[Waverley ARS] Dummy Load calculations

[Henrik Stenstrom]

Hi Adam,

Amateur Radio Magazine Jan/Feb 2007 has a good construction article for
a 200W dummy load with power meter using a complement of 2W resistors.
It would be an easy job to sub higher wattage resistors in this design.

I may be able to PDF the article, let me know.

Henrik Stenstrom VK2FHHS



-----Original Message-----
From: members-bounces at us.cactii.net
[mailto:members-bounces at us.cactii.net] On Behalf Of Adam Carmichael
Sent: Saturday, 21 April 2007 2:25 PM
To: members at vk2bv.org
Subject: [Waverley ARS] Dummy Load calculations

Hello List,

I'm reading into how to calculate the components for a 50 Ohm dummy load

- and so I decided to test myself, unfortuneately I cannot grade myself!

I was wondering if somebody would be kind enough to check what I'm doing

is correct.

The question I gave myself:

I want to build a 50 Ohm, 1kW dummy load. What resistors are required?

So I set out to find out. 1W resistors would mean I would need 1000 
resistors, which is too much soldering! 5W resistors would mean a lot of

soldering, but 200 resistors is still manageable; and so I continue.

Total Resistance required: 50 Ohms
Number of resistors: 200
Power dissipation per resistor: 5W

Remembering the equation for parallel resistances of equal resistors:

R(total) = R1 / (number of resistors)
50 Ohms = R1 / 200

Or, rearranged:

50 * 200 = R1
          = 10 000 Ohms

Therefore: 200 * 10K Ohm 5W resistors are required to build a 1kW dummy 
load.



Thanks in advance,

Adam

-- 
Adam "carneeki" Carmichael - VK2FNRD
p: +61415371990
w: http://carneeki.net
e: carneeki at carneeki.net
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