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From: John Larkin on 19 Jun 2010 12:48
On Sat, 19 Jun 2010 07:25:45 -0700 (PDT), George Herold
<ggherold(a)gmail.com> wrote:

>On Jun 19, 1:01�am, John Larkin
><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
>> On Fri, 18 Jun 2010 21:12:15 -0700 (PDT), George Herold
>>
>>
>>
>>
>>
>> <ggher...(a)gmail.com> wrote:
>>
>> >John Larkin wrote:
>> >> On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
>> >> <ggher...(a)gmail.com> wrote:
>>
>> >> >On Jun 18, 11:27?am, John Larkin
>> >> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
>> >> >> On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...(a)tpg.com.au>
>> >> >> wrote:
>>
>> >> >> >"oo pere oo"
>>
>> >> >> >> A 1K resistor will discharge a 1000uF capacitor in 5s.
>> >> >> >> A 100R resistor achieves the same in 0.5s.
>> >> >> >> Time constant is tau=R C and in 4 or 5 tau you have discharged your
>> >> >> >> capacitor.
>>
>> >> >> >> The power absorbed by the resistor at the moment of closing the circuit is
>> >> >> >> V /R. For V=63V and R=100R, this is roughly 40W. At t=0.1s voltage is just
>> >> >> >> 23V and instantaneous power absorbed 5W. Your jig will certainly survive
>> >> >> >> such short spikes. If you take 1K, a single resistor will achieve the same
>> >> >> >> at the cost of some extra time :)
>>
>> >> >> >** All perfectly correct.
>>
>> >> >> >However, a physically small resistor can only absorb so much energy in a
>> >> >> >short space of time before the conductor inside MELTS !!
>>
>> >> >> >The energy stored in a capacitor is give by the formula:
>>
>> >> >> > E = 0.5C x V squared.
>>
>> >> >> >Taking a practical, worst case example of a 1000uF cap charged to 400
>> >> >> >volts - ?the stored energy is 80 joules, most of which is dumped in the
>> >> >> >first 0.2 seconds.
>>
>> >> >> >Depends very much on the construction of the particular resistor whether it
>> >> >> >can absorb such a large energy spike or not.
>>
>> >> >> >The best type is carbon composition, then wire wound and last of all
>> >> >> >deposited film resistors.
>>
>> >> >> I'd guess that a resistor can easily absorb E joules, where E = P * T
>> >> >> and P is its power rating and T is its thermal time constant. T is
>> >> >> maybe 10 or more seconds for something like a carbon comp, maybe less
>>
>> >> >How big a carbon comp? �I would have guessed something less.. maybe 1
>> >> >second. �Hmm, just wondering if you can really define a thermal time
>> >> >constant for a resistor. �At least for a carbon comp, the heat is
>> >> >being generated everywhere inside the thing. �I think of a thermal
>> >> >time constant as heating one end of something and asking how long it
>> >> >takes the whole thing to warm up.
>>
>> >> It's interesting how many people speculate and simulate and don't
>> >> actually measure stuff.
>>
>> >>ftp://jjlarkin.lmi.net/A-B_tau.gif
>>
>> >This is sitting in air? �Aren't you measuring the convective cooling?
>>
>> Yes. At the end of the curve, the applied power is almost balanced by
>> heat losses.
>>
>> >I bet the slope at the beginning gives the heat capacity.
>>
>> Yes.
>>
>>
>>
>> >> Tau is roughly 40 seconds.
>>
>> >I'm not sure what you mean by tau? �At ten watts didn't it heat up
>> >twice as fast?
>>
>> Tau is the time it takes my graph to go from start to 63% of its
>> terminal value. It's the thermal time constant of the resistor.
>>
>> Sure, the rate of temperature rise is proportional to the applied
>> power. More watts will zoom my graph vertically. Tau won't change, at
>> least until something melts, or convection gets nonlinear maybe.
>
>Opps my mistake.. twice the heat, twice the slope, but twice the final
>temp difference so same TC. (late night mistake... ah heck I make
>mistakes all day long.)
>
>Seems like if you could estimate the temperature at which the resistor
>'craps out' then you could use the slope of your curve at the origin
>to guess at the total energy. (Sorry this is perhaps obvious to you,
>I'm just a bit slow.)

