12V cutoff circuit

Discussion in 'General Electronics Chat' started by doug08, Nov 6, 2011.

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  1. doug08

    Thread Starter Member

    Jan 30, 2011
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    I put together the circuit below. It is designed to save a 12V lead acid battery if the voltage drops to a preset level by cutting off the power. I used the exact components except for the bc548> ksp2222a I used, and for the 4.7V zener> I only had a 5.1V zener. I tried out the circuit, does not work. I know for a fact the N-Channel Mosfet works, I tested it with a tester and also by applying 3v to the gate. I also made sure of the pins.(G-D-S).The ksp2222a supposedly can be substituted for the bc548. I tested it before I soldered it in. The LM358 I pulled from scrap electronics(possibly faulty). When I apply a 12V lamp to the load, and 12V battery supply to the circuit.....nothing happens. The light does not turn on. I tried adjusting the POT in each direction and resetting the circuit with the push button switch...nothing. The soldering is perfect, no bridges...etc. Depending where the POT is set, I get 0 to 5.55V at pin 3, and 11.19V at pin 2 of the LM358. Power is not making it out of pin 1 to supply the gate voltage. Any Ideas?

    Thanks

    [​IMG]
     
  2. jimkeith

    Active Member

    Oct 26, 2011
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    Voltage at pin 2 with button pressed?
     
  3. SgtWookie

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    Jul 17, 2007
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    Sounds to me like either ZD1 is open, Q1 is open or miswired, or the 10k resistor to S1 is open, or S1 won't actually close ... or the 22k resistor to ground isn't actually 22k, it's 220 Ohms or something else low-value.

    If you can get pin 2 to drop to ~5.3v when you press S1, then all that is OK. If it goes back up above 11v when you release S1, then D1 or the 10k Ohm resistor from the output of IC1 to D1 is faulty, the output of IC1 isn't working, or your battery voltage is just too low.

    You probably need another 10k resistor between VR1 and the battery negative terminal.
     
  4. doug08

    Thread Starter Member

    Jan 30, 2011
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    Never tried w/the button pressed.
    [​IMG]
    [​IMG][​IMG]
     
  5. doug08

    Thread Starter Member

    Jan 30, 2011
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    The switch definitely works. When I press the switch, pin 3 stays the same at 5.55V or less depending where the POT is set. Pressing the switch actually makes pin 2 voltage rise by .05V. Transistor is E-B-C....I tested that with the diode setting on my dmm and got .665v form the base to one leg, and .668v from the base to the other leg. Red probe to the base. Having the 5.1V zener should be no problem....correct? Battery voltage at 12.47V. I hate when a simple circuit becomes a problem.
     
  6. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    It was 11.19V and then went to 11.24V? (+0.05V). That's the problem. The transistor is not switching, at least it's not able to pull the pin 2 down through the zener diode.
    Recheck your layout. If that's truly ok then measure the voltage on the collector of the transistor when pressing the switch.

    You could also pull down pin 2 to ground for a moment and see if the rest of the circuit works. Tell us how it behaves.
     
    Last edited: Nov 6, 2011
  7. doug08

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    Jan 30, 2011
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    while taking a reading on the collector of the transistor. The voltage rose from 5.85V to 5.9V.
     
  8. jimkeith

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    Oct 26, 2011
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    Looking at the flat with the pins down is cbe--recheck your wiring.

    Also, jumper transistor to see if your op amp circuit works.

    22K resistor in upper left may need to drop to perhaps 15K to get better range on pot.
     
  9. praondevou

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    Jul 9, 2011
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    Yeah well, that doesn't look right. Emitter is connected to ground? What's the base voltage when switch is pressed / released?
     
  10. doug08

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    Jan 30, 2011
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  11. doug08

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    Jan 30, 2011
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    At the base of the transistor.
     
  12. praondevou

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    Jul 9, 2011
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    According to datasheet Vbe(sat) is 2.6V max ---- at 50mA base current!!! Your base current is much lower than this.
    This is where you have to start. Why do you measure 4V from base to emitter?

    Did you short emitter collector as we suggested?
     
  13. SgtWookie

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    Try measuring from Q1's base to emitter, and alternately close / open S1. You should see the voltage alternate between 0v and ~0.7v, ±0.1v. If not, you have problems.

    Then try shorting between the base and the emitter of Q1, and measure from pin 2 of IC1 to ground. What do you read?

    [eta]
    Oops, I didn't see the above 2 posts before reading this. If you're seeing voltages out of the range of 0v to 0.8v on the base, the transistor is having problems. Are you certain that the 10k resistor is between S1 and the base?
     
  14. doug08

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    Jan 30, 2011
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    12v+ flows down through the 10K resistor to the base....Why wouldn't I have around just under 4V there? If you use the voltage divider calculator, the voltage at the base comes out exactly to the 3.9v that I have when the button is pressed.
     
  15. praondevou

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    No you'll have around 0.7V, as Sgt just wrote. That's your Vbe. Q1 is not correctly wired or faulty.

    Also , if you removed Q1 and D1 (that leaves only the voltage divider) you would not have 4V on the 22k resistor when the switch is pressed but 8.25V, the rest being over the 10k.
     
    Last edited: Nov 6, 2011
  16. doug08

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    Jan 30, 2011
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    why it would be only 0.7v when the voltage divider produces 3.92v? Is it because there is a voltage drop from the current being used?
     
  17. SgtWookie

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    The base-emitter junction of a transistor acts like a forward biased diode.

    That's why the voltage doesn't rise much even when the current increases significantly.

    However, with a 10k resistor between the base and +12v, you'll only get 1.2mA tops through the be junction.

    Try an experiment - take a 10k resistor and any diode you have lying around. Put the two in series across your 12v battery or supply. See what you measure across the diode when it's forward biased, and when it's reverse biased.
     
  18. doug08

    Thread Starter Member

    Jan 30, 2011
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    I was looking at what was reduced from the 12.45v using the voltage divider. So since everything is wired perfectly....I should just swap out the ksp2222a with a 2n3904. If that fails, whats next? the lm358?
     
  19. SgtWookie

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    Until you can figure out how to get roughly 5.1v to 5.3v on pin 2 when Q1 turns on, you're dead in the water.

    Fix that first.

    As I mentioned in an earlier post, you may need another resistor between VR1 and ground (the battery - terminal). A 7.5k resistor should work quite well.
     
  20. doug08

    Thread Starter Member

    Jan 30, 2011
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    Would be either the transistor or the zener diode. Correct? The 1.8k ohm is definitely correct and connected to 12v+, and the other side to pin 2 and the zener. I can try the 7.5k ohm resistor in series with the POT and neg terminal.
     
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