Need help with Amp Meter

Comments

• dre Posts: 106

  2010-06-13 21:45 edited 2010-06-13 21:45

  Patrick,
  
  Although not analog, how about using a "Kill-a-watt" meter? It has a button for displaying amperage.
  It is very useful for other metering too [noparse][[/noparse] kwatt-hrs, etc.]
  
  cheers, David
• Patrick D Posts: 62

  2010-06-13 22:34 edited 2010-06-13 22:34

  I considered that - but for aesthetic reasons I want the meter to be mounted on the controller's case.
• stamptrol Posts: 1,731

  2010-06-14 12:39 edited 2010-06-14 12:39

  You don't mention it, but is your 0-20A meter meant for AC?
  
  Usually, AC ammeters were meant to be used with current transformers and the meter itself was a 5 A movement (sometimes 1 A)
  
  As you've found, DC ammeters use a shunt which generates 50 mV ( sometimes 100 mV) when the required current flows through the shunt. As to the resistance, its just Ohm's law. V=IR.
  
  In general, the two are not interchangable.
  
  ▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
  Tom Sisk
  
  http://www.siskconsult.com
  ·
• Patrick D Posts: 62

  2010-06-14 22:07 edited 2010-06-14 22:07

  Well - to be honest I don't know if it is for AC or DC - it didn't say...
  
  Here is the meter - can you tell?
  http://www.suntekexpress.com/business-%5E_%5E-industrial/repair-%5E_%5E-tester-%5E_%5E-cleaning-tools/10007924/inventory.html
• Franklin Posts: 4,747

  2010-06-15 02:02 edited 2010-06-15 02:02

  DC
  
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  - Stephen
• Patrick D Posts: 62

  2010-06-15 02:04 edited 2010-06-15 02:04

  well - crud. Thanks for the info!
• kwinn Posts: 8,697

  2010-06-15 02:25 edited 2010-06-15 02:25

  This looks like a DC meter. You can use it with AC by adding a bridge rectifier and shunt resistor as shown in the attached schematic. The value of the shunt resistor can be determined by using a low voltage source and connecting the meter in series with a high value potentiometer.
• Patrick D Posts: 62

  2010-06-15 02:48 edited 2010-06-15 02:48

  Thank you!
  
  
  
  about determining the shunt resistor:
  "...determined by using a low voltage source and connecting the meter in series with a high value potentiometer" - I hate to be a nuisance but could you direct me to a web site where I could get more info on this? I understand Ohms law, series and parallel circuits, etc, but not sure I know how to do what you are telling me.
• kwinn Posts: 8,697

  2010-06-15 05:09 edited 2010-06-15 05:09

  There are 2 ways to approach this.
  
  Referring to CIRCUIT A in the attached diagram, connect the battery (1.2 - 1.5V), a 10K potentiometer (set to max resistance), and the meter as shown.
  
  1 - Adjust the potentiometer until the meter reads full scale and then read the voltage between points A and C. This is the full scale voltage of the meter.
  
  2 - Read the voltage between points A and B, remove the battery from the circuit, and measure the resistance between points A and B. The full scale current of the meter is V/R.
  
  3 - The shunt resistor required would be the full scale voltage of the meter (from 1) divided by 20 (from the printed meter scale).
  
  It may be difficult to source a shunt resistor of the required value so you can use a slightly higher value shunt resistor and a potentiometer as shown in CIRCUIT B. If you have an AC current meter you can connect it to the line and use it to calibrate CIRCUIT B.
• Patrick D Posts: 62

  2010-06-15 11:54 edited 2010-06-15 11:54

  Thank you very much!