Circuit to drive a 5V relay

Discussion in 'The Projects Forum' started by woon_h88, Dec 19, 2015.

  1. woon_h88

    Thread Starter Active Member

    Mar 25, 2009
    46
    2
    Hi,

    Im trying to create a circuit which when a short-circuit is found at load (RL), The sensing resistor (RS) will sense the change and amp using the LM324.
    But the output of LM324 only able to supply about 3.3V which not enough to trigger the 5V relay to break the circuit from damage. The 5V is supply by another power supply circuit which can varies from 3~16V (Fixed at 5V for now). Is there some way i can do to get an output of around 4.5V with the use of only discrete component? Tried to put a darlington transistor using 2x 2N3904 but its double all the output of the LM324 to 4.9V.

    upload_2015-12-20_0-53-49.png


    Thank.
     
  2. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,564
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    2N7000?
    Max.
     
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  3. woon_h88

    Thread Starter Active Member

    Mar 25, 2009
    46
    2
    Hi,

    Hmm, I should replace the whole darlington transistor to something like this?

    upload_2015-12-20_2-5-55.png

    I see the datasheet the 2N7000 need at at least 0.8V to the gate. So i have to re-adjust the gain for the LM 324 to fit the spec i set. Am i right?

    Thank
     
  4. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    If your relay needs less than 20 to 30 mA, you can try this. If you don't have sharp on/off transitions, you can add some positive feedback for hysteresis (not negative feedback like you had)

    image.jpg
     
  5. woon_h88

    Thread Starter Active Member

    Mar 25, 2009
    46
    2
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    In the drawing in post #4 you want to reverse the + and - inputs into the opamp or comparator. Still, the LM324 would have a hard time making 30 mA, even if it were sinking the current to ground. A better way is a small driver transistor as in post #3, either a 2N7000 FET or something like a bipolar 2N4401, 2222, or 3904. The bipolars would need a 1K base resistor. One end of the relay goes to the +5V and the other end goes to the collector. The emitter goes to GND.

    ak
     
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  7. hp1729

    Well-Known Member

    Nov 23, 2015
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  8. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Since we need it to cut out when the voltage across the sense resistor gets too high. And, the relay is connected to Vcc, the polarity in my drawing works as simulated and as shown.
     
  9. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Here is a circuit that can handle the load if your relay coil. Any NPN transistor suggested by analogkid (above) will work. Almost any NPN transistor with gain over 100 and can handle 50 mA or more will work, analog kid listed the most common types.

    Adjust the pot to have the relay open at the required voltage on the current-sensing resistor.

    NOTE: the op amp inputs are inverted from the earlier schematic because the transistor essentially inverts the logic.


    image.jpg image.jpg
     
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  10. hp1729

    Well-Known Member

    Nov 23, 2015
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    Your math may be correct. Is the objective to measure the motor current? Move that 1.3 ohm resistor over between the motor and ground.
     
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  11. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Correction per @hp1729

    Also, your relay will start chattering when you hit your current limi because the op amp will turn the transistor on as soon as the relay is cut out. You should maybe use
    - a current limiter instead of relay
    - a latching relay (so it stays open once current limit is reached

    Even with hysteresis, you will have chattering because the relay is either open or closed.

    Good luck

    image.jpg image.jpg
     
  12. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    In post #1, RL, 10 ohms, is the load. Without more information, this smells like an oscillator.

    ak
     
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  13. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Agreed and Noted in post 11. It is, however, what the OP seems to be asking for.
     
  14. woon_h88

    Thread Starter Active Member

    Mar 25, 2009
    46
    2
    Hi,

    Thank you..Just tested the circuit and the relay is working!..The RL is the load (will varies). Depend on the load, the relay will trigger to stop supplying to the load.

    Thank again..
     
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