Yes. If you extrapolate the initial slope of an exponential like this,
it intersects the asymptotic limit in 1 tau.

My initial wild-guess thinking was to guess the thermal time constant
of a resistor (semething I do have a vague feeling for) and guess its
joule capacity from its power rating and that. It is more scientific
to look at the initial slope, degc/sec per watt, pick some peak
temperature, and do the math from that.

Sort of the same thing.

So, using 150C peak, this resistor is good for about 200 joules. My
10-watt, hold-it-as-long-as-you-can estimate was similar.


"One experiment is worth a thousand expert opinions."

- Werner Von Braun


From: George Herold on 19 Jun 2010 21:58
On Jun 19, 12:48 pm, John Larkin
<jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
> On Sat, 19 Jun 2010 07:25:45 -0700 (PDT), George Herold
>
>
>
>
>
> <ggher...(a)gmail.com> wrote:
> >On Jun 19, 1:01 am, John Larkin
> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
> >> On Fri, 18 Jun 2010 21:12:15 -0700 (PDT), George Herold
>
> >> <ggher...(a)gmail.com> wrote:
>
> >> >John Larkin wrote:
> >> >> On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
> >> >> <ggher...(a)gmail.com> wrote:
>
> >> >> >On Jun 18, 11:27?am, John Larkin
> >> >> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
> >> >> >> On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...(a)tpg.com.au>
> >> >> >> wrote:
>
> >> >> >> >"oo pere oo"
>
> >> >> >> >> A 1K resistor will discharge a 1000uF capacitor in 5s.
> >> >> >> >> A 100R resistor achieves the same in 0.5s.
> >> >> >> >> Time constant is tau=R C and in 4 or 5 tau you have discharged your
> >> >> >> >> capacitor.
>
> >> >> >> >> The power absorbed by the resistor at the moment of closing the circuit is
> >> >> >> >> V /R. For V=63V and R=100R, this is roughly 40W. At t=0..1s voltage is just
> >> >> >> >> 23V and instantaneous power absorbed 5W. Your jig will certainly survive
> >> >> >> >> such short spikes. If you take 1K, a single resistor will achieve the same
> >> >> >> >> at the cost of some extra time :)
>
> >> >> >> >** All perfectly correct.
>
> >> >> >> >However, a physically small resistor can only absorb so much energy in a
> >> >> >> >short space of time before the conductor inside MELTS !!
>
> >> >> >> >The energy stored in a capacitor is give by the formula:
>
> >> >> >> > E = 0.5C x V squared.
>
> >> >> >> >Taking a practical, worst case example of a 1000uF cap charged to 400
> >> >> >> >volts - ?the stored energy is 80 joules, most of which is dumped in the
> >> >> >> >first 0.2 seconds.
>
> >> >> >> >Depends very much on the construction of the particular resistor whether it
> >> >> >> >can absorb such a large energy spike or not.
>
> >> >> >> >The best type is carbon composition, then wire wound and last of all
> >> >> >> >deposited film resistors.
>
> >> >> >> I'd guess that a resistor can easily absorb E joules, where E = P * T
> >> >> >> and P is its power rating and T is its thermal time constant. T is
> >> >> >> maybe 10 or more seconds for something like a carbon comp, maybe less
>
> >> >> >How big a carbon comp?  I would have guessed something less.. maybe 1
> >> >> >second.  Hmm, just wondering if you can really define a thermal time
> >> >> >constant for a resistor.  At least for a carbon comp, the heat is
> >> >> >being generated everywhere inside the thing.  I think of a thermal
> >> >> >time constant as heating one end of something and asking how long it
> >> >> >takes the whole thing to warm up.
>
> >> >> It's interesting how many people speculate and simulate and don't
> >> >> actually measure stuff.
>
> >> >>ftp://jjlarkin.lmi.net/A-B_tau.gif
>
> >> >This is sitting in air?  Aren't you measuring the convective cooling?
>
> >> Yes. At the end of the curve, the applied power is almost balanced by
> >> heat losses.
>
> >> >I bet the slope at the beginning gives the heat capacity.
>
> >> Yes.
>
> >> >> Tau is roughly 40 seconds.
>
> >> >I'm not sure what you mean by tau?  At ten watts didn't it heat up
> >> >twice as fast?
>
> >> Tau is the time it takes my graph to go from start to 63% of its
> >> terminal value. It's the thermal time constant of the resistor.
>
> >> Sure, the rate of temperature rise is proportional to the applied
> >> power. More watts will zoom my graph vertically. Tau won't change, at
> >> least until something melts, or convection gets nonlinear maybe.
>
> >Opps my mistake.. twice the heat, twice the slope, but twice the final
> >temp difference so same TC.  (late night mistake... ah heck I make
> >mistakes all day long.)
>
> >Seems like if you could estimate the temperature at which the resistor
> >'craps out' then you could use the slope of your curve at the origin
> >to guess at the total energy.  (Sorry this is perhaps obvious to you,
> >I'm just a bit slow.)
>
> Yes. If you extrapolate the initial slope of an exponential like this,
> it intersects the asymptotic limit in 1 tau.
>
> My initial wild-guess thinking was to guess the thermal time constant
> of a resistor (semething I do have a vague feeling for) and guess its
> joule capacity from its power rating and that. It is more scientific
> to look at the initial slope, degc/sec per watt, pick some peak
> temperature, and do the math from that.
>
> Sort of the same thing.
>
> So, using 150C peak, this resistor is good for about 200 joules. My
> 10-watt, hold-it-as-long-as-you-can estimate was similar.
>
> "One experiment is worth a thousand expert opinions."
>
> - Werner Von Braun- Hide quoted text -
>
> - Show quoted text -

