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Discussion in 'General Electronics Discussion' started by gregfox, Feb 12, 2014.

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  1. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Problem: The on/off circuit of the lab power supply is to be powered by 4 LiOn bats, 16.8V fully charged. The 4013 will not turn on at the higher voltage, but will turn on if the voltage is taken down to 13V. The chip is good for 18-20V (Digikey part MC14013BDTR2GOSCT-ND) I can’t determine what is causing this. Do you have any ideas?
     

    Attached Files:

  2. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    flip flop will not turn on

    Problem: The on/off circuit of the lab power supply is to be powered by 4 LiOn bats, 16.8V fully charged. The 4013 will not turn on at the higher voltage, but will turn on if the voltage is taken down to 13V. The chip is good for 18-20V (Digikey part MC14013BDTR2GOSCT-ND) I can’t determine what is causing this. The clock is good at the higher voltages. Do you have any ideas?
     

    Attached Files:

  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I'm not sure if there's any different information in your attachments so I merged your two threads rather than deleting one f them.
     
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Hi Greg,

    Let me get this straight. Correct me if I'm wrong.

    The first half of the 4013 (top, in the diagram) generates a positive pulse on its Q output each time the pushbutton is pressed. This pulse is supposed to turn the circuit ON if it's OFF, and OFF if it's ON.

    The second half of the 4013 controls whether the circuit is ON or OFF. It is wired as a toggle flip-flop, clocked from the output of the first half. It is used with its output "backwards", i.e. when its Q output is high, the circuit is OFF.

    When the supply voltage is 16.8V (the maximum expected), the clock pulse from the first half is working properly, but the second half never turns ON.

    According to the voltages you've shown in the table on the schematic, pin 14 voltage is only 11V when the circuit is powered from 16.0V. There's something wrong there.

    My guess for why the circuit isn't turning ON would be that pin 8 is not properly low. This voltage (the voltage on C3) should be pulled pretty close to 0V by D3 and R4 when the circuit is OFF, because in that state, pin 12 should be low.

    The voltages in your table seem to be measured with the circuit ON. This won't help us figure out why it doesn't turn ON when the supply voltage is at maximum. But there is a problem. Pin 8 should not be 3.2V when the circuit is ON, because the TL431 should be pulling it low (provided that the undervoltage cutoff trimpot is set correctly).

    That should give you a few things to check out.

    BTW the circuit would look a bit tidier if you moved D2 and C3 rightwards, to be next to R5 and the TL431. R4 can connect just above C3.
     
  5. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Hi Kris,

    You said:
    “The first half of the 4013 (top, in the diagram) generates a positive pulse on its Q output each time the pushbutton is pressed. This pulse is supposed to turn the circuit ON if it's OFF, and OFF if it's ON.

    The second half of the 4013 controls whether the circuit is ON or OFF. It is wired as a toggle flip-flop, clocked from the output of the first half. It is used with its output "backwards", i.e. when its Q output is high, the circuit is OFF.

    When the supply voltage is 16.8V (the maximum expected), the clock pulse from the first half is working properly, but the second half never turns ON.”
    You are 100 percent correct.
    I tidied up the circuit and it does look better, thanks.
    I have included a new drawing with the correct voltages and in the ‘OFF’ condition. The cutoff seems to be around 13.4V. I soldered in 3 new chips(4013), and the results were all the same. I’m wondering if it may be caused by the propagation time delay of the 4013. It’s hard to understand why it would cut off at 13.4V The Q output of the 2nd flip flop will go low for the duration of the pulse of the first flip flop, then back to high..
    If there are no solutions at hand, perhaps I should try a different manufacturer.
    Many Thanks.
     

    Attached Files:

  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Try shorting pin 8 to ground. If that allows it to turn ON, find out why pin 8 is being pulled up when the circuit tries to start up. The TL431 should be pulling it low when the switched supply rail goes higher than the low voltage cutoff.
     
  7. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    So does the circuit now work at 16 volts? What is cutting out at 13 volts?. I did notice and I think Kris mentioned it. Before switch on set and reset are both zero, is this a proper condition for the device?

    Adam
     
  8. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Sorry so long to reply, dealing with 2 feet of snow.
    Yes even with pin 8 shorted to GND the same thing happens, that is; at about up to 13.5V the second flip flop acts as expected (Q pin 13 going from logic 1 to logic 0) but at any voltage above 13.5 Q stays at logic 1.
    I don't see why this is happening, it works well as long at the voltage is kept below 13.5. I tried 4 of the same chips to no avail. I isolated the circuit and no change. I increased C2 to get a good look at it on the scope and C2 pulses for 500mS (from logic 0) while Q in the second flip flop pulses 500mS (from Logic 1)
    Somehow I have a feeling the problem is in the chip itself. I ordered new ones from a different manufacturer and will try them when they arrive.
    I don't know what else to try.
     
