On 03/19/2014 9:16 PM, wrote:
I found this circuit somewhere, (I can't remember) to replace the 1.50 volt battery in a VTVM. I have an RCA WV98C Senior Voltohmyst VTVM that is in perfect condition but doesn't get used very often. This battery retrofit circuit takes power from the 6.30 volt filament transformer and is built around an LM317. With a trim pot you set it for 1.55 volts and you never have to change a battery again. It also needs to occupy an area no larger than the battery presently does.

As much as I like keeping things original, I'm worried that the battery is going to leak and rot the inside of the unit. I was going to build the circuit but it occurred to me that I should probably be concerned about the current that the LM317 will be dissipating when the meter is used on the low ohms range. Does anyone have a feel for the size that an adequate sized heat sink should be in order to handle this? Thanks, Lenny


Hi Lenny,

I can't imagine that the 1.5V battery provided that much current for the
low ohms scale. Otherwise you would burn them out pretty quickly and
leakage would have been a real risk.

Of course you could simply put an ammeter in series with a battery and
see what is drawn, I suspect it will be under 100ma, which means you
won't need much of a heat sink - if any. 500ma would be a small finned
clip on heatsink.

If you have the schematics and remember ohms law you can figure out what
the maximum current draw is for the battery...

John :-#)#

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