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Power supply for diaphragm pump

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by ADWSystems, Aug 24, 2017.

  1. GromTag

    GromTag Active Member

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    Noise on the output is an all ways problem, tho how to filter without current loss by the supplies respects and not choke the current for the pump.
     
  2. GromTag

    GromTag Active Member

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    ... not sure, but could these things be refurbished? I have seen and bought "new" things accordingly, most for parts.
     
  3. GromTag

    GromTag Active Member

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    Looking up some info on Synchronous rectifier type supplies and the timing and function are close to the problems.

    segment.

    The DC-to-DC converter comprises a transformer, a primary switching device connected in series with a primary winding of the transformer, a control circuit for controlling turning on or off of the primary switching device, a rectifier synchronous rectification switching device, and a flywheel synchronous rectification switching device, and converts a voltage of a DC power source into another voltage. The DC-to-DC converter further comprises a rectifier rise delay inductive device connected in series with the rectifier synchronous rectification switching device and a flywheel rise delay inductive device connected in series with the flywheel synchronous rectification switching device.

    Quite touchy actually.

    http://www.google.com.pg/patents/US5726869
     
  4. dave

    Dave New Member

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  5. Pommie

    Pommie Well-Known Member Most Helpful Member

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    Something is not adding up here. This result suggests the pump has a DC resistance of 55Ω (11*5) which means it should never draw more than 1A never mind 20A. How did you have this wired?

    I suggested 0.5Ω assuming the motors DC resistance was similar and would therefore reduce the initial current to around 12A.

    Edit, the value may have to be as low as 0.25Ω as the pump draws 6A. At 6A the power would be 6^2*0.25 = 9W.

    Mike.
     
    Last edited: Aug 25, 2017
  6. ADWSystems

    ADWSystems Member

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    I absolutely agree something that doesn't make sense. If it made sense then I would not have the issue and we would not be having this conversation.

    As for wiring:

    V+ -> Pump -> Resistor -> V- and V+ -> Resistor -> Pump -> V-

    I think I need a lower value resistor. A resistor that will limit the "short circuit" current to the power supply capacity while still allowing the motor to run. 12V/20A=0.6 ohm, unfortunately it should also be 240 watts but since it is not continuous I'll aim for something large but not that large.
     
  7. JonSea

    JonSea Well-Known Member

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    Could you actually draw a sketch of what you have hooked to this power supply? Sketch it on a napkin and take a picture of it. Reading the above posts, I don't think anybody is on the same page, and I'm kind of wondering if you all are on the same planet.

    If I'm reading this correctly, I think there is some confusion between resistors in series with the motor and in parallel across the motor and/or pump.
     
  8. ADWSystems

    ADWSystems Member

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    If there's confusion then it is being over thought.

    V+ -> Pump -> Resistor -> V-

    That is all. When in series, 3 wires.

    1. V+ to Pump
    2. Pump to Resistor(s)
    3. Resistor(s) to V-

    Probably the simplest circuit ever posted on Electro-Tech. If you add a switch, then 4 wires, and caps will make 5.

    Worst case:
    1. V+ to Caps
    2. Caps to Resistor(s)
    3. Resistor(s) to Switch
    4. Switch to Pump
    5. Pump to V-

    If I can find my camera and it's charged, I'll just take a picture of the setup though it will not show anything more than the above wire list.
     
  9. JonSea

    JonSea Well-Known Member

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    Pardon me for trying to help. My mistake.
     
  10. Pommie

    Pommie Well-Known Member Most Helpful Member

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    Your wattage calculations are very wrong. It's I^2R, so at 10A it is 100 x R. If R is 0.25Ω then it's 25W. As the run current is 6A then it will be 9W.

    Mike.
     
  11. ADWSystems

    ADWSystems Member

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    Assuming the pump is a dead short when starting, the current needs to be limited to the power supply rating. 12V/20S= 0.6 Ohm. I^2R=20*20*.6=12V*20A=240W. I think it depends on the level of conservativeness you are trying to achieve. Both sets of number are correct, just that mine are worst worst case.

    Yep something didn't add up. So I started over. You can forget almost everything previously posted, expect the fact the pump doesn't work. That still applies.

    V+ -> switch -> pump -> resistor ->V-

    The power supply is dialed all the way up 12.25V no load.

    The measured resistance of the pump is 1.2 ohms.

    The measured resistance of the "big" resistor is 5.2 ohms.

    When power is applied, the current is 1.8 amps.

    The voltage across the pump is 1.1V, across the resistors is 11.1V. (looks like Kirchoff's voltage law still works) .

    The pump still draw 3A when powered directly from a car battery, reading 11.9V (it's an old battery sitting around and not in use).

    It still doesn't make sense why the motor won't start and run using a power supply.
     
  12. Pommie

    Pommie Well-Known Member Most Helpful Member

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    12V into 1.2Ω should result in 10A flowing. Something still doesn't add up. Try a 0,5 or 0.25 ohm restor.

    Mike.
     
  13. ADWSystems

    ADWSystems Member

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    I plan to, but the only one I have is 1/4W so I think it will resemble a flaming fuse.

    As I recall motors on startup draw significantly more current than their faceplate rating. The 1.2 ohms is a static reading with a voltmeter. If less than 0.6 ohms, then it would be more than the 20A power supply limit.

    I hope to get a smaller higher wattage resistor soon.
     
  14. JonSea

    JonSea Well-Known Member

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    What happens if you "jump start" it from the battery (i.e., power supply commected to pump, and both connected to bsttery), then remove the battery from the circuit? Does the motor continue to run when connected directly to the power supply?
     
  15. ADWSystems

    ADWSystems Member

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    Interesting idea. I don't know. I'm not sure if I want to or how to try. I'm not sure what would happen if I connected a 12V power supply to a 650 amp-hour lead acid battery. Sounds equal parts OK and sketchy.
     
  16. JonSea

    JonSea Well-Known Member

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    If the power supply voltage is slightly higher than the battery voltage, it will charge the battery.
     
  17. ADWSystems

    ADWSystems Member

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    For the sake of discussion, what if the pump does run after starting from the battery?
     
  18. alec_t

    alec_t Well-Known Member Most Helpful Member

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    Then you could use the pump pressure to operate a switch to disconnect the battery :D
    Switchoff.PNG
     
  19. Les Jones

    Les Jones Well-Known Member

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    You could also have a diode between the power supply and the battery and have the coil of a relay connected to the power supply side of the diode. The contacts would be is series with the motor so that the motor would only run if there was output from the power supply. It should also be possible to use the direct output of the power supply to enable a modified version of the original driver circuit which would remove the need for a relay.

    Les.
     
  20. JonSea

    JonSea Well-Known Member

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    If the pump runs with only the power supply but only starts with the battery connected, the power supply can't handle the starting current. Plain and simple.

    If the pump will handle slightly more voltage, an easy answer is to use a power supply of slightly higher voltage in parallel with a smallish battery. The power supply float charges the battery, and the battey provides the higher starting current.

    Norhing else would be required. Power supply and battery in parallel, connected to the pump via a single switch.
     
  21. ADWSystems

    ADWSystems Member

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    I think I would still like to limit the starting current with a small value high wattage resistor.

    I am not sure I am comfortable connecting the power supply and battery. How about connecting them via 1N4004 diodes? Battery & diode in parallel with power supply & diode, both diodes then connected to the pump. The diodes only needs to last long enough for for the one test.
     

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