Capacitor discharge probes

[Shaun]

Hello,

I recently bought a Blue ESR meter and assembled it and it is working well.
In the instructions it mentions making a jig with a 100 ohm 5 watt resistor
attached to some probes for discharging capacitors before testing them with
the ESR meter.

How does one calculate the required resistor wattage for different capacitor
values. Say I wanted to discharge a 63 volt cap at 500 microfarads, would 5
watts be enough. Right now I've made a probe jig with four 100 ohm 3 watt
resistors for a total of 100 ohms at 12 watts.

thanks in advance,

Shaun

 

[Tom Biasi]

"Shaun" <rowl@nomail.com> wrote in message
news:yuzSn.62258$mi.39337@newsfe01.iad...

Hello,

I recently bought a Blue ESR meter and assembled it and it is working
well. In the instructions it mentions making a jig with a 100 ohm 5 watt
resistor attached to some probes for discharging capacitors before testing
them with the ESR meter.

How does one calculate the required resistor wattage for different
capacitor values. Say I wanted to discharge a 63 volt cap at 500
microfarads, would 5 watts be enough. Right now I've made a probe jig
with four 100 ohm 3 watt resistors for a total of 100 ohms at 12 watts.

thanks in advance,

Shaun

Hi Shaun,

I could say a few words here but Sam can provide you with much more info.
See here: http://www.repairfaq.org/sam/captest.htm

Tom

 

[Shaun]

"Tom Biasi" <tombiasi@optonline.net> wrote in message
news:4c1acb9e$0$5016$607ed4bc@cv.net...

"Shaun" <rowl@nomail.com> wrote in message
news:yuzSn.62258$mi.39337@newsfe01.iad...
Hello,

I recently bought a Blue ESR meter and assembled it and it is working
well. In the instructions it mentions making a jig with a 100 ohm 5 watt
resistor attached to some probes for discharging capacitors before
testing them with the ESR meter.

How does one calculate the required resistor wattage for different
capacitor values. Say I wanted to discharge a 63 volt cap at 500
microfarads, would 5 watts be enough. Right now I've made a probe jig
with four 100 ohm 3 watt resistors for a total of 100 ohms at 12 watts.

thanks in advance,

Shaun


Hi Shaun,

I could say a few words here but Sam can provide you with much more info.
See here: http://www.repairfaq.org/sam/captest.htm

Tom
Thanks for the link, there is alot of information there but not what I was

asking for.

How do you calculate resistor power rating required to discharge a capacitor
of x volts and y microfarads?

Shaun

 

[Phil Allison]

"Shaun"

How do you calculate resistor power rating required to discharge a
capacitor of x volts and y microfarads?

** You are labouring under a mis-apprehension.

Any resistor connected across a capacitors will EVENTUALLY discharge it.

The only things to worry abut are how long will it take and might the
resistor become damaged by too much current .

A 100 ohm, 5W *wire wound* resistor will discharge almost any electro ( to a
safe voltage) in no more than a few seconds - however a very large, high
voltage electro might have enough energy to destroy it.


..... Phil

 

[Phil Allison]

<stratus46@yahoo.com>

"Shaun"

I recently bought a Blue ESR meter and assembled it and it is working

The vast majority of capacitors will be discharged by the time you get
your test probes ready to use. I've been doing ESR stuff several years
and only once did I damage a meter by sending power into a Capacitor
Wizard. It was a small value (around 10 ohms) resistor that fried and
only took a few minutes to replace. Or do you just want formulas?


** The " Blue ESR Meter " will be damaged by voltages more than about 50
volts DC

http://members.ozemail.com.au/~bobpar/esrmeter.htm


...... Phil

 

On Jun 17, 7:22 pm, "Shaun" <r...@nomail.com> wrote:

"Tom Biasi" <tombi...@optonline.net> wrote in message

news:4c1acb9e$0$5016$607ed4bc@cv.net...





"Shaun" <r...@nomail.com> wrote in message
news:yuzSn.62258$mi.39337@newsfe01.iad...
Hello,

I recently bought a Blue ESR meter and assembled it and it is working
well. In the instructions it mentions making a jig with a 100 ohm 5 watt
resistor attached to some probes for discharging capacitors before
testing them with the ESR meter.

How does one calculate the required resistor wattage for different
capacitor values.  Say I wanted to discharge a 63 volt cap at 500
microfarads, would 5 watts be enough.  Right now I've made a probe jig
with four 100 ohm 3 watt resistors for a total of  100 ohms at 12 watts.

thanks in advance,

Shaun

Hi Shaun,

I could say a few words here but Sam can provide you with much more info.
See here:http://www.repairfaq.org/sam/captest.htm

Tom

 Thanks for the link, there is alot of information there but not what I was
asking for.

