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Discussion in 'General Electronics Discussion' started by Urleigh, Jan 11, 2020.

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  1. Urleigh

    Urleigh

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    Jan 11, 2020
    I'm a mechanical engineer suddenly drawn to the dark side. I'm learning a lot with my eBay purchased Radio Shack electronics learning lab and I'm intrigued by an effect I've noticed while playing around with the "Two Transistor Dual-LED Flasher" circuit (page 60 of Workbook 1: I'm assuming there are a few of these things about). I'm probing around the circuit with a volt meter, surprised to see 10V when the supply voltage is 7V. I'm thinking that there must be an effect similar to a "Charge Pump" going on due to the circuit's pairs of capacitors and transistors? Another head-scratcher is that the voltage measured across the capacitors has it's polarity reversed e.g. the terminal identified as positive on the capacitor case (which is consistent with the circuit schematic) measures negative relative to the other terminal. If anyone is familiar with this Radio Shack kit I'd be grateful for any insights into what's going on. Not being able to figure it out is driving me mildly nuts.
     
  2. Hunter64

    Hunter64

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    Nov 20, 2018
    There are probably a lot of people not familiar with this kit but who may be able to help if you post at least a schematic.
     
    HellasTechn and davenn like this.
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I think you refer to the kit described in this manual. Next time please add a schematic diagram of your circuit (e.g. a screenshot or a pdf).
    The operation of a transistor astable multivibrator (that's what this circuit is) is described e.g. here or probably better here, alas for NPN transistors (which is the way more common circuit). But the circuit works in the same way, only polarities are inversed.

    Assume Q1 is on (LED 1 is on) and Q2 is off. The anode of C1 is at +6 V (neglecting emitter collector voltage of Q1), the cathode of C1 is charged by R3 to a voltage towarss 0 V. The cathode of C2 is therefore at ~5.4 V (assuming a base-emitter voltage of 0.6 V for Q1). Once the voltage on the cathode of C1 falls below ~5.4 V (again assuming a base-emitter voltage of 0.6 V this time for Q2), Q2 will start to conduct. This puts the anode of C2 to + 6 V (again neglecting the emitter collector voltage this time of Q2). Since the voltage across C2 can not change instantaneously, the cathode of C2, the voltage on the cathode rises to a voltage considerably higher than the 6 V operating voltage. That is (imho) what you measure.
     
    hevans1944 likes this.
  4. Urleigh

    Urleigh

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    Jan 11, 2020
    Thanks Hunter 64 and Harald. Harald has provided an explanation it's going to take me a little while to digest and he apparently has found a copy of the learning lab manual.
    A bit late on my part, but point taken, I have added the schematic with a few of the voltage readings I am seeing.

    Clipboard01.jpg


    [MOD Edited - put image inline so people don't need to open a pfd]
     
    Last edited by a moderator: Jan 11, 2020
  5. Bluejets

    Bluejets

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    Oct 5, 2014
    While you are out searching on Google, bring up voltage doubler for another to keep you going.o_O
     
  6. davenn

    davenn Moderator

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    This is the very old Astable Multivibrator, do some googling on those to find out how they operate :)
    I remember building one some 50 years ago :)
     
  7. duke37

    duke37

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    Jan 9, 2011
    Those are pnp transistors and if they are germanium will withstand 7V. Silicon transistors should not have more than 6V on the base/emmitter junction so in this case limit the power supply voltage to 6V.
     
  8. Cannonball

    Cannonball

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    May 6, 2017
    Your caps are in backwards. It takes a positive voltage on the base of each transistor to turn them off.
     
  9. Urleigh

    Urleigh

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    Jan 11, 2020
    All I can say is that the schematic is copied from the workbook and the circuit does work. Having said that it did occur to me that, maybe, the capacitors should be reversed. As I tried to indicate with my additions to the schematic the anode of the capacitor is always negative with respect to the cathode alternating between 0V and -4.7V. With respect to ground the cathode of the capacitor (and hence the base of the transistor) alternates between +6V & +11.5V.

    I know that electrolytic capacitors are polarized, but apparently they will pass an AC current, so doesn't that mean that the voltage across them can also be reversed?
     
  10. Urleigh

    Urleigh

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    Jan 11, 2020
    Thanks for the warning. The learning lab kit runs off AA batteries which are nominally 1.5V , so with 6 batteries in series it provides 1.5V, 3V, 4.5V, 6V, etc. The fresh Lithium batteries I'm using actually provide 1.8V, so I'm a bit over-powered.
     
  11. Audioguru

    Audioguru

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    Sep 24, 2016
    I agree that the polarity of the capacitors is backwards.
    Here is a well-copied schematic on many internet sites showing NPN transistors and the supply voltage reversed.
    The capacitor polarity is correct but the circuit in the kit has the same polarity which is backwards.
     

