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# Power amplifier specification determination

#### dolgaleb

##### Member level 1
I have to determine the specifications of a power amplifier that runs off a battery. The battery is capable of delivering 10 W of power. The power amplifier has to produce 6W of output power. Running the module without the amplifier requires 2W from the other components in the module. How do I determine the DC requirements for the PA?

Not much headroom for inefficiency.
10W - 2W - 6W = 2W max loss or 6/(6+2) *100%= 75% efficiency minimum.
If audio, then Class D is required.
If RF, then Class E is needed.,

Suppose the output is a speaker with resistance 4 ohms. Then going by Ohm's law, 6W through 4Ω calculates as approximately 5V average volts. V=√(PR)=√24.

To apply this as bipolar AC audio can be done by a full H-bridge, whose power supply is greater than 7V, since peaks of 5V rated sine waves are 7V amplitude.

Hi,

since the OP posted in the RF section ... I don´t think it´s audio.

Klaus

Not much headroom for inefficiency.
10W - 2W - 6W = 2W max loss or 6/(6+2) *100%= 75% efficiency minimum.
If audio, then Class D is required.
If RF, then Class E is needed.,
10W - 2W - 6W = 2W max loss or 6/(6+2) *100%= 75% efficiency minimum.

Thanks for replying to my post. However, I don’t understand your analysis.

The battery is capable of 10 W.

The DC consumed w/o the PA is 2W.

Based on this then we have

10W – 2W – PA_DC = 0 or a low number for margin

The PA_DC is what I am trying to determine. You used 6W which is the output power of the PA not its DC consumption.

Also,

Efficiency = (Po – Pin)/DC power ~ (Po)/DC power = 6/ PA_DC where PA_DC is an unknown and we are trying to figure out.

Could you explain the logic behind your calculations?

10W – 2W – PA_DC = 0
Correct.
Split PA_DC into to parts. RF output and dissipation i.e. waste heat
RF output = 6W, what you want, this and the 2W needed for the rest of the system can not be changed.
10W = 2W + 6W + PA dissipation
This allows a maximum of 2W for dissipation
PA total DC input = 8W maximum. RF output = 6W. Amplifier efficiency = 100 * 6/8 =75%. minimum