Transistor Switch Circuit Problem

Discussion in 'The Projects Forum' started by julchak, Dec 18, 2014.

  1. julchak

    Thread Starter New Member

    Dec 18, 2014
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    Hi everyone,
    I'm very new to electric circuit design, a mechanical engineer here so excuse me for my lack of knowledge on the matter. I worked with another coworker to put together this circuit however am running into issues which neither of us can figure out and was hoping that I could use your guys' knowledge to help guide me through this troubleshooting, as I have a feeling it is probably a rather simple problem that us MechE's are just overlooking. Note: I've linked each of the components listed to their respective datasheets for reference, and vendor part numbers.

    The circuit is a control circuit which I am putting into a testing device in order to ensure that the lithium ion battery does not succumb to a low voltage which would damage it - the circuit's intention is to provide 2 set points which will trigger different battery protection actions:
    1. Signalling a low battery red LED, turning on the LED at a battery level of ~7.3V
    2. Signalling a relay to switch off and kill the system when the battery level reaches ~6.3V
    We've put together this control circuit using a NO pushbutton (SB4011NOH-2F) for the power on button, a NC button (SB4011NCH-2C) for the off button, a Omron G5V01 SPDT Relay (653-G5V-1-DC6), two transistors (ON Semiconductor TIP31AG - 863-TIP31AG), resistors, an LED (67-1148-ND) and a fuse (486-2007-ND).

    I tried to draw a diagram of how the circuit is arranged:
    [​IMG]
    With this set up the transistors and everything works just fine at ambient temperature (~25°C), however when I've tested it in a freezer at about -10°C the transistors act really weird, prematurely giving off signals which light up the red LED at a voltage higher than the 7.3V set point, and will get to a point where the system shutdown transistor is pretty much flipping the relay, such that if you hit the NC off pushbutton the system will shut down and not turn on again (as the transistor is not permitting it). You are able to simply put your finger on either transistor and you can notice either the relay coil voltage getting further away from the trip voltage, or the red led fade away....it seems as though this is a temperature issue? I used to have a 10kOhm where the 1.05 kOhm is now and a 72kOhm where the 10kOhm is - we changed these to lesser resistances thinking this would fix the problem to no avail.

    The only thing is that these transistors are rated to -65°C so we are just completely flabbergasted at the whole problem.

    Is there something simple in the design of the circuit that we are overlooking? Are we using the transistors in an appropriate manner for this control circuit?

    ANY help would be GREATLY appreciated...I've been hitting my head against the wall for weeks over this problem.


    THANKS,
    Jeff
     
  2. MikeML

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    The way you are detecting the two trip voltages is very primitive: very temperature dependent, and very dependent on the β of the transistors. Do you want a suggestion that will provide much more precise and stable trip points, using a device that costs pennies?
     
  3. Alec_t

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    The base-emitter voltage of a transistor reduces by about 2mV per degree Celsius increase, leading to drift of your 'reference' voltages. Better to use a proper stable reference voltage generator and a couple of comparators to control the LED and relay.
     
  4. julchak

    Thread Starter New Member

    Dec 18, 2014
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    Yes Mike, I'm very open to any suggestions. Any information you can give me on how to set up a better circuit would be appreciated - very new to this all so a straightfoward pathfoward would be ideal.

    Ah well there we go...How would I go about this stable reference voltage generator and comparator control circuit?
     
  5. MikeML

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    The 150mW that the relay coil draws from the battery could be a significant part of the battery's total load? What load does the battery/relay ultimately supply?
     
  6. julchak

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    Dec 18, 2014
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  7. MikeML

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  8. julchak

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    Dec 18, 2014
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    Hmm, yeah, I didn't realize this. This was the only way we found to be able to make this battery level indicator/shutdown circuit with our knowledge. What is the alternate layout you are thinking of?
     
  9. MikeML

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    Using a PFET. Has an on-resistance of a few mΩ, and takes zero power to switch it. You still want separate NO and NC momentary switches to Start and Stop, or would a simple ON-OFF toggle switch do (as long as the battery drain goes to zero after it discharges below 6.2V)?

    After recharging the battery, you would have to move the toggle switch to OFF momentarily, and then back to ON to begin a new discharge cycle. The toggle switch has the advantage that it mechanically indicates the state, while the latching relay does not.
     
  10. julchak

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    Dec 18, 2014
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    We don't want to drain the battery below the low set point of 6.3 V, for design purposes it would be easiest to keep the pushbuttons (already have enclosures made up), but if this toggle method would be more efficient/practical then I could make the switch to it.

    Not sure if I fully understand how it would work though? Does it mechanically flip to OFF at the setpoint of 6.2V? Is there a way to keep the system off when in storage? If you accidentally flipped it to off momentarily would it just immediately begin discharging until the setpoint is reached? If so then this is not ideal for the application. We want to be able to Turn the system on, run it till the 6.3V kill setpoint, recharge it, use it again, or be able to just turn the system off anytime there is still battery left.
     