  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Hmm, that's strange.

    Perhaps the second flip-flop is clocking again on the falling edge of its clock input because of some noise or poorly defined voltage change.

    Try this. Lift pin 9 and tack a 10k resistor between the lifted IC pin and pin 7. Press and hold the pin onto the pad, power up the circuit, and make sure it's OFF.

    Then lift the pin off the pad and push the power button once. See whether it powers up.

    If it does, there may be some kind of glitch on the falling edge of the signal from pin 1. Try inserting a 1k resistor in series with D1.
     
    Last edited: Feb 14, 2014
  10. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Well, how do you like that! I was just about to try your experiment, when the new 4013 came in. The new chips are TI CD4013BPWR (digikey 296-14089-1-ND ) I quickly soldered one in and presto… it worked across the range up to +20V (the high end of its range).
    So ON=no work; TI=works!
    I’ve been looking at the two datasheets and found just a few differences, but it was evidently a problem with ON semi. Who would-a-thunk? but why?
    Thanks for all your help!
     
  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Interesting. I think it's possible that the TI part has a Schmitt trigger on the clock input. The data sheet doesn't show one; it shows a standard gate, but it's a very old data sheet (a scan, says it was acquired from Harris) and it would be possible to add a Schmitt trigger and still comply with the old specifications, so it's a possibility.

    If you had tried the test I suggested in post #9 before you tried the new 4013, we might know whether that's the explanation or not. I would be interested to know, but that would require you to remove the new 4013 and put the old one back, and I wouldn't ask you to do that.
     
  12. gregfox

    gregfox

    149
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    Mar 25, 2013
    I will try the test. I have a heat station so removing the part is no problem.
     
  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK cool. You could also try connecting a 0.1 µF ceramic or multi-layer ceramic capacitor directly from pin 14 to pin 7 with the leads cut as short as possible.
     
  14. Electrobrains

    Electrobrains

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    Jan 2, 2012
    Hello Greg
    I did not look into your full circuit concerning the logical function.

    I just quickly saw the polarity protection section of the diagram has an error!

    You have a P-MOSFET there (2SJ334). It has the Drain and the Source interchanged.
    You must put the Source on the input side (and the cathode of the Z-Diode should follow the source)!
     
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    No, the reverse polarity protection circuit is OK. Here's how it works.

    When the battery is connected the right way round, current flows through Q2's body diode and powers up the circuit connected to its source. The gate is pulled negative by R3 and Q2 conducts, eliminating the voltage drop of the body diode.

    If the battery is connected backwards, the body diode does not conduct. Q2 has a negative drain-to-source voltage, which is normal for a P-channel MOSFET. Any leakage through Q2 will pull the source slightly negative relative to Q2's gate which is the opposite polarity to that which would turn Q2 ON.

    You could argue that a backwards connected diode should be added across the circuitry because of possible leakage currents.

    BTW, it's nice to see you back on Electronics Point, Electrobrains :)
     
  16. Electrobrains

    Electrobrains

    259
    5
    Jan 2, 2012
    Hi Chris! Yes it's nice to see you all here again.
    I like this site. There are often very good discussions here.

    About the polarity protection: Sorry, yes you are right! I was too quick with my comment.
    In fact I have used that circuit several times (but mostly protecting on the minus side with an N-MOSFET).
     
  17. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    OH well, I was removing the part for the 4th time, and this time most all the pads on this TSSOP part lifted, as a consequence I will not be able to perform the test. I guess I need a new board, it was time anyway, too much experimentation.
    As far as the cap goes, I had a bypass .01 cap in the circuit.
    The polarity protection Mosfet confused me at first, as I kept looking at it and wondering if I placed it correctly, then finally figured it out.
    However, to have a complete Lab supply with polarity protection, automatic low battery cutoff, current limiting, and less then 10mV of ripple on a 50X50 mm board blows my mind. The only thing I regret is the combo voltage current meter. It’s very nice and quite accurate enough, but when you press the button to see the internal battery condition, it disconnects the power to the load.
    Search ebay.com for VOLT-Amp-Combo-MeterRegards, (nice meter).
    Regards
    Greg
     

    Attached Files:

  18. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Right, I'm not surprised that the pads lifted off after four removals of the component!

    You have a decoupling capacitor, but it's not very close to the 4013 and it's connected through thin (inductive) tracks. I think it's worth trying a cap straight across the pins.
     
  19. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Right you are, since I need a new board, I'll put the cap where it should be.
    Thanks
     
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