Yeah I sorta realized you'd done the same thing, But those were
thoughts on the stairs.

I'd still like to try dumping 100’s of Joules in one second into a
resistor.

George H.
From: Cydrome Leader on 20 Jun 2010 02:29
In sci.electronics.basics George Herold <ggherold(a)gmail.com> wrote:
> On Jun 19, 12:48?pm, John Larkin
> <jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
>> On Sat, 19 Jun 2010 07:25:45 -0700 (PDT), George Herold
>>
>>
>>
>>
>>
>> <ggher...(a)gmail.com> wrote:
>> >On Jun 19, 1:01?am, John Larkin
>> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
>> >> On Fri, 18 Jun 2010 21:12:15 -0700 (PDT), George Herold
>>
>> >> <ggher...(a)gmail.com> wrote:
>>
>> >> >John Larkin wrote:
>> >> >> On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
>> >> >> <ggher...(a)gmail.com> wrote:
>>
>> >> >> >On Jun 18, 11:27?am, John Larkin
>> >> >> ><jjlar...(a)highNOTlandTHIStechnologyPART.com> wrote:
>> >> >> >> On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...(a)tpg.com.au>
>> >> >> >> wrote:
>>
>> >> >> >> >"oo pere oo"
>>
>> >> >> >> >> A 1K resistor will discharge a 1000uF capacitor in 5s.
>> >> >> >> >> A 100R resistor achieves the same in 0.5s.
>> >> >> >> >> Time constant is tau=R C and in 4 or 5 tau you have discharged your
>> >> >> >> >> capacitor.
>>
>> >> >> >> >> The power absorbed by the resistor at the moment of closing the circuit is
>> >> >> >> >> V /R. For V=63V and R=100R, this is roughly 40W. At t=0.1s voltage is just
>> >> >> >> >> 23V and instantaneous power absorbed 5W. Your jig will certainly survive
>> >> >> >> >> such short spikes. If you take 1K, a single resistor will achieve the same
>> >> >> >> >> at the cost of some extra time :)
>>
>> >> >> >> >** All perfectly correct.
>>
>> >> >> >> >However, a physically small resistor can only absorb so much energy in a
>> >> >> >> >short space of time before the conductor inside MELTS !!
>>
>> >> >> >> >The energy stored in a capacitor is give by the formula:
>>
>> >> >> >> > E = 0.5C x V squared.
>>
>> >> >> >> >Taking a practical, worst case example of a 1000uF cap charged to 400
>> >> >> >> >volts - ?the stored energy is 80 joules, most of which is dumped in the
>> >> >> >> >first 0.2 seconds.
>>
>> >> >> >> >Depends very much on the construction of the particular resistor whether it
>> >> >> >> >can absorb such a large energy spike or not.
>>
>> >> >> >> >The best type is carbon composition, then wire wound and last of all
>> >> >> >> >deposited film resistors.
>>
>> >> >> >> I'd guess that a resistor can easily absorb E joules, where E = P * T
>> >> >> >> and P is its power rating and T is its thermal time constant. T is
>> >> >> >> maybe 10 or more seconds for something like a carbon comp, maybe less
>>
>> >> >> >How big a carbon comp? ?I would have guessed something less.. maybe 1