  11. Alec_t

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    I agree with Mike that a solid state switch would be preferable to a relay. However, if you wanted to retain the relay, here's one option:-
    BatteryMonitor.gif
     
  12. MikeML

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    Here is my version. I am using a PFET as the latch M1. V1 is the battery. As before, there is a normally-closed OFF button S1, and a normally-open ON button S2. I am using two LM431/TL431s as voltage comparators, comparing an external voltage to their internal 2.495V reference. U1 switches at 6.29V and U2 switches at 7.31V. U1 lights the Led(s) at 7.3V and everything goes off at 6.29V. R6 simulates a load (not part of the circuit).

    247a.gif

    Follow the traces in the simulation. V(on) and V(off) define the times when their respective push buttons are asserted. The latch turns on V(load) jumps to V(bat) at 1s and at 5s. At 3s, the Off switch is pushed, turning off V(load). V(load) also turns off at about 18s because the battery voltage reached 6.29V. As the battery discharges from 8V down to 6V, the LED turns on at 7.3V and is turned off automatically by U2...

    Why are there two LEDs in series? When the cathode of a TL431 switches to the low state, the voltage between cathode and anode is ~2.2V, which is higher than the Vf of a Red LED. By putting two Red LEDs in series, the combined Vf of two guarantees that they turn off when the TL431 switches low. If you used a single White LED with a Vf of about 3V, then you only need one LED. You can hide one of the LEDs inside the box. Think of it as a Zener diode with a Zener voltage of ~1.8V...

    When the latch is on, the current from the battery is the load current, plus about 1mA for U2, plus about 4mA for U1 (including the LED) , plus about 400uA for the R1-3 voltage divider. When the latch turns off, the leakage current from the battery will be about 1uA.
     
  13. julchak

    Thread Starter New Member

    Dec 18, 2014
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    Alec and Mike, thanks for your replies!

    I am game to change the relay - as it looks I'll have to get new PCB's anyways so I can fit the new components/layout; if taking the relay out saves 15% battery I'm game.

    Mike,
    As for your recommended circuit, I do want to make sure that the rest of my circuit won't affect what you've made up and that everything will in fact work like we believe.

    The confusion with the 2 LEDs with series doesn't quite make sense to me, but I think that I may have been unclear about the 2 LEDs....one is a green led to show the on condition which powers down when the 6.3V setpoint is hit (with the rest of the system), the other is the red led to alert of a low battery at 7.3V. I've added on the rest of the circuit which I didn't show here (as I didn't think it mattered in this equation, I may be wrong) - you will see a 24V voltage regulator (OBQ24SC0512), a green LED (SSI-LXR1612GD), a signal conditioner (ICA4H), a 3.6 V voltage regulator (LDO03C), and a data logger (PROCESS101A).

    Let me know if this throws any foreseeable curveballs into the circuit you are showing me here?

    Also, Why does the battery drain when in the off position? Is this unavoidable?
    [​IMG]
     
  14. Alec_t

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    Here's a totally different approach, relying on the inherent switching threshold of a CMOS logic gate changing as its supply voltage Vdd changes. An LT1634 is used as a 4.096V reference, because it can function (if the sim can be believed) with only a few uA current. Standby current after the battery drops below the 6.3V cutoff point is only ~6.5uA. Sim shows the switching points are stable over a -20C to +40C temperature range :-
    BatteryMonitor3.gif
     
  15. julchak

    Thread Starter New Member

    Dec 18, 2014
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    Hmm...I like this type of approach with the comparator that you guys are doing, and the minimal current draw on the battery shouldn't be a problem I would hope with the battery dropping to a dangerous level (~6V) unless it were to sit on the shelf for years.

    Just one thing - with the new diagram I posted up above I show the rest of the circuit to which this battery monitor circuit feeds power to. I have a 3.6V voltage regulator (LDO03C) and a 24V regulator (OBQ24SC0512). Would it be possible/advantageous/beneficial in any way to use one of these as my comparative reference voltage supply?
     
  16. MikeML

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    Here is a mod to my circuit which improves the LED usefulness. Per your request, the circuit now has two LEDs. The Green one is on any time the Latch is On. Initially while the battery is fully charged, the Red LED is off. As the battery discharges through 7.3V, the Red LED turns on, and the Green LED gets slightly dimmer, but stays on. When the battery discharges through 6.3V, the latch is cleared and everything shuts off.

    With the latch on, the total battery discharge current is the load current plus about 5mA (most of which flows through the LED(s)). With the latch off, less than 2uA of leakage.

    249b.gif
     
    Last edited: Dec 19, 2014
  17. julchak

    Thread Starter New Member

    Dec 18, 2014
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    Alright awesome. And there shouldn't be any effect from temperatures between -40°C and 100°C on this circuit?
     
  18. julchak

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    Dec 18, 2014
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    I noticed that you used different LED's. Is there any way that I would be able to use the same LEDs I had used before? Again I had designed the enclosure with ports for those LEDs and have them installed (more of a panel mount type of an LED)?
     
  19. MikeML

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    The effect of temperature on both switching points. Green is -40C, 100C is dark Blue. Steps are 20C each.

    247t.gif
     
  20. MikeML

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    As long as the ones you are using do not have an internal resistor, like most panel mount LEDs have. To work as simulated, the Vf for the Red and Green LED should be between 1.8V and 2.2V.
     
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