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DC Amp Meter Connection

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I'mClueless

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Hello,

I have a new Simpson analog DC amp meter and I need help with the connection. Simpson's web sight calls for a 10 Ohms impedance. The attached picture is from Simpsons meter connection document. I certainly don't believe that it should be an integral link in the service line. The illustration is too vague for me to comprehend.

Yep,
I'mClueless

**broken link removed**

**broken link removed**
 
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The typical meter movement used this way is a voltmeter with 50mV full scale. To figure out what shunt resistance you should use, take the full scale voltage of the meter movement and divide that by the full scale current you want to read.

e.g. to read 1A, Rshunt = 0.05/1 = 0.05Ω = 50mΩ

to read 50A, Rshunt = 0.05/50 = 1mΩ
 
The meter for use with external shunts has a sensitivity of 50mV full scale, as Mike ML stated, and has a 10 ohm meter resistance. That may seem low for a meter resistance but it's connected in parallel with the shunt, which is in the milliohm region, depending upon the full-scale current sensitivity desired. Thus the meter resistance is much higher than the shunt where it measures the 50mV voltage.
 
The meter data sheet you present is one of these as you know. The meter scale is 0 to 300 Amps and that meter is designed to work with one of these shunts and to be specific the 0 to 300 Amp shunt.

Pretty much as covered. So the output of the shunt is 0 to 50 mV based on 0 to 300 Amps flowing through the shunt. Thus even though the meter is a 0 to 50 mV meter movement the actual scale of the meter face is 0 to 300 amps. The meter is scaled to provide what it reads based on the output of the shunt it is used with. If you wanted to test just the meter movement you would apply 50 mV to drive it to full scale.

Ron
 
If you wanted to test just the meter movement you would apply 50 mV to drive it to full scale.
Since applying such a small voltage as 50mV can be difficult with typical power supplies, it would be safer to add a series resistor and use a higher voltage to test the meter. Since the meter resistance is 10Ω, the full scale meter current is 5mA. Thus 5V through a 1kΩ resistor in series with the meter will give a full-scale meter reading.
 
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Gentlemen,

Thanks all of you for your input. Electronics has always facsinated me, however I never took the time to learn the science. For 45+ years I was a machinist in every imaginable aspect of the trade and when I retired I decided to learn how to write software for CNC machine tool controllers. Having mastered that goal I'm now back to messing around with electronics.

It's hard to ask questions because it's rather revealing, however I'm sure glad that I ask for help prior to burning up my new amp meter.

Now as I understand the situation the shunt is actually an integral link in the service line and based on my meter I should use the one that I have highlighted.

**broken link removed**

If that is correct then is the 10 Ohm resistor in the line between the shunt and the meter POS post?

Yep,
I'mClueless
 
If that is correct then is the 10 Ohm resistor in the line between the shunt and the meter POS post?
You don't need a resistor. The 10 ohms is the resistance of the meter movement and it's given for information only. You just connect the shunt directly to the meter terminals.
 
crutschow,

So I will have a service cable with a shunt installed at some point that goes to the load and a common cable that returns to the power source. Correct?

Now is the shunt actually a coil of wire around the service cable similar to the handheld amp meter like I used to check automotive starters with?

Yep,
I'mClueless
 
No, the shunt is a resistor, through which the line current flows. You have to break a connection to insert the shunt in series with the line being measured.

Be damn careful, because if you install the shunt in say a line carrying 120VAC, then the two wires running from the shunt to a remoted meter will be carrying the 120V line voltage, too. (Just like the picture you posted shows).
 
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Some actual shunts are pictured in the link I provided and that should give you an idea of what they look like. Shunts and amp meter configurations like this are generally used to measure DC current, there are better means out there to measure AC current. Also when measuring current where higher voltages are involved the shunt is generally placed in the neutral or return line side. Much as MikeMl points out. What I am saying is that if someone wanted to measure AC mains current in a residence that a shunt like this would not be a wise choice in my opinion.


