Reducing Relay Power Consumption - Holding Current versus Pickup Current

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ความคิดเห็น • 9

  • @SimonALogan
    @SimonALogan 2 หลายเดือนก่อน

    Very helpful tutorial, thanks Simon

  • @mmghv
    @mmghv 4 ปีที่แล้ว

    I saw a simpler variant of this circuit which doesn't use the transistor, just the 100uf capacitor directly across the resistor, I tested it and it works great.

    • @simoncarter568
      @simoncarter568  4 ปีที่แล้ว +1

      Interesting! I’ll have to try that

  • @bgable7707
    @bgable7707 3 ปีที่แล้ว

    Simon, great explanation, thank you. Question, I'm doing a DIY yard sprinkler system proj. w the standard cheap NC relays and wiring the values to the NO terminals when the relays are not energized. My concern/interest is what the power draw for the relay is when in the NC / default position. It appears the LED are lit in this position. Is there any other draw or charge what would cause these to fail over time that you can think of?

    • @simoncarter568
      @simoncarter568  3 ปีที่แล้ว

      If the circuit is energized there will be a current draw by the relay coil and resistor in normal operating mode. Current draw will be power supply voltage/(K1 coil resistance + R1). No other draw.

  • @pmacgowan
    @pmacgowan 9 หลายเดือนก่อน

    If I have a 12v relay using drawing 80ma and still works down to 6.5v (drawing 46ma) , what values of resistors & cap would you suggest in your circuit ?, Also how would you handle the transistor base voltage at 12v ?

    • @simoncarter568
      @simoncarter568  9 หลายเดือนก่อน

      You should only need to change R1 value. Probably a value around 150 ohms would work.

    • @simoncarter568
      @simoncarter568  9 หลายเดือนก่อน

      If 150 does not work, try slightly lower value around 120 ohms.