11V-14.8V cutoff circuit

Discussion in 'General Electronics Chat' started by hazim, Sep 12, 2011.

  1. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Hi
    I want to design a circuit that is a low battery cutoff and overvoltage cutoff or in other words it has an output connected to the input supply voltage which is a 12V battery, when the battery's voltage is either less than 11V or over 14.8V then the output disconnects from input... I used LM358 and tried to build a circuit that does what I've described, I thought it would be easy but when I cam to practical I faced some problems.. Anyway I want your ideas/recommendations/help/guidance... :)

    In the attachment the lamp represents a relay.

    Regards,
    Hazim
     
  2. eblc1388

    Senior Member

    Nov 28, 2008
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    Starts by connecting the LM358 positive supply to 12V and not +5V to increase its common mode voltage input range.

    However, you should really consider using the real LM393 comparator to replace the opamp LM358, which you have wired as a comparator.

    Edited: Added simulation of Opamp power from +5V Or from battery. R8 or R14 is the relay. The cutoff limit is approximately correct to OP's requirement

    [​IMG]
     
    Last edited: Sep 12, 2011
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  3. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I didn't get LM393 comparators yet. Using LM358 with positive supply connected to the +12V source, the circuit worked fine but after replacing the base transistor of 2N2906 with 47k one because when the battery's voltage decreases under 11V, the voltage across the relay decreases to about 1.5V. At 1.5V the relay's contacts still not released (i.e. the relay doesn't switch back off...). Replacing R2 (attachment in post #1) solved the problem. I have a lot of LM358 and I need to make about ten of this circuit for someone and for me also. I still need to add hysteresis.
     
  4. eblc1388

    Senior Member

    Nov 28, 2008
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    Yes, you do definitely. It is not easily done in your present design.

    The battery voltage will most likely recover to almost 12V when its load is cutoff.
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    Is your battery a lead-acid type? As that's about what the voltage range seems to be.

    If you drain a lead-acid battery down that low (11v), the battery will not last very long, as that is a very deep discharge. You should really disconnect the load once the battery is discharged to just above 12v.
     
  6. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    I knew that you will tell me that. You always say this thing and you told me that before one or two times. I want to tell you that I knew about that, but in real life things may go far from theory and especially here in Lebanon where everything is contrary and degeneration everywhere and billions of dollars where spent for electricity service in that small country and this service is still 12/24... Here every home uses big lead-acid batteries, there is UPS and inverters everywhere, the UPS/inverters are designed to work until the battery's voltage drops to 10.5V. Usually batteries are lasting from 2 to 4 years. If a battery on a certain load will drop to 12V after 2 hours and 10.5V after 5 hours, and if it lasts 5 years when not draining it under 12V and will last only 2 years when draining it till 10.5V, then everyone here will choose the second option.
    The circuit will stop draining the battery actually under 10.5V and not 11V. I'm building it for a local ISP that uses batteries for wireless stations, without it believe that the batteries will drain much lower than 10V.

    BTW, I've made few changes in the circuit and it's working well now. I may make one more editing in it. I'll post the final circuit when finished.

    Regards,
    Hazim
     
  7. Smoke_Maker

    Active Member

    Sep 24, 2007
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    Hazim

    Your right, a 12 volt lead acid battery is dead when voltage drops to 10.5 volts "under load" at what ever that load is, if you decrease the load you will be able to get more energy out of the battery. When you discharge a battery at the rated discharge current to 10.5 amps and then disconnect the battery it will read 12 volts, a battery fully charged is 12.6 volts and when completely dead it is 12.0 volts.

    11 volts is a good low point for under load, for a deep cycle battery.
     
  8. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    Ok, this is my final design for this circuit. I have some notes on it..

    First, using LM393 is better than LM358 because LM358 is not designed to be used as comparator but op-amp, while LM393 is a real comparator. Using LM393 will require some resistors' values to be changed.

    Using the exact resistor values in the circuit, the load will be turned off at battery voltage less than approx. 10.5 or greater than approx. 14.8V.

    Of-course a relay should be used for loads that draw more than 0.5A.

    I small problem still in the circuit but isn't a problem for me; when the voltage increases above 10.5V, at approx. 11V the relay chatters a little if the load is connected and the increase of the voltage is very slow.

    I thought by adding the MOSFET that this will protect the circuit from reverse polarity connecting. But I tried that connecting the power supply in reverse and the IC LM358 was burnt out.

    Any idea about the last two points?

    Regards,
    Hazim
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    LM393 comparators have open-collector outputs; they can sink current, but cannot source it. This will cause a problem with U1B, as it needs to be able to source current in order to turn on the NPN transistor. Usually, pull-up resistors are used to source current, but that would goof up your hysteresis feedback.

