While building my bench PSU I found some DC milliammeters for under $10 and decided to convert them to read the units I was concerned with. The PSU is the Elliot lab supply. Figuring out the size of the series resistor was easy to convert one ammeter to a voltmeter. However, I am not sure I have the correct plan for converting the 0-10 milliammeter to a 0-1 ammeter. Here are my thoughts (per Rod's site and an old physics book):
To convert an ammeter to read a different scale, simply determine the current required to move full scale and the meter's resistance. Put the ammeter in parallel with a shunt resistor sized to allow the desired current to flow through the meter. Specifically:
(Ifs*Rmeter)/(Iscale-Ifs)
where:
Ifs is current required for full scale movement of meter
Rmeter is the meters resistance
Iscale is the desired scale
Plugging in values obtained from a gage:
Ifs=.01 A
Rmeter=240 Ohm
and my desired scale, Iscale, of 1 A
I obviously get a 2.4 Ohm resistor. Now, running .99 A through a 2.4 Ohm resistor yields a voltage drop nominally of 2.4 V and a power dissipation of nominally 2.4 W.
Do these values seem reasonable? I was not planning on having a 2.4 voltage drop per rail due to meters. I also was not intending to dissipate 2.4 W of additional heat through a resistor. These numbers aren't unacceptable, just unexpected.
Is it common for an ammeter to cause this magnitude of loss. The example on Rod's website showed an ammeter spec'd at 58 Ohm and .001A full deflection. This obviously results in a tiny shunt resistor (.058 Ohm) and an equally low Voltage drop (.1 V) and very small power consumption. I have not, however, seen any ammeters from places like Mouser or Digikey that are spec'd similarly.
Any thoughts, suggestions or additional insight is appreciated.
Sandy.
PS: I should add that I have not received the actual meters I ordered yet. I was simply wanting to get the calculation format together so when I did, I could get the actual specs and purchase the correct shunt resistor/trimpot.
To convert an ammeter to read a different scale, simply determine the current required to move full scale and the meter's resistance. Put the ammeter in parallel with a shunt resistor sized to allow the desired current to flow through the meter. Specifically:
(Ifs*Rmeter)/(Iscale-Ifs)
where:
Ifs is current required for full scale movement of meter
Rmeter is the meters resistance
Iscale is the desired scale
Plugging in values obtained from a gage:
Ifs=.01 A
Rmeter=240 Ohm
and my desired scale, Iscale, of 1 A
I obviously get a 2.4 Ohm resistor. Now, running .99 A through a 2.4 Ohm resistor yields a voltage drop nominally of 2.4 V and a power dissipation of nominally 2.4 W.
Do these values seem reasonable? I was not planning on having a 2.4 voltage drop per rail due to meters. I also was not intending to dissipate 2.4 W of additional heat through a resistor. These numbers aren't unacceptable, just unexpected.Is it common for an ammeter to cause this magnitude of loss. The example on Rod's website showed an ammeter spec'd at 58 Ohm and .001A full deflection. This obviously results in a tiny shunt resistor (.058 Ohm) and an equally low Voltage drop (.1 V) and very small power consumption. I have not, however, seen any ammeters from places like Mouser or Digikey that are spec'd similarly.
Any thoughts, suggestions or additional insight is appreciated.
Sandy.
PS: I should add that I have not received the actual meters I ordered yet. I was simply wanting to get the calculation format together so when I did, I could get the actual specs and purchase the correct shunt resistor/trimpot.
you have to determine the internal resistance of the meter, then you can determine the value of the shunt. I put the procedure up at www.tech-diy.com/metershunt.gif
Thanks Jack!
Thanks for posting the measurement method for determining internal resistance of the meter. I was intending to simply use my multimeter, which the article explicitly states not to do. You saved one of my meters for sure.
As to the point of the internal resistance, Rod indicated that for a meter he had on hand, the internal resistance was 58 ohms for full-scale deflection of .001A. Mouser indicated that the internal resistance was around 240 ohms for a 0-10 mA meter. Other similar meters also had a ratio of current flow to resistance which would yield a 2.4 V drop.
When the actual meter I purchased arrives, I will use the method you offered to determine what its internal resistance is. I am surprised, though that the analog meters Mouser offers would result in such a high loss. It seems like for many applications, this would be a significant downfall. Are you aware of any technical reason that a 2.4 V loss would make the meter more useful than a much lower loss?
Thanks again!
Sandy.
Thanks for posting the measurement method for determining internal resistance of the meter. I was intending to simply use my multimeter, which the article explicitly states not to do. You saved one of my meters for sure.
As to the point of the internal resistance, Rod indicated that for a meter he had on hand, the internal resistance was 58 ohms for full-scale deflection of .001A. Mouser indicated that the internal resistance was around 240 ohms for a 0-10 mA meter. Other similar meters also had a ratio of current flow to resistance which would yield a 2.4 V drop.
When the actual meter I purchased arrives, I will use the method you offered to determine what its internal resistance is. I am surprised, though that the analog meters Mouser offers would result in such a high loss. It seems like for many applications, this would be a significant downfall. Are you aware of any technical reason that a 2.4 V loss would make the meter more useful than a much lower loss?
Thanks again!
Sandy.
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