The AM Forum

Just want to make sure I'm doing this correctly. I have a Simpson NOS meter made for Motorola. It's 3,000 VDC. Meter says "full-scale 8 milliamps." I'll actually use it for a supply of around 2.2 KV. Using Ohm's law, the value of the dropping resistor should be 3,000/.008 or 375K Ohms; correct? As for wattage, it would be 2200 volts X .008 for 17.6 so 20 watts should do the job. Are my calculations correct? Mike W9ZSL

8K is a pretty hefty burden for a high V meter.  Design has always gravitated to either 1mA or 500uA movements, so as to use simpler & lesser Wattage resistors, i.e. 1 Meg/kV.

73DG

Hi Mike,

I've found from experience that running a resistor at or near its full wattage rating makes it run hot.  It varies with the type of resistor and the manufacturer's method of rating,  but by going 100% (or higher) in power rating is a good practice to keep them just warm to the touch. Heat can affect resistor longevity, adjacent components, circuits boards or even solder joints if pushed to the limit.

T

Full scale on the meter is .008 amps... 8 milliamps.

I double-checked and full-scale is 8 ma.

Are you sure you want to use this meter?  Another with the traditional movement would be much easier to implement if the style was the same.

73DG

375K at 8ma dissipates 24 watts.  You would need about 75-100watt resistor to keep it running cool.  If your HV PS is going to have bleeder resistors the meter can be used at the ground end of the bleeders.  This is most likely how the meter was intended to be used,  that's why the odd 8ma meter movement.  Lets say 25 ma of bleeder current,  use the meter at the ground end of the bleeders.  Add a shunt resistor across the meter to bypass 17 ma of current.  Use the full scale 3KV to do the calculations.

Fred

Your calculations are right. It's true the high current will be a nuisance and a little cost because of the higher wattage multiplier. Go ahead and use it if you like it. do what you want that's my motto!

Because of the high current and high voltage requirement, a good choice is tubular ceramic resistors with wattage ratings x2 to x4 so the value won't change due to heat and so that they will last forever. Tubular ceramic resistors should usually handle as much voltage as it takes to dissipate the rated wattage, another reason for 2x higher watts -to give voltage margin.

It's easy to mount the resistors to ceramic standoffs, or just use the metal standoff mounts that come with tubular resistors and mount them to a fiberglass or phenolic sort of insulating board.

If the board is vertical the heat will pass up across all the resistors and should not cook the board over the years. Mounting tubular resistors vertically also lets convection air pass through them.

three 100K and one 75K units, each 10 Watts each should be fine -a x2 margin on watts. Or 20 watt resistors for an 80W string -some extra margin on top of it. If nothing else it's an extra bleeder for safety in case the main one quits.

The picture is something like that -the phenolic? board. Probably you have all this stuff in the ol' junkbox!

I have a 2KV Simpson with a matching dropping resistor but the plate supply will probably be just over that voltage. Under load, it would come close but I don't want to pin the meter. My plate iron is 4600 VCT.

I considered all the possibilities with this and went back to eBay. Found a NOS matching 3KV Simpson with dropping resistor for $24.95. I'll use my 2KV for the modulator supply. The is the meter panel with a 750 volt Simpson. I'll replace it with the 3KV. All's well that ends well! Thanks / 73!