[lou-ace]

Ok here’s the scoop. I bought a 12v analogue volt meter and want to use it for my 120v pack. According to my “ meter” book to find the value of a meter one divides the face value movement of the meter by it’s resistance than multiplies this by the desired use. The resistance of this meter is about 500 ohms and the swing is 18 hence 500/18=27.77… x 120volts=3333.3330hms ( or3.3k) – 500 ohms for the internal resistance of the meter =2.8 kohms needed to balance the meter. This I did with 5 wtt resistors and hooked the meter to the pack ( my DVM read 152v) and to my delight the 12v meter hung at about 16…… for about 5 seconds than poof!!! Up went the resistors. Fried to a crisp. What gives! Any ideas? I don’t have the electrical training to figure this out, and I’m delighted with the presentation of the meter in my dash……. Any help on how to make this work would be great…….. tesaract, Major, come on guys……….. thanks. Lou-ace

 

[Roy Von Rogers]

Ok here’s the scoop. I bought a 12v analogue volt meter and want to use it for my 120v pack. According to my “ meter” book to find the value of a meter one divides the face value movement of the meter by it’s resistance than multiplies this by the desired use. The resistance of this meter is about 500 ohms and the swing is 18 hence 500/18=27.77… x 120volts=3333.3330hms ( or3.3k) – 500 ohms for the internal resistance of the meter =2.8 kohms needed to balance the meter. This I did with 5 wtt resistors and hooked the meter to the pack ( my DVM read 152v) and to my delight the 12v meter hung at about 16…… for about 5 seconds than poof!!! Up went the resistors. Fried to a crisp. What gives! Any ideas? I don’t have the electrical training to figure this out, and I’m delighted with the presentation of the meter in my dash……. Any help on how to make this work would be great…….. tesaract, Major, come on guys……….. thanks. Lou-ace

Try reading this and see where you made your mistake.

http://sound.westhost.com/articles/meters.htm

Roy

 

[major]

This I did with 5 wtt resistors and hooked the meter to the pack ( my DVM read 152v) and to my delight the 12v meter hung at about 16…… for about 5 seconds than poof!!! Up went the resistors. Fried to a crisp. What gives! Any ideas?

Hi ace,

My ohmslawmath says about 2.9 watts in the resistor. Maybe a bad resistor??? Was it rated for that high a voltage?

major

 

[Nathan219]

I went about this a bit differently 18V normal deflection 500 ohm resistance gives (18V/500ohm = 0.036 Amps), (120V / 0.036Amp = 3333.33 ohms). So I agree with your math. 0.036Amp * 120V = 4.32 Watts. Pretty close to 5 Watts. Give yourself a little larger safety margin on the power 10 watt or heat sink your resistor.

 

[tomofreno]

0.036Amp * 120V = 4.32 Watts.

Actually about 3.7W (and about 0.66W in the meter resistance) since there is about 102V across the 2800 Ohm resistor due to resistive voltage division with the 500 Ohm in the meter. Why did your DVM read 152V if it is a 120V pack? A voltage of 152V would give about 5.9W dissipation in the resistor explaining why it blew up. Sounds like you need larger resistance if the pack will be at 152V.

Voltage across external R = external R*pack voltage/(external R + meter R)

Power dissipated in external R = voltage across external R squared/external R

 

[major]

Actually about 3.7W.....

O.K. Maybe we can get this straight. He has a 500Ω meter, 18V full scale. Meaning movement needs 18V / 500Ω = 0.036A at full scale deflection. He then put a 2800Ω resistor in series with the meter, connects to a battery and saw 16V on the meter. So the meter current was 16/18 * 0.036A = 0.032A. The power in the resistor is I²R = 0.032A² * 2800Ω = 2.8672W.

Is that not correct

edit: see post #11

 

[Nathan219]

 

[lou-ace]

great responces guys and an excellent article Roy! I originally wired the resistors like this


+-1.0 kohms--1.80 kohms-----meter---500 ohms
_ ------------------------------------------


I think i have left over 5 wtt resistors and I'll try a parallel/series arangement like this.

+1.0 k----1.8k--------meter/500 ohm.
+1.0 k----1.8k

- -----------------------------------

hopefully this will split the load? I'll let you know. thanks again.

 

[lou-ace]

Actually about 3.7W (and about 0.66W in the meter resistance) since there is about 102V across the 2800 Ohm resistor due to resistive voltage division with the 500 Ohm in the meter. Why did your DVM read 152V if it is a 120V pack? A voltage of 152V would give about 5.9W dissipation in the resistor explaining why it blew up. Sounds like you need larger resistance if the pack will be at 152V.

Voltage across external R = external R*pack voltage/(external R + meter R)

Power dissipated in external R = voltage across external R squared/external R

the system was being charger at the time

 

[Nathan219]

Remember the resistor value is divided by the number of parallel legs. you would need something like this.
---[-----2.8K----2.8K-----]-------500ohm meter-------
[-----2.8K----2.8K-----]
this would allow you to use 5 watt resistors.
Or go to Radio Shack and get 3 packs of 33K ohm ½ watt resistors and connect 12 of them in parallel this gives you
----2.75K-----------------------500ohm meter-----
It is 10% off but pretty close for being able to buy the parts locally.
Put a 500mA fuse on that to save your meter.

 

[major]

O.K. Maybe we can get this straight. He has a 500Ω meter, 18V full scale. Meaning movement needs 18V / 500Ω = 0.036A at full scale deflection. He then put a 2800Ω resistor in series with the meter, connects to a battery and saw 16V on the meter. So the meter current was 16/18 * 0.036A = 0.032A. The power in the resistor is I²R = 0.032A² * 2800Ω = 2.8672W.

Is that not correct

O.K. This bugs me. Something isn't right here. If he had 152V with a load of 3300Ω (2800Ω in series with a 500Ω meter) then the current would be 0.046A. And 5.94W in the resistor. But that would mean that the meter was seeing 23V, not about 16V as he said.

I'm guessing this is the case and the resistor failed because 5.94W is above the 5W rating. Maybe the meter was pegged and bounced back to 16

 

[Nathan219]

Lou,
It looks like you are going to have to just build the meter and see if it blows up. It is confusing because the meter goes up to 18V, and the meter apparently has a resistance of 500 ohms. This gives 18V/ 500 ohms = 0.036 Amps for full scale deflection. To achieve full scale deflection with 120V you would need a 2.8K resistance in series with the meters 500ohms this gives 120V/ 3300 ohms = 0.036Amp flowing through the meter circuit. If you want to measure 180V you would need 5K of resistance to achieve 180V/ 5000ohms = 0.036Amps, so 4.5k in series with the 500 ohms of the meter. It is a voltage divider 90% of the voltage is dropped a crossed the resistor 10% a crossed the meter. Sorry for the confusion, but even us monkeys figure it out if you get enough of us together.
Maybe through a picture of your meter up?

 

[lou-ace]

no you guys will love this, after reviewing my stuff, I was using 1/2 ( oops read the fine print 0.5 watt not 5 watt!!!) watt resistors.... what a DICK, I've since ordered 10 watt resistors from Norvac, these should work fine..... the fuse is a great Idea. thanks again all.

 

[EVfun]

Who'd of thunk one little dot could matter so much