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NNTP-Posting-Date: Wed, 15 Jan 2003 10:17:41 -0600
Date: Wed, 15 Jan 2003 16:17:41 GMT
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Subject: Re: RF-Amplifiers (again)
From: Joe McElvenney
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> A couple of days ago I asked you, where to buy an amplifier.
> Thanks to all of you!! I think I found what i want. but there are still
> some questions about it.
> The amplifiers are specified for certain bandwidth.
> The Output power is related to the output impedance (mostly 50 Ohms)
The 50 ohms specified is most probably the load impedance at which it
delivers the stated power. The average 50 watt RF power amplifier does
not have a 50 ohm source impedance otherwise it could never give
efficiencies greater than 50 percent.
> But, what I need is a statement about the output voltage.
> So, my thinking of calculating the output voltage is:
> 1. R = U/I
> 2. P = R*I^2 ; P=50Watts , R=50Ohm (Output Inpedance)
> 2.a I = 1 Amp
> Inserting I into 1. and R = 50Ohms I get the output voltage:
> U = 50 V (for a high impedance load)
This would be correct for a 50 ohm load!
> If I have a 50 Ohm load the maximum output voltage drops to 25V,
> because of the impedance matching.
> Is that right??
Probably NOT! If the manufacturer specifies a 50 source, then yes it
would be but see earlier in this posting. In common usage the word
'matching' really has two meanings -
Firstly, where you match the load to the source impedance to get
maximum transfer of power and secondly, where the load is chosen to
'match' to a requirement to deliver a specified power with due regard to
efficiency and distortion. Certain items such as test gear and low-power
amplifiers are usually in the first category while power amplifiers are
in the second.
Cheers - Joe
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