Boy you sure will end up being a Dummy Load if you forgot it is Mother's Day. Maybe its not too late to get a card for the XYL. (If you have one, in addition to the XYL, don't forget the Girl Friend.)
So, why do we have Dummy Loads (DL)? There are many good reasons but primarily it is to set a benchmark of how one of our transmitter configurations performs into a theoretical and idealized load. Often this is overlooked as a 1st step in a testing process. It is also useful to check out our rigs into a dummy load to not put signals into the air that are one-way communications. Both Critical and Important!
One of the more famous dummy loads is the Heathkit Cantenna. This is a slick device where a nominal sized pure carbon resistor (read non-inductive) say around 100 watts is immersed into a gallon paint can size of transformer oil. This now takes the rating up to a 1 KW. If you can find one -- snag it. MFJ (the company now a SK) and others made similar dummy loads. But my trusty Heathkit stands out for me.
At QRP power levels something smaller than a Cantenna might be in order. This leads us to homebrewing a dummy load. The design should be such that it would present 50 Ohms and for safety (no smoked parts) a 15 watt rating. Now I have seen some quick and dirty Dummy Loads built using ten 470 Ohm 1/2 watt resistors. When paralleled that would come to 47 Ohms at 5 watts.
Or does it, especially if you used 10% resistors? Your Dummy Load could be in the range from 42.3 to 51.7 ohms. Thus your power calculations could be way off. If you measured 20Volts Peak to Peak across 50 Ohms that is 1 watt or 1000 milli-watts.
So, lets convert 20Volts Peak To Peak to Root Mean Square (RMS) where the Peak value is 10 and the RMS is .707 X 10, or 7.07-Volts RMS. We square that and get 49.9849 and divide that by 50 Ohms we get 0.999698 watts or 999.698 milliwatts. Call it 1 watt!
Taking our 7.07 Vrms squared as 49.9849 and divide that by 42.3 we get 1.1817 watts or 1182 Milliwatts. If we take the same 49.9849 and divide that by 51.7 then we get 0.9668 watts or 966.8 milliwatts. The % swing is significant at QRP levels. The closer you can get to 50 Ohms the more accurate the readings. I would look at 510 Ohms versus 470 Ohms.
The best way to connect 10 resistors in parallel is to get two pieces of PC Board about 2X2 inches. Stack the two boards and drill 10 Holes in a Square pattern allowing space between the holes, A square pattern about 1.25 inches on a side will take 8 resistors and two on a diagonal fits ten 1/2 watt resistors. Insert the resistors in the matching holes and then solder away. The PC Board also acts as a heat radiator.
Now even though you have an extra class license, such a project may seem too complex and then you could just buy this jewel.
Caddock Electronics up in Oregon has the product, a 1% 50 Ohm 15-watt non-inductive resistor that does the trick for you and on the back side is a ceramic area that can be mounted to a heat sink.
For those who lack look up skills the DigiKey P/N is on the photo. The cost is about $7 but sure beats having to exercise the grey matter or finding two pieces of PC Board and drilling 10 holes and soldering 22 connections and then worrying about was that 9 or 10 resistors and are there any bad solder joints. The extra two connections are for the BNC connector to connect to coax.
With the Caddock Device you will need to drill one hole to mount the unit and you will need to solder two connections for a mating connector ((BNC)
While I am poking some fun about building a dummy load -- there is more truth than fiction to what I have stated.
Always ahead of the curve.
73's
Pete N6QW