Ron
 
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MikeMl, Reloadron,

If the shunt is a resistor then it certainly cannot be a integral link in the service line. So it must be like this:

**broken link removed**

CORRECT?

Yep,
I'mClueless
 
MikeMl, Reloadron,

If the shunt is a resistor then it certainly cannot be a integral link in the service line. So it must be like this:

**broken link removed**

CORRECT?

Yep,
I'mClueless


NO NO NO! That connection will instantly blow up the power supply. Remember that the resistance of the shunt resistor is like 0.001Ω (1mΩ). If connected across the output of a 10V supply, it would draw a current of I = E/R = 10/0.001 = 10,000A!!!!!.

A resistive shunt MUST be connected in SERIES between the power supply and the load, just like the picture in your first post showed!
 
MikeMl,

My project is a welder and the circuit has the potential of 270A/100VDC. I'm having trouble understanding how there can be a resistor placed in the circuit without having an effect on the service.

Will not the added resistance will have an effect on the service?

Yep,
I'mClueless
 
You purchased a 300A shunt, right? That means it will have a resistance of 0.05/300 = 0.00016667Ω = 167uΩ.

At 270A flowing from your welder, the voltage drop across the shunt will be E = IR = 270*0.00016667 = 0.045V = 45mV.

So instead of delivering 100V to your tip, with the shunt installed, the voltage at the tip will actually be (100-0.045) = 99.955V.

I wouldn't loose much sleep over it, considering that the resistance of your cable is likely about 100 times higher than 167uΩ.
 
MikeMl,

My project is a welder and the circuit has the potential of 270A/100VDC. I'm having trouble understanding how there can be a resistor placed in the circuit without having an effect on the service.

Will not the added resistance will have an effect on the service?

Yep,
I'mClueless

Yes, it will and no it won't. Yes, the shunt can be seen as a resistance in series with the load (the arc) but it will be negliable. Remember the resistance of these shunts is measured in milli and micro ohms unlike seeing a common resistor.

Many of the older large welding machines where I work used current shunts to measure the weld current. In some cases we used shunts to not just measure the weld current but to record it as part of the weld process. Things like the arc voltage, current, part travel were all recorded.

The shunt is placed in series with the load (the arc) to measure the weld current.

Ron
 
MikeMl,

My project is a welder and the circuit has the potential of 270A/100VDC. I'm having trouble understanding how there can be a resistor placed in the circuit without having an effect on the service.

Will not the added resistance will have an effect on the service?

Yep,
I'mClueless

hi,
The 10R you keep quoting is the Impedance when used on AC circuits.

Attached is the Simpson datasheet.

The actual resistance of this type of shunt is milliohms [ ie: 0.001R], so if you used a 300A DC shunt that developed 50mV that would mean that you would only 'lose' 0.05V drop in the 100V DC voltage in the circuit.

Its perfectly OK.

View attachment SIMPSON-ELECTRIC_IN&.PDF AAesp03..gif
 
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Ron,

I greatly appreciate your information and patience. This is exactly what I was looking for and I understand why it took so long to get it. I was the A lead in the machine shop at Boeing Military Aircraft and it took awhile to figure out why those with less experience had such a hard time understanding seemingly drop dead simple explanations. Once that I learned that they lacked the foundation knowledge that I possessed and took for granite then I was able to explain in such a way that they could understand.

You guys just didn't realize I'mClueless
 
Ron,

I greatly appreciate your information and patience. This is exactly what I was looking for and I understand why it took so long to get it. I was the A lead in the machine shop at Boeing Military Aircraft and it took awhile to figure out why those with less experience had such a hard time understanding seemingly drop dead simple explanations. Once that I learned that they lacked the foundation knowledge that I possessed and took for granite then I was able to explain in such a way that they could understand.

You guys just didn't realize I'mClueless

Everyone explained things but when you actually got to the application then the lights came on. I heven't been over to our welding department in awhile but years ago I did some development work out there for a cladding process we do nuclear welding and every weldment is recorded.

Ron
 
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