I have a Simpson Center Zero DC Amp Meter that needs a shunt. How do I size the shunt ?
This Simpson Amp Meter Model 29 is a center zeroed meter rated in 'Microamps' and goes from 0 to 500 microamps DC.
It has an internal resistance of 362 ohms.
How big would the external shunt have to be to have the meter handle up to 5 amps @ 13.8 volts ?
Full deflection is @ 180 milivolts.
It has an internal resistance of 362 ohms.
How big would the external shunt have to be to have the meter handle up to 5 amps @ 13.8 volts ?
Full deflection is @ 180 milivolts.
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Oct 17, 2011. 9:48 PMdrbill (author)
says:
Being that Both Jack A Lopez and iceng have really good answers, I cannot award both as Best Answer, and the fact that I have learned a lot here today I have to say that
BOTH Are Best Answer because they both present different ways to do the same thing. One is presented so I only have to but the resistors for the shunt and the other for a way to correct [?] a coil resistance error endemic to analog meters.
Thank You Both For Your Individual Answers.
73's
DrBill
BOTH Are Best Answer because they both present different ways to do the same thing. One is presented so I only have to but the resistors for the shunt and the other for a way to correct [?] a coil resistance error endemic to analog meters.
Thank You Both For Your Individual Answers.
73's
DrBill
Thank you..... you are a rare bird :-)
I haven't had the pleasure of signing off for 30 years
73's
ICeng
I haven't had the pleasure of signing off for 30 years
73's
ICeng
For full deflection at 5 amperes,
R = V / I = 0.18v / 5a = o.036 ohms
Use Nichrome wire.
A
R = V / I = 0.18v / 5a = o.036 ohms
Use Nichrome wire.
A
Oct 17, 2011. 4:33 PMdrbill (author)
says:
Don't have Nichrome Wire.
I do have high wattage low ohms cement resistors....
This is starting to sound like a Dummy Load.
I do have high wattage low ohms cement resistors....
This is starting to sound like a Dummy Load.
Neither, I was just pointing out something that caused me a difficulty
long before the advent of digital meters.
I recalled an analog meter meter resistance is not a constant over the
full meter needle pointer swing but I don't remember the variation value.
Then I pointed to your center zero Simpson and suggested what ever the variation might be, it should be half as large for a half swing left or right.
I had an analog current meter ( they were expensive pre-digital days )
that provided Pricey hand wound external plug in shunts to allow on meter
to measure different current levels as you are trying to do .
I never spoke of watts, But if you wind a .o.036 ohm resistor the power
will be I2R = 5 * 5 * .036 = 0.9 watts.
If You use Jack's o.1 ohm shunt approach I2R = 5 * 5 * 0.1 = 2.5 Watts.
Both solutions work, but the Jack A Lopez solution is a simpler because
it allows you to buy the shunt resistor instead of DIY.
Oct 17, 2011. 9:35 PMdrbill (author)
says:
Thank you for making that clear to me. I was kind of confused.
What is Nichrome Wire and where can I get some?
The meter need not be all that accurate. All I need it for is to be able to see approximations of charging or dis-charging a tractor battery. I would use an automotive ammeter except they are not readily available on Oahu and I really don't want to send for one because then I don't learn anything. I am then just an appliance operator that blindly replaces parts without knowing Why after they burn up.
Here I have found 2 ways of doing the same thing. I have learned a crapload here today. Thank You for your input.
There is a place out here that sells digital meters but they want $450.00 for them.
A bit out of my price range for a thing I am building.
What is Nichrome Wire and where can I get some?
The meter need not be all that accurate. All I need it for is to be able to see approximations of charging or dis-charging a tractor battery. I would use an automotive ammeter except they are not readily available on Oahu and I really don't want to send for one because then I don't learn anything. I am then just an appliance operator that blindly replaces parts without knowing Why after they burn up.
Here I have found 2 ways of doing the same thing. I have learned a crapload here today. Thank You for your input.
There is a place out here that sells digital meters but they want $450.00 for them.
A bit out of my price range for a thing I am building.
I suggest using two resistors:
The first is 638 ohm resistor, or a potentiometer which can be adjusted to this value. This resistor goes in series with the old ammeter to give a sum resistance of 638 + 362 = 1000 ohms for both in series.
