DCC and amp meter

hpdrifter Dec 20, 2022

  1. hpdrifter

    hpdrifter TrainBoard Member

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    Evidently you can't send encoded power thru a amp meter.....too well. That was gonna be my question, however I did a search and found the "evidently" part....not too well.

    Everything I found on diy amp meter for such use is a dead link...."hmmm, site not found"

    Anybody have any insight to the building of a amp meter explicitly for dcc use?

    Thanks
     
  2. wvgca

    wvgca TrainBoard Member

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    use a bridge to get proper dc and then add 1.4 volts to the voltage readout ..
    will be close enough ..
     
  3. hpdrifter

    hpdrifter TrainBoard Member

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    for amps?
     
  4. wvgca

    wvgca TrainBoard Member

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    same thing, add bridge to convert to DC [ordinary] voltage
     
  5. hpdrifter

    hpdrifter TrainBoard Member

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    if I do that, doesn't it kill the pwm signals and then no DCC signal to the track since an amp meter has to go in series?
     
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  6. Sumner

    Sumner TrainBoard Member

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  7. wvgca

    wvgca TrainBoard Member

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    on the DCC amp meter you would run it off the track itself, not in series with the power to : leads ..
    easy to do, just watch where it's hooked up ..
     
  8. hpdrifter

    hpdrifter TrainBoard Member

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    never mind. I'll just run em DC, get the amps from that and expect it's close in DCC.
     
  9. Leo Bicknell

    Leo Bicknell TrainBoard Member

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  10. hpdrifter

    hpdrifter TrainBoard Member

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  11. hpdrifter

    hpdrifter TrainBoard Member

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    Leo, I have searched all the parts up, except the .1 ohm resistor. I found a .2 5W resistor. Seems the .1's are out of stock.
    I know I can put two of those parallel and get .1 ohm, but just curious; since the IC is measuring the minute voltage drop across the resistor, would a .2 double the output(or some sine/cosine factoro_O) or BLOW EVERYTHING UP?:(

    I made a couple of assumptions there; that you may have built one of these or know a bit about the electronics.
    I'm no EE and can't cipher all the info on the data sheets.
    Another assumption there; bigger voltage drop greater output; hope it's not inverse.:oops:
     
    Last edited: Dec 21, 2022
  12. ajkochev

    ajkochev TrainBoard Member

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    I posted this same question on FB a couple of days ago! I got a lot of responses, a few recommended adding a common DC amp meter just before your track power source goes into your DCC station while it is still DC. Since I plan on using DCC-EX my input voltage will be 12 volts and only dedicated to the motor shield going to the track this will be a easy inexpensive setup, just hope it works and gives accurate readings. Quite a few folks have done this and got good results. Since those amp meters are not too much figured I'd give it a try.
     
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  13. hpdrifter

    hpdrifter TrainBoard Member

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    I can see that working, but I already have a Digitrax DSC51 and don't plan to update.

    Thanks.
     
  14. Mark Ricci

    Mark Ricci TrainBoard Member

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    Measuring the total current drawn from the shield's 12 Vin both before running and during. Subtract the 2 to eliminate the shields "idling" current might provide a closer value.
     
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  15. vasilis

    vasilis TrainBoard Member

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    hi, put two 0.2ohm in parallel
     
  16. Leo Bicknell

    Leo Bicknell TrainBoard Member

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    0.2 won’t blow anything up. I’m not sure it will just change the reading a by a factor of 2 though. The formulas are in the data sheet for the measurement IC. Shouldn’t be too hard to figure it out.
     
  17. hpdrifter

    hpdrifter TrainBoard Member

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    Thanks, I had that already figured out.

     
  18. Leo Bicknell

    Leo Bicknell TrainBoard Member

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    I'm back at a real computer now, so....
    https://www.diodes.com/assets/Datasheets/ZXCT1009.pdf

    That's the IC that's doing the measurement. It says:

    VSENSE = VIN - VLOAD
    VOUT = 0.01 x VSENSE x ROUT

    A 1A current is to be represented by a 100mV output voltage:

    1) Choose the value of RSENSE to give 50mV > VSENSE > 500mV at full load.
    For example VSENSE = 100mV at 1.0A. RSENSE = 0.1/1.0 => 0.1Ω.

    2) Choose ROUT to give VOUT = 100mV, when VSENSE = 100mV.
    Rearranging 1 for Rout gives:
    ROUT = VOUT /(VSENSE x 0.01)
    ROUT = 0.1 / (0.1 x 0.01) = 100Ω

    Assuming the parts were originally picked correctly, doubling Sense to 0.2Ohm is going to double Vsense to 200mV. That is still in the 50-500mV range, but will reduce the highest amperage the chip can read by half. It says it's designed for 5A, so this is going to make it accurate only to 2.5 amps or so. FWIW, I'm not wild about doubling up the 0.2Ohm here in parallel. Accuracy is somewhat important here, and if the resistors are a bit off, paralleling them could amplify that error.

    The correct parts are available, and can be found for an OK price:

    0.1 Ohm 7W through hole: https://www.mouser.com/ProductDetail/Ohmite/17FPR100E

    0.1 Ohm 7W SMD: https://www.mouser.com/ProductDetail/KOA-Speer/SLN5TTEDR100F

    0.1 Ohm 5W through hole: https://www.digikey.com/en/products/detail/ohmite/WHER10FET/3114531

    0.1 Ohm 5W through hole: https://www.digikey.com/en/products/detail/riedon/UB5C-0R1F1/2176618

    0.1 Ohm 5W SMD: https://www.digikey.com/en/products/detail/vishay-dale/WSR5R1000DEK/9757249

    Keep in mind when you search that Mouser puts these in a separate section called "Current Sense Resistors". Also, in the search they will be 100mOhm, not 0.1 Ohm.

    I would probably buy the WHER10FET from Digikey if I was making one.
     
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  19. hpdrifter

    hpdrifter TrainBoard Member

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    thanks. I guess I gave up too quickly. i searched 3 or 4 and they were all sold out. Went to Allied and they were all sold out plus had minimum order also. Didn't check Mouser.
     

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