I have an old analog AC panel meter. The meter has different ranges for example the meter at full scale deflection can be 30 volts, 100v, 50v or 25v. My question is how can setup the meter to measure for example at the 100v full deflection. Do I need to add a shunt resistor or a series multiplier resistor ?

I looked on the internet and it talks about how you can change an ammeter by adding a shunt or series resistor but I dont see anything on how to work with this AC volt meter that I have.

Any help would be appreciated !!

Carlos

Any meter is fundamentally a current meter----e.g., without any series or shunt resistance, the movement might be 50 uA full scale. To calculate the required shunt or series resistance, you would also need to know the resistance of the meter. Let's GUESS that it is 100 ohms.

2 examples of how to use it:

1. To make a 500mA meter, add a shunt such that the parallel combination is(0.05/500) * times the 100 ohms---i.e. 0.01 ohms

2. To make a 500 volt meter, add a SERIES resistor such that the total is (500/0.05) ** K ohms = 10 K ohms


*The ratio of the currents

**500 volts / 0.05 mA = 10K ohms

Righe on the money as usual, remember to hook up a set of diodes back to back to somewhat guard against blowing out the meter movement.

That's an odd meter if it has both 25V and 30V scales. But yes, you can make it read anything you like by adding a multiplier resistor. If the multiplier comes out to a fairly low value, you'll want to reduce it by the internal resistance of the meter itself, so the overall resistance is correct.

here is picture of my ac volts meter, if i dont add any resistor at all what would this meter read at full deflection ?

No way to answer that---why not try it?----you can hook it up to a Variac (best) or to one or more small filament transformers.

Look at my example of how to do the math---There might be a resistor inside, but you can still treat it like a "black box".

No 30-volt scale---just 25, 50, and 100

I tried connecting the meter to my variac and as soon as i applied the ac voltage on the meter the needle started shaking so I took it apart and found that it has a resistor inside. Below is a picture, can someone tell me if it is connected in series or in parallel ?? I am still confused with all this and would really appreciate if one of you can explain what is going on. I measured the resistor with my fluke meter and it is 108.2 oms.

Thanks,
Carlos

As found, what was your F.S. reading on 60 Hz ?


Forget the math. Look at the beginning of the scale. Rectifier meters are mostly
bridging meters which means adding them to a circuit doesn't change the circuit
voltage. I would just use 3 pots. Tweak for best tracking between the 5 and 25 mark,
on each range.


I found some Tripplett like that in the Radio Electronic Master 1965.

Your instrument will probably come out to :

at 25, 50,100 V f.s. the 'impedance' will be:
20, 50, 91 ohms per volt , 25-133 Hz.

The meter looks to be a moving coil type and will only work at DC. If you put AC on it will just quiver. The resistor in the meter looks to be a swamp, a low value put in series with the meter coil to protect it. It meter probably came out of some equipment which contained the rectifier and some resistors to switch the ranges, when used on AC.

I would not connect this meter to a Variac just yet, you could blow it up.

Use another meter to find out what the resistance of the meter and swamp is first. It might be around 1000 ohms or so. Then use a Pot, say 5K Ohms, and connect it to the meter via a 1.5 volt battery and measure the current to get full scale deflection while turning the pot. Then you will find out what meter you have. It might be a 50uA or a 1mA or what ever FSD. Then you can begin to to make it into an AC meter if you want

So what can I do to make the meter an "AC Voltage Meter" ?

Thanks,
Carlos

Carlos,

After you have determined the movement's basic full-scale sensitivity (in ua and/or mv) that others have explained above..then you have several different ways to go, and all of them require rectifiers (probably two silicon) and from one to three resistors (and additional resistors in the series chain to get higher ACV ranges). Your meter is an older Triplett, so they may have used the AC bridge that is pictured in your TTT630 Red-Dot (your oldest vertical turret unit with the 6.8K and 11.1K resistors in the two rectifier bridge, shown on the circuit diagram), which is a "meter-grade" 1/2 wave bridge, even though it has two rectifiers. Or you could use the type of balanced-bridge (I think they are full-wave) design shown in any of the Simpson 260 (series 3 and later) VOM manuals which I know you have; or the modified full-wave (again "meter-grade") bridge shown in a Triplett 310-C (Type 2 or later) VOM manual. You don't want to damage the movement, so work starting with the higher Ohms resistors until you get the ACV readings needed. The Simpson 260 manuals explain how to calibrate the bridge for both low and higher ACV readings. My preferred bridge circuit is the type shown in the Simpson 260 VOM manuals (found at http://www.simpson260.com ), which is a balanced bridge with a 5K resistor on each leg and easier (in my opinion) to understand. In fact, Triplett, in their later 60 series analog VOM's (after mid-70's) used this (Simpson 260) type of ACV bridge possibly because it is very dependable and easier to service/calibrate and understand than the ACV bridges found on their earlier model 630's. Be sure to study your Triplett and Simpson VOM manuals for how they use the same bridge circuit and just add in series additional "range" resistors to get higher ACV ranges.

Fred

There are many designs for AC Voltmeters on the web. Here is one, you will need to experiment with the value of R

1N4148 diodes are only spec'd for a reverse voltage of 75, not enough for a 100 VAC meter.

Use 1N4004 or better (1N4005, 1N4006, 1N4007), rated for a reverse voltage of 400.

100 VAC * 1.414 * 2 = 283 volts

If you put the multiplier resistors ahead of the diodes they won't see much reverse voltage. And I think you have to wire it that way, since all the scales have the same nonlinearity.

The scales might be inaccurate down at the low end however, since the old scale probably assumed a copper-oxide rectifier.

Germanium or Schottky diodes might match the scale better, but I've never tried it.

Good point, thanks. I didn't think it through.

I put a 1.5 volt battery across the meter along with a 25K variable resistor and my VOM placed in the 1.2 DCMA range in series. I tuned the variable resistor to achieve full scale deflection and my VOM said it was .4 mA. I then placed another variable resistor in parallel with the meter and tuned this pot to get 1/2 full scale deflection. I then measured the resistance on this second variable resistor and it said 140 ohms.

By the way I also tried measuring a 9 volt battery and the needle pegged all the way to the right.

Carlos

Good point, use 1N4005-7

The resistor in the above diagram would be about 220K Ohms for the 100 Volt AC range, and assuming you used 1N4007 diodes.

Please read all the replies - there is not 100v across the diodes as the series resistor has dropped most of the input voltage. There is so little voltage across the diodes that they have not properly turned 'on' until about 5v indicated as evidenced by the non linear scale below 5v.

The most important parameter measured by the OP is the 400uA for full scale. To pass 400uA from 100v you need a dropper of 100/400 Megohms = 250kohms - the internal 140ohms of the meter movement is negligible compared to the external resistor. Similarly you'll need 50/400Meg=125k for 50v range and 25/400=62k for the 25v range. A string of series resistors with a three position switch should be easy.

However given that anything to do with electronics and radio is subject to a tolerance of +/-20% your best bet is to adjust an external potentiometer for FSD then read off the value with your DMM!