    Attached Files:

    Cannonball likes this.
  12. Cannonball

    Cannonball

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    May 6, 2017
    In a pnp transistor to turn it on the base has to be at least .6-.7 volts lower than the emitter. To turn it off the base and the emitter must be equal or the base to have a higher voltage than the emitter. Turn your caps around and see what happens to the voltages that you recorded.
     
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I bet not much.
     
  14. Cannonball

    Cannonball

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    May 6, 2017
    Me too. When installing an electrolytic capacitor into a circuit the + side goes to the higher voltage point and the - side goes to the lower side point. This will not hurt the capacitor unless you exceed the working voltage of the capacitor.
     
  15. shrtrnd

    shrtrnd

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    Jan 15, 2010
    This has nothing to do with the expert advice given above.
    I just want to say that when I started-out many years ago with learning circuits, I found it very important to make sure
    I was using the correct 'ground' when using a voltmeter. Nothing to do with your circuit here, I just mention it because
    it sounds like you'll be doing a lot more tinkering with electronics, and measuring voltages across the correct circuit
    paths can often come into play when making those measurements, and not seeing the voltages you expect.
    ... end of unsolicited advice.
     
  16. Urleigh

    Urleigh

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    Jan 11, 2020
    Thank you everyone for the interest and input. I did try reversing the polarity of the capacitors in the circuit and it did not make any difference to the operation. I was also able to record the voltages at the points indicated in the schematic on post #4, above, and they are shown below: -

    [​IMG]
    And here is my attempt at explaining what is going on, and in particular how it is I am measuring 10V in a circuit with a 7V supply voltage: -
    • At “A” the voltage at the base of Q2 is 6.5V and since the voltage across C1 is 0V the voltage at the collector of Q1 must also be 6.5V.
    • At “A”, or shortly after, the voltage at the Q2 Base is low enough to cause Q2 to switch on. Current from the collector of Q2 flows thru C2, restricted by the R2 resistor, the voltage at the Q1 collector increases such that Q1 turns off.
    • When Q1 turns off the voltage at the collector of Q1 falls rapidly, so while the voltage at the C1 cathode remains at about 6V the voltage difference across C1 increases then levels off at 4.5V
    • At “B” C2 is charged to the extent that the current passing through it and R2, and therefore the voltage at the base of Q1, reduce so that Q1 switches on. (This bit is sketchy, but I’m thinking that, with Q1 on, the charge flowing into the C1 anode boosts the voltage at the cathode by a corresponding amount hence the voltage at the cathode (and Q2 Base) increases from 6V to 10V and Q2 turns off).
    • Between “B” & “C” the excess charge on the C1 cathode bleeds away through R3 and the cycle repeats.
    I'd be amazed if I haven't got something wrong, so I'm happy to be corrected.
     
    Last edited: Jan 15, 2020
  17. Cannonball

    Cannonball

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    May 6, 2017
    You are on the right track.
     
    Audioguru likes this.
  18. Audioguru

    Audioguru

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    Sep 24, 2016
    We need to know the part number of the transistors you are using. Then we can lookup the maximum allowed reverse voltage emitter-base breakdown voltage since your supply voltage is high enough to cause it.

    Each capacitor charges to about 4V backwards (the + end goes negative since the - end is positive). Backwards electrolytic capacitors conduct when they should not conduct.Then the collector of the transistor connected to the charged capacitor turns on, which drives the other end of the capacitor (connected to the base of the other transistor) to a positive voltage that is 6.5V higher than the 4V. Then the base-emitter voltage of the transistor is reverse-biased and is much higher than the maximum allowed 5V to 6V.

    A transistor is slowly damaged by repeated breakdown of its reverse-biased emitter-base junction so usually a lower supply voltage is used or a protection diode is added in series with each base to prevent the breakdown.

    The backwards capacitors conducting and the emitter-base breakdowns change the calculated frequency of the oscillator.
     
  19. Urleigh

    Urleigh

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    Jan 11, 2020
    The transistors are marked "9015". Should I reverse the capacitors (i.e. anode to transistor base) and record the voltages again? In terms of the LEDs flashing on and off that seems to make no difference, but it might make a difference to the voltages measured across the various components.
     
  20. Audioguru

    Audioguru

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    Sep 24, 2016
    Of course you should reverse the backwards capacitors so that their polarity is correct.
    The datasheet for the transistors says the absolute maximum allowed reverse emitter to base voltage is 5V and without protection diodes your circuit gives them much more, which will slowly destroy them.
     
    Cannonball likes this.
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