the input power":

$A_p = P_L/P_{in} = \frac{(V_L)^2/R_L}{(V_{in})^2/R_{in}} = (A_v)^2(\frac{R_{in}}{R_L})$.

Am I right in thinking that these are two different versions of power gain and are not equivalent? The first one takes into account the power dissipated by $R_C$ while the second one does not?

That leads me to the output power. Above they defined output power as "power delivered to the load", but in the next section they write it as $P_{out}=V_{c(max)}I_{c(max)}$ which, for the case where the Q-point is at the center of the ac load line, will be equal to $P_{out}=(0.707V_{CEQ})(0.707I_{CQ})=0.5I_{CQ}V_{CEQ}$. But this second one takes into account the power dissipated by $R_C$, so it is not just "the power delivered to the load"?

Finally, they use this last defintion of output power to obtain the maximum effciency (when the Q-point is at the center of the ac load line) as

$n_{max}=P_{out}/P_{DC} = \frac{0.5I_{CQ}V_{CEQ}}{I_{CC}V_{CC}} = \frac{0.5I_{CQ}V_{CEQ}}{I_{CQ}*2V_{CEQ}} = 0.25$.

But this seems a strange way to define efficiency, since you are including the power dissipated by $R_C$ and not just the power dissipated by the load $R_L$ (i.e. the thing you actually want to power!). Why do it like this?

Thanks for any help!

Last edited: Apr 2, 2012
2. bountyhunter Well-Known Member

Sep 7, 2009
2,498
511
Efficiency is always defined as the ratio of power delivered to the load divided by the total power drawn off the input source that powers the whole circuit.

3. epsilonjon Thread Starter Member

Feb 15, 2011
65
1
When you say "power delivered to the load" do you mean the power delivered to $R_L$ or to $R_L||R_C$?

4. bountyhunter Well-Known Member

Sep 7, 2009
2,498
511
Whatever the load is you are driving. From what you posted, it looks like RL.

5. epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Okay thanks.

So for example, I have this question in the book i'm reading:

"Design a swamped class A power amplifier that will operate from a dc supply of +15 V with an approximate voltage gain of 50. The quiescent collector current should be approximately 500 mA, and the total dc current from the supply should not exceed 750 mA. The output power must be at least 1 W."

It doesn't say anything about the load resistor $R_L$ which the amplifier will be driving, so in this situation would it be fair to assume that the 1W is being delivered to $R_C$ and the swamping resistor? Then if I do connect a load resistor it will decrease the output power delivered to all these, and also decrease the voltage gain?

Thanks again.

6. Audioguru Expert

Dec 20, 2007
10,959
1,252
Why heat your home in summer with a class-A amplifier?
Use a cooler class-AB amplifier instead.

7. Adjuster Late Member

Dec 26, 2010
2,147
302
Perhaps the exercise is designed to illustrate the point that the efficiency of a class A amplifier is low, particularly when the DC path is via a resistor than say an output transformer.

Transformer or choke coupling can raise the efficiency to approach 50%, maximum, but of course the quiescent current is still enormous. Using a dumb resistor (Rc as in your diagram) for the DC path makes things twice as bad.

Even Great Grandpa's tube radio or hearing aid headphones did better than that if they were run straight in series with the anode supply, but transformer coupling or even the lossy resistor load and blocking capacitor required less bravery on the part of the old fellow. They also kept the DC out of the phones, which might avoid deflecting the diaphragm too much in one direction, or in an extreme case overheating the coils. I think the anode currents were normally too low for the latter though.

Last edited: Apr 5, 2012