Power Ratings of RF Amplifiers

[nauman]

Hi all

I am studying RF amplifiers data sheets and there is some confusion i am facing in its specifications. One such amplifier claims Power Output at 50 ohm load with 20% duty cycle to be 2 KW. However, the minimum gain it specifies is 63 dB and maximum input it claims to be 10 dBm (around 0.707 Vrms for 50 ohm ). Using this gain and input, maximum output voltage is around 990 Vrms @ 50 ohm load. Using simple power formula i.e. V^2/R, the output power comes to be around 19 KW @ 50 ohm load?.

Can anyone kindly tell me where i went wrong in my calculations?

Thanks

 

[tech99]

I think the 10dBm max input power is what the amplifier can survive and not its input power for 2kW.

 

[nauman]

10 dBm is maximum RF input drive.

 

[davenn]

Using this gain and input, maximum output voltage is around 990 Vrms @ 50 ohm load. Using simple power formula i.e. V^2/R, the output power comes to be around 19 KW @ 50 ohm load?.

Can anyone kindly tell me where i went wrong in my calculations?

show ALL your working

and don't forget the 63 dB gain is a power gain NOT a voltage gain

 

[nauman]

Hi
Thanks for help. You mean if i apply 0.223 Vrms (corresponding to 0dBm) to power amplifier input, i will not necessarily get around 315 Vrms output from this power amplifier at 50 ohm load?
Kindly educate me about the difference b/w voltage and power gain in contest of what i mentioned above.
Thanks

 

[tech99]

Hi
Thanks for help. You mean if i apply 0.223 Vrms (corresponding to 0dBm) to power amplifier input, i will not necessarily get around 315 Vrms output from this power amplifier at 50 ohm load?
Kindly educate me about the difference b/w voltage and power gain in contest of what i mentioned above.
Thanks

If the amplifier has a gain of 63dB then for 2kW output the required drive will be as follows:-
2kW = +33dBW = +63dBm.
Drive power = output power - gain = +63 - 63 = 0dBm
Full output is obtained with 0dBm drive but the amplifier can servive over driving up to +10dBm. In these circumstances it will be driven into saturation.

 

[nauman]

2kW = +33dBW = +63dBm.

May i ask what is the reason to consider 63 dB as 63 dBm? As clearly indicated in data sheet, it is dB only (even in case of control voltage, it mentions 10 dB variation). Only in case of RF drive, it mentions dBm specifically.

 

[tech99]

May i ask what is the reason to consider 63 dB as 63 dBm? As clearly indicated in data sheet, it is dB only (even in case of control voltage, it mentions 10 dB variation). Only in case of RF drive, it mentions dBm specifically.

Sorry, I think I have confused you. The output power is 2kW which is 10 log 2000/1 = 10 x 3.3 = 33dBW. This is 63dBm. The gain is 63dB.

 

[Tom.G]

@nauman The dB is a way of representing a ratio between two numbers. When talking about Gain, no units are required for the dB scale because Gain is already a ratio. When talking about Power levels the dB is suffixed with, as in these posts, with a 'w' for Watts or an 'm' for milliWatts. That gives you a base to reference the dB value to.

As @tech99 pointed out, when referring to a Power ratio the formula for dB is 10 log (<Power_in_question>/<reference_Power>). That's where he got the

10 log 2000/1 = 10 x 3.3 = 33dBW

If you want the Power (2000W) referred to 1 milliWatt (0.001W), you obtain your output power as 10 log (2000/0.001) = 63dBbm.

Since the amplifier has a Power gain of 63dB (or 2 000 000), the input Power needed is 63dBm - 63dB = 1milliWatt... or equivalently, (<output_Power> /<Power_gain>) = 2000/2,000,000 = 1milliWatt.

Wikipedia has a long entry on the decibel at: https://en.wikipedia.org/wiki/Decibel
There is also a more involved article at: http://www.animations.physics.unsw.edu.au/jw/dB.htm

Extra Background Info
BTW, the 10 log(...) ratio works for Power ratios. For Voltage or Current ratios the math works out to 20 log(...). The Bel was originally defined by Alexander Graham Bell, the guy that invented the telephone, as an easy way to represent the apparent loudness of sound. The response of the Human ear is logarithmic, hence the log term in the formula. When the concept started being used for Electronics, etc., it was felt that the Bel was too big a unit for convenient usage, so the deciBel was introduced as 1/10 of a Bel. Eventually the capitalilized 'B' became lower case in the word, but the proper abbreviation is still considered to be 'dB'. In common usage, even this nicety is being lost and we commonly see 'db' these days.

Hope this helps!
Tom

 

[nauman]

Sorry, I think I have confused you. The output power is 2kW which is 10 log 2000/1 = 10 x 3.3 = 33dBW. This is 63dBm. The gain is 63dB.

Thanks for clarity

 

[nauman]

@nauman

If you want the Power (2000W) referred to 1 milliWatt (0.001W), you obtain your output power as 10 log (2000/0.001) = 63dBbm.

Since the amplifier has a Power gain of 63dB (or 2 000 000), the input Power needed is 63dBm - 63dB = 1milliWatt... or equivalently, (<output_Power> /<Power_gain>) = 2000/2,000,000 = 1milliWatt.

Tom

Thanks