How output power is calculated
时间：04-04
整理：3721RD
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Hi,
I am trying to understand how power is quoted on rf transmitter. I am referring here to a portable device that would be operated from a battery.
If a manufacturer is claiming a 5W output power, then I'll make few assumptions around it, and would like to validate it. To achieve 5W, I assume the device is expecting a 50 ohm load. So, P=V2/R. So, to achieve 5W that is claimed, and taking into consideration Vrms=Vpp*0.707, Vpp would be SQR(5*50)/0.707 = 22V. I would assume then that the supply voltage of this device should be either 22V or higher. One more assumption I would like to validate is that to drive a 50 ohm load at maximum power transfer, the output impedance of the output stage would be 50 ohm as well, so may be the device would need double the supply voltage of the output stage if he wants to have 22V PP across the output load.
Assume that all my above assumptions are true, which I doubt, does this mean that such devices have ways to generate much higher DC voltage that the available battery voltages being 9V or 12V to achieve such output power levels?
I am trying to understand how power is quoted on rf transmitter. I am referring here to a portable device that would be operated from a battery.
If a manufacturer is claiming a 5W output power, then I'll make few assumptions around it, and would like to validate it. To achieve 5W, I assume the device is expecting a 50 ohm load. So, P=V2/R. So, to achieve 5W that is claimed, and taking into consideration Vrms=Vpp*0.707, Vpp would be SQR(5*50)/0.707 = 22V. I would assume then that the supply voltage of this device should be either 22V or higher. One more assumption I would like to validate is that to drive a 50 ohm load at maximum power transfer, the output impedance of the output stage would be 50 ohm as well, so may be the device would need double the supply voltage of the output stage if he wants to have 22V PP across the output load.
Assume that all my above assumptions are true, which I doubt, does this mean that such devices have ways to generate much higher DC voltage that the available battery voltages being 9V or 12V to achieve such output power levels?
Hi,
This is not correct. Either:
* Vrms=Vp*0.707, or
* Vrms=Vpp*0.707* 0,.5
Not necessarily.
Yes. Either before or after the amplifier. After the amplifier it could be a resonant circuit or some kind of transformer.
Klaus
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