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1. moved_sdf_NYbill (movedsdfnybill@sn.gunmonkeynet.net)'s status on Saturday, 04-Feb-2017 19:49:10 UTC moved_sdf_NYbill
   Charged a 63v 22000uf cap up to 30v last Friday for kicks. It still has 13v on it.
   • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca (bobjonkmanformer@sn.jonkman.ca)'s status on Saturday, 04-Feb-2017 22:46:04 UTC Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca

     in reply to
     Remind me of the math. How much current can you pull from a 22000uF capacitor with 64V on it?
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Saturday, 04-Feb-2017 22:53:45 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     The current will be given by I = I_0 (1-e^{t/(RC)}), where I_0 is the maximum current that can be delivered by the capacitor when it's discharged at time zero. So the current maximum will depend on not just C, but the resistance in the circuit. What's the resistance?
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Saturday, 04-Feb-2017 22:54:14 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     CORRECTION: I = I_0 e^{t/(RC)}
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Saturday, 04-Feb-2017 22:55:42 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     And you'll also need Ohm's Law, since V_0 = I_0 R will give you the voltage-current relationship at time zero, when V_0 = 64V.
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Saturday, 04-Feb-2017 22:57:31 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     Oh, and C=22,000uF.
   • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca (bobjonkmanformer@sn.jonkman.ca)'s status on Saturday, 04-Feb-2017 23:25:01 UTC Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca

     in reply to

     • Stephen Sekula
     I remember learning that in the Magnetism and Electricity course from Uni, but I don't remember what I actually learned. http://sn.jonkman.ca/attachment/49136
     Stephen Sekula and @mcscx@quitter.se like this.
   • meece (meece@sealion.club)'s status on Saturday, 04-Feb-2017 23:42:16 UTC meece

     in reply to

     • Stephen Sekula
     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     @bobjonkman @steve I guess you could say it didn't resonate with you.
     Chazcon likes this.
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Sunday, 05-Feb-2017 00:21:23 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     • meece
     @bobjonkman, BAH-DUM-BUM! :-)
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Sunday, 05-Feb-2017 00:23:17 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     Actually, now that I see the original @nybill post I realize we can estimate the resistance in the circuit using the fact that he have us the start day of his charge and the current voltage. :-) I love this stuff.
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Sunday, 05-Feb-2017 00:52:56 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     Thanks for a fun distraction tonight, @nybill and @bobjonkman! Attached is a calculation using some of the RC equations that, given the original post, estimates the resistance through which charge in the capacitor leaked over the 8 days and 1.5 hours it was sitting. Enjoy! https://chirp.cooleysekula.net/attachment/21662
   • moved_sdf_NYbill (movedsdfnybill@sn.gunmonkeynet.net)'s status on Sunday, 05-Feb-2017 01:08:55 UTC moved_sdf_NYbill

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     @bobjonkman I was getting stuff from Bangood and saw it. I have no plans for it. It just made me laugh so I bought it. My power supply only goes to 30v DC. (although, I think I read I can gang the outputs. I've never had plans to.)
     Stephen Sekula likes this.
   • moved_sdf_NYbill (movedsdfnybill@sn.gunmonkeynet.net)'s status on Sunday, 05-Feb-2017 01:09:29 UTC moved_sdf_NYbill

     in reply to

     • Stephen Sekula
     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     @bobjonkman @steve Here is the beast. As big as a D cell battery. :P http://media.gunmonkeynet.net/u/nybill/m/added-this-to-a-recent-order/
     Stephen Sekula likes this.
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Sunday, 05-Feb-2017 01:16:07 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     @bobjonkman, That is a beaut!
   • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca (bobjonkmanformer@sn.jonkman.ca)'s status on Sunday, 05-Feb-2017 01:20:15 UTC Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca

     in reply to

     • Stephen Sekula
     @steve@chirp.cooleysekula.net That's a lovely analysis! If the leakage could be due to the dielectric's resistance, wouldn't the observed leakage be due to *both* that and the leakage through air? Then the calculated resistance would be that of dielectric and air in parallel...
     LinuxWalt (@lnxw48a1) {3EB165E0-5BB1-45D2-9E7D-93B31821F864} likes this.
   • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca (bobjonkmanformer@sn.jonkman.ca)'s status on Sunday, 05-Feb-2017 01:31:17 UTC Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca

     in reply to
     And to answer my own question, with R=3.8E7 then the current at the start was 30V/R = about 0.8 uA, and at the end 13V/R = about 0.3 uA
   • Stephen Sekula (steve@chirp.cooleysekula.net)'s status on Sunday, 05-Feb-2017 03:27:04 UTC Stephen Sekula

     in reply to

     • Former Bob Jonkman -- Please use the new server at https://gs.jonkman.ca
     Exactly. After seeing that picture, I suspect the culprit is the dielectric inside the capacitor, as there is so much surface area tucked into that cylinder and, with age, I am sure the dielectric is more leaky than it was originally - though I am sure it's always been leaky. Still, 8 days and it only lost about half its stored charge... not bad!