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[AMPS] Power Handling of Resistors

To: <amps@contesting.com>
Subject: [AMPS] Power Handling of Resistors
From: jono@enteract.com (Jon Ogden)
Date: Fri, 10 Mar 2000 07:26:49 -0600
on 3/10/00 12:29 AM, Ian Roberts at itr@nanoteq.co.za wrote:

> Jon Ogden wrote:
>> Capacitors discharge at an exponential rate and cannot deliver an
>> instantaenous spike.
> 
> Speaking from memory, I recall that a 1 micro size cap can provide a
> peak current of 20 amps, therefore 32 micros can provide...?

It's still not an INSTANTENOUS function.  A cap doesn't discharge all of
it's energy at once.  And 20 amps from 1 uF sounds like an awful lot.

Let's see....

Let's say we have 32 uF at 4000 volts as in the example all along.

The stored energy is:

J = (32E-6 * 16,000,000)/2
J = 512/2

J = 256 Joules

 
Now,  if 1 uF can deliver 20 amps then at 32 uF, we have 640 Amps peak
capability.

So at 4000 Volts and 640 Amps, we get 2,560,000 Watts.

Is the typical ham amplifier power supply really capable of delivering a
peak power of 2.56 MegaWatts?

Knowing that we have 256 Joules of available energy and that a Joule is a
Watt second, this means that the cap would be discharged in 0.39 uSec.

Let's say we have connected the supply output to a 50 Ohm resistor connected
to ground.

We know that capacitor discharge is expressed by the formula:

V = E *(e^(-t/RC)) 

where E is capacitor voltage and e is the natural logarithmic base (2.718).

Let's see how much voltage our 32 uF capacitor bank has left after 0.39
uSec.

V = 4000*(e^(-0.39E-6/(50*32E-6)))

V = 3999.025

So....I don't think a 1 uF cap can provide a peak current of 20 Amps.

The amount of time for the capacitor bank to discharge to 0.5 volts (since
you can't mathematically take the log of 0) is:

0.5 =4000* e^(-t/RC)

1.25E-4 = e^(-t/RC)

-8.987 = -t/RC

8.987 = t/(50*32E-6)


t = 14.3792 milliseconds.

There for we have approximately 256 Joules of energy released in 14.3792
mSec.  This would be 17,803 Watts delivered to the resistor in that period
of time.

A lot less than 2.56 MW.

We also know the resistor can handle 64 kW at 5 msec.  At 15 msec (close to
the 14.3792 millisecond number) the power handling is 64kW/3 or 21.333 KW.

The resistors will NOT be destroyed.

There!

73,

Jon
KE9NA

-------------------------------------
Jon Ogden
KE9NA

Member:  ARRL, AMSAT, DXCC, NRA

http://www.qsl.net/ke9na

"A life lived in fear is a life half lived."


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