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  • mecanicafina's instructable Safe Capacitor Discharge Tool's weekly stats: 1 month ago
    • Safe Capacitor Discharge Tool
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      119 favorites
      18 comments
  • mecanicafina commented on mecanicafina's instructable Safe Capacitor Discharge Tool1 month ago
    Safe Capacitor Discharge Tool

    As Fezder already pointed out, 1 mF = 1 milliFarad = 1000 uF. With a 1k resistor the 3RC discharge time is two and half minutes, if you are happy with that 5 watt is plenty as Wo is just under 4W.

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  • mecanicafina commented on mecanicafina's instructable Safe Capacitor Discharge Tool1 month ago
    Safe Capacitor Discharge Tool

    Thanks for your question. The good news is, you can use the circuit as is, but there is a catch. I suggest you optimize it for your application, here is why and how. For a 50 mF capacitor charged at 63 V the circuit as it stands will work. In fact, the max input voltage is 63 V, the max input current is 29 mA, and the max instantaneous power dissipated on the resistor is 1,80 W, all well within the component specifications. However the discharge time constant RC is 110 sec, so for the voltage to drop to 5% of 63V you will have to wait 3RC = 330 sec. That is way too long if you do this on a regular basis. To drop RC the only option we have is to lower the value of R, which shortens the safe discharge time, but also increases the power dissipation in the resistor. Sizing the resistor is a...see more »Thanks for your question. The good news is, you can use the circuit as is, but there is a catch. I suggest you optimize it for your application, here is why and how. For a 50 mF capacitor charged at 63 V the circuit as it stands will work. In fact, the max input voltage is 63 V, the max input current is 29 mA, and the max instantaneous power dissipated on the resistor is 1,80 W, all well within the component specifications. However the discharge time constant RC is 110 sec, so for the voltage to drop to 5% of 63V you will have to wait 3RC = 330 sec. That is way too long if you do this on a regular basis. To drop RC the only option we have is to lower the value of R, which shortens the safe discharge time, but also increases the power dissipation in the resistor. Sizing the resistor is a bit tricky because the dissipated power decreases exponentially from its initial maximum value. If we were to size the resistor as if the power dissipated was constant and equal to the initial value, we would oversize it quite a bit. Here is what I mean.The instantaneous power dissipated in the resistor is given by W(t)=Wo exp (-2t/RC), where Wo=Vo^2/R is the instantaneous power dissipation when the discharge process is started. Let’s say we want to drop the safe discharge time to 30 sec. That means RC has to be about 10 sec, this gives a resistance R of 200 Ohm. Now Wo is almost 20 W, that is a bit of a resistor! Math comes to help, because if we plot the curve W(t) for 0<t<30 we see that the instantaneous power dissipated in the resistor drops under 10 W already after about 3 sec. and under 5 W after about 7 sec. This suggests that we can go quite lower than 20 W. A jellybean 200 Ohm 10W resistor mounted on a heat sink will happily handle this (you will need to use an enclosure big enough for the circuit to accommodate the resistor and heat sink). I am thinking here about the jellybean power resistors with aluminum body, and they need a heat sink. You could probably go even lower in terms of wattage but then you would really need to be picky about the construction of the power resistor you choose, for example try a vetroceramic power resistor. They can glow red for a few seconds and live happily even after. I would not go this route as those components are quite specialized and may not be easy to source. In summary, swap the 2200 Ohm 3W resistor for a 200 Ohm 10W with an adequate heat sink and you’ll be fine.