>> >> >> >second. ?Hmm, just wondering if you can really define a thermal time
>> >> >> >constant for a resistor. ?At least for a carbon comp, the heat is
>> >> >> >being generated everywhere inside the thing. ?I think of a thermal
>> >> >> >time constant as heating one end of something and asking how long it
>> >> >> >takes the whole thing to warm up.
>>
>> >> >> It's interesting how many people speculate and simulate and don't
>> >> >> actually measure stuff.
>>
>> >> >>ftp://jjlarkin.lmi.net/A-B_tau.gif
>>
>> >> >This is sitting in air? ?Aren't you measuring the convective cooling?
>>
>> >> Yes. At the end of the curve, the applied power is almost balanced by
>> >> heat losses.
>>
>> >> >I bet the slope at the beginning gives the heat capacity.
>>
>> >> Yes.
>>
>> >> >> Tau is roughly 40 seconds.
>>
>> >> >I'm not sure what you mean by tau? ?At ten watts didn't it heat up
>> >> >twice as fast?
>>
>> >> Tau is the time it takes my graph to go from start to 63% of its
>> >> terminal value. It's the thermal time constant of the resistor.
>>
>> >> Sure, the rate of temperature rise is proportional to the applied
>> >> power. More watts will zoom my graph vertically. Tau won't change, at
>> >> least until something melts, or convection gets nonlinear maybe.
>>
>> >Opps my mistake.. twice the heat, twice the slope, but twice the final
>> >temp difference so same TC. ?(late night mistake... ah heck I make
>> >mistakes all day long.)
>>
>> >Seems like if you could estimate the temperature at which the resistor
>> >'craps out' then you could use the slope of your curve at the origin
>> >to guess at the total energy. ?(Sorry this is perhaps obvious to you,
>> >I'm just a bit slow.)
>>
>> Yes. If you extrapolate the initial slope of an exponential like this,
>> it intersects the asymptotic limit in 1 tau.
>>
>> My initial wild-guess thinking was to guess the thermal time constant
>> of a resistor (semething I do have a vague feeling for) and guess its
>> joule capacity from its power rating and that. It is more scientific
>> to look at the initial slope, degc/sec per watt, pick some peak
>> temperature, and do the math from that.
>>
>> Sort of the same thing.
>>
>> So, using 150C peak, this resistor is good for about 200 joules. My
>> 10-watt, hold-it-as-long-as-you-can estimate was similar.
>>
>> "One experiment is worth a thousand expert opinions."
>>
>> - Werner Von Braun- Hide quoted text -
>>
>> - Show quoted text -
>
> Yeah I sorta realized you'd done the same thing, But those were
> thoughts on the stairs.
>
> I'd still like to try dumping 100?s of Joules in one second into a
> resistor.
>
> George H.

be careful with that.

50J exploded 20 guage magnet wire like a fuse and blew a piece into my
hand once.

I wasn't expecting that to happen at all.
 | 
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