Then in parallel with the resistor and ammeter, place a 0.1 ohm resistor for the shunt. This resistor should have a power rating greater than 2.5 watts, which is how much power it would dissipate if 5 amperes were flowing through it. since I2*R = 5*5*0.1 = 2.5
So basically you've got a 0.1 ohm resistor, and a 1000 ohm resistor in parallel with each other. The ratio of current flowing through the shunt compared to that flowing through the meter is close to 10000:1, and that's what you want, so that when 5 A is flowing through the shunt, 5*10-4 A = 500*10-6 A is flowing through the meter.
Anyway, I think an approach like this, using two resistors will give you more freedom in your choice of resistors. It will definitely be easier than trying to find that one perfect shunt resistor that is (5/4.9995)*10-4 * 362 ohms = 0.0362 ohms
Of course you can probably figure this monkey-junk out yourself, for whatever two resistors you want, using Kirkoff's laws. That's how I'm looking at it. A picture of this is attached.
http://www.instructables.com/file/F5QQF4YGTQVOY3E/
The first is 638 ohm resistor, or a potentiometer which can be adjusted to this value. This resistor goes in series with the old ammeter to give a sum resistance of 638 + 362 = 1000 ohms for both in series.
Then in parallel with the resistor and ammeter, place a 0.1 ohm resistor for the shunt. This resistor should have a power rating greater than 2.5 watts, which is how much power it would dissipate if 5 amperes were flowing through it. since I2*R = 5*5*0.1 = 2.5
So basically you've got a 0.1 ohm resistor, and a 1000 ohm resistor in parallel with each other. The ratio of current flowing through the shunt compared to that flowing through the meter is close to 10000:1, and that's what you want, so that when 5 A is flowing through the shunt, 5*10-4 A = 500*10-6 A is flowing through the meter.
Anyway, I think an approach like this, using two resistors will give you more freedom in your choice of resistors. It will definitely be easier than trying to find that one perfect shunt resistor that is (5/4.9995)*10-4 * 362 ohms = 0.0362 ohms
Of course you can probably figure this monkey-junk out yourself, for whatever two resistors you want, using Kirkoff's laws. That's how I'm looking at it. A picture of this is attached.
http://www.instructables.com/file/F5QQF4YGTQVOY3E/
Your math is good, you came very close to my o.036 ohm shunt value.
Your solution however depends on the current from the ratio of R1 & R2.
I'm concerned about the fact that the meter movement apparent
resistance is not a constant across the the swing of the needle.
Regrettably I don't remember the change ( it was small ) and on a
centered meter this deviation would be half as much and explains why,
plug in shunt current meters of old had pricey wound shunts to change
a meter's current range.
A
Your solution however depends on the current from the ratio of R1 & R2.
I'm concerned about the fact that the meter movement apparent
resistance is not a constant across the the swing of the needle.
Regrettably I don't remember the change ( it was small ) and on a
centered meter this deviation would be half as much and explains why,
plug in shunt current meters of old had pricey wound shunts to change
a meter's current range.
A
Oct 17, 2011. 5:10 PMdrbill (author)
says:
@iceng -
How many watts do these resistors have to be? 1/8 W, 1/4 W, 70 W?
"I'm concerned about the fact that the meter movement apparent
resistance is not a constant across the the swing of the needle."
What does this mean? How does it change?
"a centered meter this deviation would be half as much".
The 180 mV is off the spec sheet I found for this meter, are you saying the meter will only use 90 mV because it is a center zero meter?
How many watts do these resistors have to be? 1/8 W, 1/4 W, 70 W?
"I'm concerned about the fact that the meter movement apparent
resistance is not a constant across the the swing of the needle."
What does this mean? How does it change?
"a centered meter this deviation would be half as much".
The 180 mV is off the spec sheet I found for this meter, are you saying the meter will only use 90 mV because it is a center zero meter?
ICeng under a watt
Jack A Lopez 2½ watts
+90 mV swing Right / -90 mV swing left
90 - ( -90 ) = 90 + 90 = 180
Jack A Lopez 2½ watts
+90 mV swing Right / -90 mV swing left
90 - ( -90 ) = 90 + 90 = 180
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