    Thanks for your question. The good news is, you canuse the circuit as is, but there is a catch. I suggest you optimize it for yourapplication, here is why and how. For a 50 mF capacitor charged at 63 V thecircuit as it stands will work. In fact, the max input voltage is 63 V, the maxinput current is 29 mA, and the max instantaneous power dissipated on theresistor is 1,80 W, all well within the component specifications. However thedischarge time constant RC is 110 sec, so for the voltage to drop to 5% of 63Vyou will have to wait 3RC = 330 sec. That is way too long if you do this on aregular basis. To drop RC the only option we have is to lower the value of R,which shortens the safe discharge time, but also increases the powerdissipation in the resistor. Sizing the resistor is a bit trick...see more »Thanks for your question. The good news is, you canuse the circuit as is, but there is a catch. I suggest you optimize it for yourapplication, here is why and how. For a 50 mF capacitor charged at 63 V thecircuit as it stands will work. In fact, the max input voltage is 63 V, the maxinput current is 29 mA, and the max instantaneous power dissipated on theresistor is 1,80 W, all well within the component specifications. However thedischarge time constant RC is 110 sec, so for the voltage to drop to 5% of 63Vyou will have to wait 3RC = 330 sec. That is way too long if you do this on aregular basis. To drop RC the only option we have is to lower the value of R,which shortens the safe discharge time, but also increases the powerdissipation in the resistor. Sizing the resistor is a bit tricky because the dissipatedpower decreases exponentially from its initial maximum value. If we were tosize the resistor as if the power dissipated was constant and equal to theinitial value, we would oversize it quite a bit. Here is what I mean.The instantaneous power dissipated in the resistoris given by W(t)=Wo exp (-2t/RC), where Wo=Vo^2/R is the instantaneous powerdissipation when the discharge process is started. Let’s say we want to dropthe safe discharge time to 30 sec. That means RC has to be about 10 sec, thisgives a resistance R of 200 Ohm. Now Wo is almost 20 W, that is a bit of aresistor! Math comes to help, because if we plot the curveW(t) for 0<t<30 we see that the instantaneous power dissipated in theresistor drops under 10 W already after about 3 sec. and under 5 W after about7 sec. This suggests that we can go quite lower than 20 W. A jellybean 200 Ohm 10W resistor mounted on a heatsink will happily handle this (you will need to use an enclosure big enough forthe circuit to accommodate the resistor and heat sink). I am thinking hereabout the jellybean power resistors with aluminum body, and they need a heatsink. You could probably go even lower in terms of wattage but then you wouldreally need to be picky about the construction of the power resistor youchoose, for example try a vetroceramic power resistor. They can glow red for afew seconds and live happily even after. I would not go this route as thosecomponents are quite specialized and may not be easy to source. In summary, swap the 2200 Ohm 3W resistor for a 200Ohm 10W with an adequate heat sink and you’ll be fine.

    Well spotted and corrected, thank you.

    Funny, same resistance I picked up for generic usage. I amsuggesting a lower resistance value in my answer to SPenrodbelow because his caps are 50 mF,so I feel the discharge time will be too long. Actually, in your case the discharge time to 5% of the initial 63V should notbe 160 sec, but about 100 sec.

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  • mecanicafina commented on mecanicafina's instructable Safe Capacitor Discharge Tool1 month ago
    Safe Capacitor Discharge Tool

    You are welcome, thanks for your feedback

    Good point. The resistor fits quite snugly in that milled slot, and I added the braid (from a piece of coax) to improve heat transfer while padding the fit a bit (the resistor body is not a cylinder). I have added some silicone like sealant afterwards anyway to fix everything in place even better. Unfortunately I had already taken the pictures, but I will add a note in the text.

    The full voltage could appear at the LED terminals if one of the 1N4007 diodes would fail with an open junction. In that case one of the LED’s would see the full input voltage applied as direct voltage, the other one as reverse voltage. For an input voltage of 300 V one of the LED (hard to say which one) would blow up, however I am not sure it would explode.One could add some sort of overvoltage protection to the circuit of course, but maybe the simplest course of action is to just modify the case using LED mounting holes that are not pass through. The material is “transparent” enough that you would still see the LED lighting up.

    he full voltage could appear at the LED terminals if one of the 1N4007 diodes would fail with an open junction. In that case one of the LED’s would see the full input voltage applied as direct voltage, the other one as reverse voltage. For an input voltage of 300 V one of the LED (hard to say which one) would blow up, however I am not sure it would explode.One could add some sort of overvoltage protection to the circuit of course, but maybe the simplest course of action is to just modify the case using LED mounting holes that are not pass through. The material is “transparent” enough that you would still see the LED lighting up.

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