From: Crusty Curmudgeon
Subject: Re: That's really interesting!?!?!?
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Date: Thu, 16 Jan 2003 15:18:59 GMT
NNTP-Posting-Date: Thu, 16 Jan 2003 09:18:59 CST
Volts (RMS) = square root of (power times impedance)
So for 1 watt in a 50 ohm system, the voltage is 7.07 volts AC RMS.
For 50 watts in a 50 ohm system, the voltage is 50 volts AC RMS.
For 500 watts in a 50 ohm system, the voltage is 158 volts AC RMS.
But, there are other things that can make the voltage much higher or lower for a
given power. Things RF engineers study. Impedance mismatch is a very important
factor in RF engineering. It happens... a lot. And if you make decisions about
equipment choices without regarding impedance mismatch effects, well, be
prepared for difficulties.
Impedance mismatch is a difficult thing to explain without getting involved in
standing wave ratios (SWR), transmission line length and transmission line
impedance. Ideally, when the amplififer output impedance is the same as the load
impedance, the formula I gave above holds true. But, if the load impedance is
different, like the 25 ohms you mentioned in your post, well, the answer isn't
simple. It now depends on the transmission line between the load and the
amplifier, the design of the amplifier and where you measure the voltage along
the transmission line. About the only way to simply estimate the voltages are to
calculate the maximum and minimums it could be, with the caveat the amplifier's
power output remains the same. (Certainly no guarantee of that!)
Assuming the load impedance is a pure resistance, and if the transmission line's
characteric impedance is 50 ohms then you can approximate the minimum and
maximum voltages by multiplying the voltages from the above formula by 50/25 and
25/50. So for the 1 watt case, somewhere along the transmission line the voltage
could be as high as 14.14 volts AC RMS or as low as 3.535 volts AC RMS. Where
these voltage maximums and minimums occur is based upon the transmission line's
length, velocity factor, and the frequency of the power from the amplifier.
See how complex the answer to your seemingly simple question becomes when real
world effects are involved? Most people don't want to write a chapter length
answer, especially when, for complete understanding, the answer could (and
does!) fill a book!
On 16 Jan 2003 14:12:29 GMT, Karsten wrote:
>First of all Thanks to all of your spending
>lifetime for questions "where no question's
>If I've a problem, there is always a question
>about it, too.
>But, It's very interesting for me to see how
>different answers can be.
>Jean said: It's impossible to calculate it !!
> I should figure out the driving
> voltage at the output stage.
> There is question at all!!! Just
> convert dbm -> Watt -> Volt.
> Ok, I've a question for you. I've a link
> for a datasheet right here:
> May I ask you, where I can find the
> dbm Statement for the output power???
>All statements are so different that it is still not
>clear to me.
>I'm sorry and thanks all
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spoke at the annual peace assembly at Community of Christ headquarters, May 2002.