    Your 7805 & 1k resistor are a problem. Since a 7805 sources 5mA to 5.5ma from the GND pin, that will increase the output of the 7805 by 5v to 5.5v for a total of 10v to 10.5v. However, the 78xx series has a dropout voltage of ~2v; so when your battery is nearly dead, your reference voltage divider will no longer be accurate. As a matter of fact, your battery has to be at 12.5v in order for the regulator to output 10.5v. Below that, the regulator output won't be regulated anymore.

    In order to get a regulated output, you'll need to go no higher than 8v. This means that your R8 needs to be somewhere around 510 Ohms to 540 Ohms.

    Since 7805 regulators have a minimum ~5mA current, it will continue to drain the battery after the load is removed.

    Q3/R11 won't work as you have discovered. The reason for this is the MOSFETs' body diode; it conducts when the source is ~0.7V more positive than the drain. If you want to protect against reversing the power connections, then use a diode (Schottky preferred) in the positive lead.
     
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  10. Smoke_Maker

    Active Member

    Sep 24, 2007
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    Hazim,

    Put a small cap across R9 and R10 to slow down the feed back change.
     
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  11. tom66

    Senior Member

    May 9, 2009
    2,613
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    It is possible to use that circuit as a reverse polarity protection device. You need a PMOS though; the IRFZ44 is an NMOS device. http://www.geofex.com/article_folders/mosswitch/mosswitch.htm
     
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  12. hazim

    Thread Starter Active Member

    Jan 3, 2008
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    SgtWookie, thank you for the information about LM393. What you said about the voltage regulator is right, I even know it, that's why in the first circuit in post one I put 470 Ohm resistor instead of the 1k one. I used 1k because the output voltage was low with 470 Ohm, it was was about 6V only. At first, I found contradiction between what I've tested and theoretically, I then found that the voltage regulator I used has lower quiescent current. It is actually 78L05 means it is TO-92 package. The quiescent current is 2mA, 2mA x 1kOhm = 2V :). 5+2=7V, the difference with 10.5V is 3.5V. It's my fault since I didn't mention it's 78L05 and not 7805.

    Thanks tom66 for the link. I always say there should be some way to make a reverse polarity protection using a MOSFET and get red of the voltage drop of a protection diode. I was thinking of using a p-channel MOSFET but not is reverse direction as in the link, in this picture:
    [​IMG]
    I'll try it tomorrow.
     
  13. hazim

    Thread Starter Active Member

    Jan 3, 2008
    419
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    Hi again.
    This is the final design of the circuit. by the two potentiometers I set the upper and lower voltages. the MOSFET reverse battery protection is working great. On thing still to fix is the problem I've talked about before. The problem is as follows: when the voltage of the battery is increasing slowly from say 10V to 12V, and because of the hysteresis the relay should switch at say 12V, what happens occurs only when the voltage increase is slow, and the load is connected, the relay doesn't switch normally but chatters till the voltage is increased a little more. I tried to add a capacitor in parallel with the hysteresis and tried a variety of capacitor values but it didn't work. I tried to but a capacitor between pin 7 and ground and also tried different values, with big capacitors the chattering becomes slow but doesn't stop.

    Chattering is happening only when load is connected and it's obvious that when the relay switch on the load will be connected to battery and thus decreasing it's voltage and so the relay switch back off and then the voltage rises an so on...

    What should I do to solve this problem
     
  14. oldtech33709

    New Member

    Sep 24, 2011
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    0
    Try moving Q3 after comparator return, U2 may be oscillating due to the changing "on" resistance of Q3 in a linear state.
     
  15. hazim

    Thread Starter Active Member

    Jan 3, 2008
    419
    13
    I tried connecting directly without Q3 but didn't solve the problem
     
  16. Smoke_Maker

    Active Member

    Sep 24, 2007
    126
    15
    Your are going to need a diode before R6 in think, your reference keeps bouncing.
     
  17. oldtech33709

    New Member

    Sep 24, 2011
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    0
    If the regulator is being used for 5V then R8 is not needed, tie common of regulator return. There should be a .01uf cap across the output of the regulator for stablity. If R8 was being used to ajust the regulator output then the bottom legs of R1 and R3 should be tied to the top leg of R8 and not return.
     
  18. Smoke_Maker

    Active Member

    Sep 24, 2007
    126
    15

    Wooops, move that 914 down to the positive post of the circuit, relay contacts on different wires.

    before you change it put a scope on it and see how much the comparator reference voltage is jumping when the relay chatterers.
     
  19. hazim

    Thread Starter Active Member

    Jan 3, 2008
    419
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    I tried what you said but that didn't solve the problem. Smoke Maker I didn't understand your last post, what is the 914 you want me to move?
     
  20. hazim

    Thread Starter Active Member

    Jan 3, 2008
    419
    13
    tom66, SgtWookie, eblc1388... where are you? I believe you have the solution :)
     
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