100ua panel meters

If you're in the Boston area, go to the MIT Flea Market which is held on the third Sunday of each month. There is a guy there who has zillions of meters for sale. I bought one a while back and it was good.

Al

Hw about these...

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in the 
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster 
it goes.

Thanks for those Dave, If nothing smaller comes up I'll go with the BGMicro meter. Mike

The problem is that there are almost no meters available at that price. Relax the specs to any surplus meter in that price *range* .OR. similar in a surplus POJ that you can dismantle. Thenput an opamp around it for the desired sensitivity(ies).

Thanks Robert, but actually there are a lot of meters available under $12.00. I'm just haveing trouble finding a meter smaller than 2-1/4" When I get the meters I want, I'll add resistors to make one a 60 volt FS meter and the other a 200 amp FS meter. I'm also looking into these digital meters to see if I can modify the current shunt so it will measure higher current.

Mike

I sent an email to BGMicro and ask about the internal resistance of this

50ua meter.

I got a response that it is 1.95K ohms. That would mean (1,950 x .0005= .975 volts). So, the meter is 1 volt FS. Does 1,950 ohms seem right? If I tried to use that as a 200 amp meter, I would need a .005 ohm 200 watt resistor. I guess I need a 50mv or 100mv meter. Mike

Nope.. you missed the decimal point placement in your math. 50 uA is 0.00005 A. The meter would drop 0.0975V, making the 20A shunt 0.0005 ohms @ approx 50W

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in the 
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster 
it goes.

Thanks Dave, That makes a whole lot more sense. I need to look back I think I did that same mistake on some other calculations. And then also note; it's a 200 amp meter, so .0975V / 200A=.0005 ohms. Then .975V x 200A= 20Watts. A 20 watt shunt will work. Thanks again, Mike Hope I'm right this time :-)

(Just noticed you screwed up your arithmetic -= 50 uA is .00005 ;-)

You can determine the internal resistance by experiment. Since it's a 50 uA meter, put a resistor in series that with your power supply provides

50 uA (20K/1V,100K/5v, etc), and adjust it for full-scale. Then, put another pot (maybe 10K) in parallel with the meter, and adjust it until the meter shows exactly half-scale. Dismantle your setup, and measure the resistance of the pot. That will be equal to the meter's internal resistance (they're in parallel, so half the current each, right?) :-)

As I mentioned, you can make a voltmeter out of an ammeter, but it makes more sense to just get a current shunt; yes, they make .005 ohm 200 watt "resistors" - they're a calibrated brass bar between two brass terminals mounted on some nonconductive plate, with separate connections for volts and amps:

It says 50 mV - Hmm - when I check your math, I get 97.5 mV - you'd have to get a 100 mV shunt, or put two in series (at probably $20 a pop!), then add a little series resistance to bring up your meter to 100 mV FS.

Have Fun! Rich

Yea, thanks Rich. I can do all that but I don't have the meters and I'm trying to purchase the one I can make work with my first purchase.

It seems I only need a 20 Watt shunt if I use a 100 mv meter.

I'm not sure what it is you're saying is 50mv, if it's the meter referenced then it is 50ua. BGMicro has some shunts for $7.95.

Thanks, Mike

When I said, "It says 50 mV", I was referring to the web page of that shunt - I'm usre they come in different values (i,e, you could get one that's 100mV/200A), And to turn a 50uA meter into a 0-100 mV meter, you'd need a total series resistance of 2K, which if your meter is really

1,950 ohms, then you'd need 50 ohms in series with it. You can get 49.9 ohm precision resistors at any distributor - heh - just looked up precision resistors - 2X 100K in parallel should be dandy. :-)

Cheers! Rich

So Rich, heard any good jokes, I've always enjoyed your humor. You haven't been here as much lately, are you busy earning capital to pay your share of the bailout. Mike

Hmmm. That is 10 K per volt. Perhaps you could live with either 100 k per volt (10uA) or 1 k per volt (1 mA) full scale movements. A small circuit change could accommodate either.

What actually determines deflection (in d'Arsonval and taut band movements) is current, it is a physics thing. Is the current you are trying to measure AC or DC, it makes some differences.

If you had said so in the first place you would have gotten much better results.