555 timer burning out, transistor base resistor calculations, PEMF

Discussion in 'The Projects Forum' started by Rittter, Jun 12, 2016.

  1. Rittter

    Thread Starter New Member

    Dec 5, 2015
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    Attached is a schematic of my circuit.
    1) I am burning out 555 timers, 9 to be exact. I am using a 12Vdc/3Amp wall wart as a supply. Objective is to maximize the voltage and current to a coil to generate an EMF. The supply voltage is well within the specs. In addition, there is a massive actual voltage drop vs calculated drop to the transistor. Any ideas would be greatly appreciated.
    2) Also on the schematic are three methods I have found to calculate the value for the base resistor of a transistor. Which one if any is of any use ?PEMF Transistor 6121612062016.pdf
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
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    Don't see offhand why timers are burning out.

    What is the purpose of the diode in series with the base resistor? It isn't needed.

    You should use the second method for calculating the base resistor value. With 6.9K and the diode, there isn't enough base current to saturate the transistor. Remove the diode and use 1K.

    What diodes do you have in parallel with the inductor? Is your objective to have a large EMF when the transistor is switched off? I'd expect the transistor to be the component burning out.
     
  3. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Your snubber looks a little strange. Are you sure it is correct? If so I think the 2n2222 is also in danger.
    But having said that add some decoupling to the 12 volt power supply close to the 555. Better yet add a small resistor between the +12 and the 555 then add the decoupling.
    Maybe 10 ohms and 100ufd.
     
  4. Rittter

    Thread Starter New Member

    Dec 5, 2015
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    Thank you. The purpose of the diode is to protect the timer.

    I will remove the diode. Originally, it was not there. After burnouts, I added it. I agree, but isn't that an excessive amount of voltage drop vs what one would calculate ?

    I am using C30 MIC 30v diodes. My objective is to maximize the emf when the transistor is off. I agree, the transistor should be burning out but is not heating up at all. These burn outs take time (5-15min.). The 555 is cool until it isn't !

    I used this snubber set up to hopefully give the coil an additional kick. 99% of the schematics out there use one diode. I was thinking of going to one diode, an FR20 MIC. I still don't see why flyback voltage, in such an explosive manner, can decide to take a dirt road through a diode and not head for the transistor, arc the transistor and toast everything. I have thought about a resistor at the supply, but I am not familiar with "decoupling". I will check it out.
     
    Last edited by a moderator: Jun 15, 2016
  5. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    How did you measure it? A bog-standard DMM won't give an accurate reading when measuring pulses. It gives a 'sort of average' reading.
    It's probably voltage spikes which are getting into the 555 supply and killing it. Hence the need for decoupling the supply bettter, as others have stated.
     
  6. AlbertHall

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    Jun 4, 2014
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    Agree about 1k base resistor.
    The snubber circuit isn't actually 'snubbing' as there is no discharge path for the 1uF capacitor. It will charge to the peak back emf generated. What voltage do you read across that capacitor?
     
  7. AnalogKid

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    Aug 1, 2013
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    Method #1 is the classic way to go. Of your three, it is the only one that takes into account all of the factors that affect the base current. Once you have the theoretical minimum base current, you can adjust it for hard saturation. "Adjust" is where there are lots of opinions. The most common is to ignore the theoretical required base current, and just divide the collector current by 10. I think this leads to beating up the transistor with excessive base current. For modern, well understood parts such as the 2N2222, I prefer dividing the collector current by 20. The difference in saturation voltage is literally unmeasurable with a 3-digit meter.

    For the maximum generated EM field, the shortest possible transistor turn-off time is critical. This means you want the 555 to suck excessive base charge out of the transistor when turning it off, which means the base diode isn't just unnecessary - it is exactly the wrong thing to put between the 555 output and the base. One possible reason for why the transistor is not failing is that with no base DC path to anywhere when the 555 output is low, it is turning off to slowly to generate a really high inductive kick. Deleting the diode fixes this. For an even faster turn-off, put a small capacitor in parallel with the base resistor.

    For the next part, assume that the forward voltage for the zener diodes is approx. 1.0 V. As a guess, this is more accurate for power diodes than the standard 0.6 V.

    Separate from that, note that if the two diodes across the coil are 30 V zeners, they are not zening when the coil flying back. (Horrible tech-talk, I know, but you get the idea.) When the transistor turns off and the inductive kick happens, the collector goes positive *above* Vcc, and the two diodes forward conduct (ignoring the cap for a moment). So they are clamping the peak coil voltage at about 2 V above Vcc, not 60 V, and dumping the coil current onto the 12 V supply rail. The resulting voltage spike (no decoupling shown) can be large enough to damage the 555 over time, but not high enough to damage the 60 V transistor. But you can't reverse the diodes, because then you have a 2 V clamp from Vcc to GND. With the cap in there, charged to 12 V when the transistor is off, the output now clamps at 14 V above Vcc, still not 60 V. What to do...

    One approach is to reverse one of the zeners and increase both to 60 V. Now you have a series diode pair that clamps at 61 V in either direction.

    There is a another way. If the idea is to clamp the peak collector voltage at 60 V, or the peak voltage across the inductor at 72 V, or whatever, another way to do this is to place one zener in parallel with the transistor, not in parallel with the coil. This means the clamping voltage is independent of Vcc (which may or may not be a good thing, up to you). Also, the clamping current now is returned to GND directly.

    ak
     
  8. gerty

    AAC Fanatic!

