Gary,
Great piece ... thanks! I always wondered about the actual numbers. Talked
to some goofball at a hamfest a while back who was telling me about his pair
of 3-500Z with 5KV on them, and how they would "put out more" than an 8877.
I said that would be great if he can keep all those watts spewing out on the
frequency he wanted. Got a blank stare from him on that one.
73 - Theo K4MO
-----Original Message-----
From: amps-bounces@contesting.com [mailto:amps-bounces@contesting.com]On
Behalf Of Gary Schafer
Sent: Tuesday, May 16, 2006 12:39 AM
To: amps@contesting.com
Subject: [Amps] How intermod limits your PEP
HOW INTERMOD LIMITS YOUR PEP.
Some interesting notes on PEP and intermod.
With 1000 watts PEP output and 3rd order IM products down -30 db from PEP
(-24 db from from one tone of a two tone test ) the useful PEP output is
only 873.6 watts. The other 126 watts PEP is wasted power.
At -20 db 3rd order products from PEP (-14 db from 1 of two tones) the
useful PEP output power is only 636 watts.
Here?s how it works:
P = E squared / R.
1000 watts PEP into 50 ohms is around 223 volts at 50 ohms.
-30 db down from 1000 watts is 1 watt. 1 watt is about 7 volts at 50 ohms.
With a two tone test signal there are two signals that are 6 db down from
the PEP power. 6 db down is ½ the voltage so each of those signals will have
111.5 volts. (111.5 volts is = ~ 250 watts average power in each signal.)
To find PEP all voltages of all parts of the signal must first be added
together. Then squared and divided by 50 ohms. Thus if we add 111.5 + 111.5
we get 223 volts. That squared and divided by 50 ohms gives us the 1000
watts PEP.
But we also have our 3rd order distortion products present and it is not
just one signal that makes it up but two! We have 1 watt (-30db) signal on
the high side of the original two tones and we have another 1 watt (-30db)
product on the low side of the original two tones. Each of those 1 watt
signals contains 7 volts of signal.
When we figure PEP we need to add all the voltages together first. So with 7
volts in each that makes 14 volts of distortion products.
223 volts in the two tone signals plus 14 volts in the two 3rd order IM
signals gives us 237 volts. Square that and divide by 50 ohms gives us 1123
watts PEP.
If we limit the amplifier power output to 1000 watts PEP (maybe that?s all
it is capable of) we first subtract the distortion products to find useful
PEP output.
Since our amplifier is limited to 1000 watts PEP output we must subtract the
14 volts from the 223 volts that make up the 1000 watt signal. That leaves
209 volts for both of the two tone signals, as high as we can run them.
Square 209 and divide by 50 ohms leaves us with 873 watts PEP of useful
power for our 1000 watt amplifier.
This leaves us with 418.5 watts of average power (½ of PEP) rather than 500
watts average power output that the amplifier would be capable of. Even
though the IM products are only 2 watts average power out the amplifier must
be throttled back to limit the peak envelope power to the amplifier
capability.
At -20 db 3rd order below PEP (-14 db below 1 of two tones) we have 10 watts
in each 3rd order product. That gives us 1432 watts PEP output with only
1000 watts of useful power out.
Or if we limit the amplifier to 1000 watts PEP output the useful power will
be around 636 watts PEP from our 1000 watt amplifier.
Now if we add in the 5th order products things get even worse.
But you can see how even moderate levels of IM products add to the peak
envelope power of an amplifier which can give a very false impression of
what is really useful power.
Since we are limited by the FCC to peak envelope power it is to our
advantage to limit distortion products to the lowest levels in order to not
hit the PEP limit too soon.
73
Gary K4FMX
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