Re: Low Cost Dummy Load


Don Wilhelm <w3fpr@...>
 

A quick approximation is that the 2133 ohm resistor will carry approximately 1/40th of the total power - so for a full 100 watt rating a 2.5 watt resistor will do.
5 426 ohm 1/2 watt resistors will do the job fine.  You might just want to use 5 470 ohm resistors in series and call it 'good enough' (50.13333 ohms).

73,
Don W3FPR


On 5/11/2015 7:28 PM, 'dekle' dekle@... [4sqrp] wrote:
 



Darryl:,
 
Let 'x' = the value of a resistor in parallel with a 51.2 Ohm resistor to yield 50 ohms equivalent resistance.
 
50 = 51.2 x / (51.2 + x)     (Product over the sum for 2 resistors in parallel.)
2560  +  50 x = 51.2 x       (Multiply both sides by (51.2 + x)).
2560 = 1.2 x                      (Subtract 50 x from both sides.)
x = 2133.3 ohms                (Divide both sides by 1.2)
 
Power rating coming soon........
 
73
Bill
KV6Z
 
 
 
 
----- Original Message -----
Sent: Monday, May 11, 2015 1:49 PM
Subject: [4sqrp] Low Cost Dummy Load [1 Attachment]

 

This is my new low cost dummy load, made with a Caddock thick film non-inductive resistor. This one is 50 ohms at 1%, and rated at 100 watts, and costs about $11.00 from Mouser or Digikey. I realize the current heat sink will not be adequate for 100 watts, but it will suffice for the 5 and 50 watt calibrations of my K3, it does get warm to the touch. Only problem is resistance measures 51.2 ohms. What value of a parallel resistor would bring that down to 50 ohms? I calculate 1234.56 ohms or 1.2 K. Is that right? Also what would the power rating of the resistor need to be? Thanks.

Darryl, KK5IB


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