    Aug 30, 2007
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    You probably have done this, but.. Have you measured the 12 volts from the wall wart ? Unregulated ones can go as high as 20 volts, which would eat your 555
     
  9. EM Fields

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    Jun 8, 2016
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    1. One of the problems with bipolar 555s is that there's a large shoot-through current in the output stage when it transitions in either direction. The cure for that is to connect a 100nF ceramic capacitor directly across pins 1 and 8.

    2. You don't need the diode on the 555's output, and for switching applications like this one I like to force the transistor's beta to 10 by running 1/10th of the collector current into the base. Since you're running about 100 mA of collector current, that would mean forcing 10mA into the base. With a drop of about 0.7 V across the base-to-emitter junction and a volt across the 555's totem-pole, that would leave you with about 10.3 volts to drop at 10mA, so your base resistor should be about 1000 ohms.

    3. It wouldn't hurt to connect 100μF or so from +12V to GND close to the coil.
     
    Last edited: Jun 14, 2016
  10. AlbertHall

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    Jun 4, 2014
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    I don't agree with most of that. First there is a capacitor so you cannot ignore it but more on that later. When the kick happens the current through the diodes is not dumped onto the 12V rail but instead it is flowing in the coil and the diodes (and it is charging the capacitor). The capacitor is connected via a rectifier (albeit two diodes in series) to the coil. For each cycle the energy stored in the inductor (100mA, 148mH) is 0.67mJ. A 1uF with that much energy stored in it would have about 36V across it. Each cycle the voltage across the capacitor will increase until there is some discharge path.

    During the time the transistor is switched on, there is 12V across the coil and the capacitor is connected across the coil by the two diodes, now reverse biased. So when the capacitor voltage reaches 60V (the two zener voltages) minus 12V across the coil, the capacitor can discharge back into the coil. Thus I would expect the capacitor voltage to be 48V - very, very close to its maximum rating.

    Hence my question about the voltage across that capacitor - that's the real measure of whether the circuit is working or not.

    I can see no reason, from the schematic, for the destruction of the '555.
     
  11. AlbertHall

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    But that would kill the back EMF, and getting that higher voltage is the point of the circuit (I think, if not what else is it doing).
     
  12. AnalogKid

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    Aug 1, 2013
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    AH - You are right about the inductor discharge path not going through the +12. I had 2 references to that, killed one, missed one.

    If, for the sake of analysis, you reduce the snubber/clamper to one perfect 60 V zener (Vf = 0 V, Vz = 60 V) and the cap, then when the transistor is on the first time the cap sits with 0 V on it, its terminal voltage unchanging, and there is zero diode current. When the transistor turns off the first time the inductor kicks up to Vcc + 0 V, the diode forward conducts, and then the cap starts charging. Back to the real world, the charging current is limited by the inductor's inductance, Q, and resistance, and the diode's actual forward voltage and bulk resistance. So the cap charges up some with each cycle, but at the start it is a very low voltage and the zener is forward biased, so I don't see any big voltage spike until the cap charges up. Eventually the diode's zener voltage limits the peak voltage across the cap - to 48 V. Sounds like areement.

    So, there is only one source of high voltage/high current, and the part rated for 18 V is failing before the part rated for 40 V. On the surface that isn't a mystery, except for the lack of a clear energy path into the 555. Ah, but is the 555 actually "burning up", or just failing? A blown reset transistor would prevent output transitions, and could happen with a stray-capacitance/bad decoupling transient on Vcc. Hmmm...

    ak
     
  13. Rittter

    Thread Starter New Member

    Dec 5, 2015
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    The wall wart is putting out 12.3vdc per my meter.
     
  14. dl324

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    Is the voltage regulated? Could be a cheap wall wart that's just a rectifier and filter cap. Your voltmeter wouldn't pick up peak voltage in that case.
     
  15. Rittter

    Thread Starter New Member

    Dec 5, 2015
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    When the 555 sends up smoke signals, I conclude it is toast (electronic technical term). Maybe it would be safer to eliminate the cap and go with an FR 207 in parallel with the coil. Thank You for the input !

    This wart is a TRIAD Model WSU120-3000 made by some of the best people in China. I got it from Digikey and it was not cheap but I understand that does not mean anything. This all argues for decoupling in various places....

    If I am understanding you correctly, putting a zenner to ground between the coil and the transistor would eliminate any questions as to where the flyback is going. I don't understand the benefit of a small cap in parallel with the base resistor.
     
    Last edited by a moderator: Jun 15, 2016
  16. EM Fields

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    Jun 8, 2016
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    You're right; I missed that part.
    What I'd do then would be to add a Zener with a reverse voltage rating somewhat below the transistor's VCE(O) in series with the catch diode, the Zener's cathode connected to the transistor 's collector, and its anode connected to the catch diode's anode.
     
  17. AlbertHall

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    Jun 4, 2014
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    To generate the highest back emf the transistor should be switched off as fast as possible. While the transistor is switched on the forward conducting base emitter will store charge and there will be some stray capacitance which will also store some charge. With the resistor to the '555 output that stored charge has to work its way through that resistor. If you put a capacitor across that resistor this provides a low impedance discharge path and hence faster turn off time = higher voltage.
     
  18. Rittter

    Thread Starter New Member

    Dec 5, 2015
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    Thank you. Someone else suggested this and I now realize this benefit. But which direction is the cap's anode ? Can't get my head around that one.
     
  19. AlbertHall

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    The anode of a capacitor? The capacitor only needs to be fairly small, say 100pF - 470pF.
     
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