How do you calculate resistor power rating required to discharge a capacitor
of x volts and y microfarads?

Shaun

The vast majority of capacitors will be discharged by the time you get
your test probes ready to use. I've been doing ESR stuff several years
and only once did I damage a meter by sending power into a Capacitor
Wizard. It was a small value (around 10 ohms) resistor that fried and
only took a few minutes to replace. Or do you just want formulas?

G˛

 

[George Herold]

Shaun wrote:

"Tom Biasi" <tombiasi@optonline.net> wrote in message
news:4c1acb9e$0$5016$607ed4bc@cv.net...

"Shaun" <rowl@nomail.com> wrote in message
news:yuzSn.62258$mi.39337@newsfe01.iad...
Hello,

I recently bought a Blue ESR meter and assembled it and it is working
well. In the instructions it mentions making a jig with a 100 ohm 5 watt
resistor attached to some probes for discharging capacitors before
testing them with the ESR meter.

How does one calculate the required resistor wattage for different
capacitor values. Say I wanted to discharge a 63 volt cap at 500
microfarads, would 5 watts be enough. Right now I've made a probe jig
with four 100 ohm 3 watt resistors for a total of 100 ohms at 12 watts.

thanks in advance,

Shaun


Hi Shaun,

I could say a few words here but Sam can provide you with much more info.
See here: http://www.repairfaq.org/sam/captest.htm

Tom
Thanks for the link, there is alot of information there but not what I was
asking for.

How do you calculate resistor power rating required to discharge a capacitor
of x volts and y microfarads?

That's a bit of a weird question. Most times caps aren't charged
(much) just sitting on the bench. So any old resistor will do. (Is
there some danger just using a peice of wire?)

The energy stored in a cap is 1/2 CV^2. If you discharge it with a
resistor R it'll take something like the time constant RC. So the
power is energy/time ~ V^2/R. (Hmm the C goes away?) anyway, if the
RC time is short the resistor won't have time to heat up much and you
could use a smaller value. So what's a short time? at a guess
anything shorter than 1 s or maybe 0.1 s.

George H.

Shaun

 

[oo pere oo]

Shaun wrote:

Hello,

I recently bought a Blue ESR meter and assembled it and it is working well.
In the instructions it mentions making a jig with a 100 ohm 5 watt resistor
attached to some probes for discharging capacitors before testing them with
the ESR meter.

How does one calculate the required resistor wattage for different capacitor
values. Say I wanted to discharge a 63 volt cap at 500 microfarads, would 5
watts be enough. Right now I've made a probe jig with four 100 ohm 3 watt
resistors for a total of 100 ohms at 12 watts.

thanks in advance,

Shaun

A 1K resistor will discharge a 1000uF capacitor in 5s.
A 100R resistor achieves the same in 0.5s.
Time constant is tau=RˇC 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

Pere

 

[Phil Allison]

"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=RˇC 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.



..... Phil

 

[oo pere oo]

Phil Allison 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=RˇC 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.



.... Phil

Just made a quick search on some Farnell 100R resistors. It is
interesting to see that a 1W CCR resistor such as CCR1100RKTB (Tyco) has
a pulse limiting power of 600W for single 0.1s pulses.

Pere

 

[John Larkin]

On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phil_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=RˇC 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
for a wirewound or axial carbon film, a fraction of a second for a
surface-mount thickfilm.

So it would probably be OK for a 2 watt axial resistor to absorb 20
joules. I've seen specs for 5-watt pulse-rated resistors that are
claimed to absorb over 100 joules.

Somebody should destroy some resistors and publish the results. I
exploded some 7.5 ohm 1206 thickfilm resistors. It takes about 60
watts in 1 millisecond. Pulsed, they look like LEDs in the infrared.

John

 

[George Herold]

On Jun 18, 11:27 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...@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.

Perhaps it's the heat capacity that determines the maximum energy it
can dissipate. E = C delta T.

George H.

for a wirewound or axial carbon film, a fraction of a second for a
surface-mount thickfilm.

So it would probably be OK for a 2 watt axial resistor to absorb 20
joules. I've seen specs for 5-watt pulse-rated resistors that are
claimed to absorb over 100 joules.

Somebody should destroy some resistors and publish the results. I
exploded some 7.5 ohm 1206 thickfilm resistors. It takes about 60
watts in 1 millisecond. Pulsed, they look like LEDs in the infrared.

John- Hide quoted text -

- Show quoted text -

 

[amdx]

"Phil Allison" <phil_a@tpg.com.au> wrote in message
news:hvfoel$njb$1@news-01.bur.connect.com.au...

"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=RˇC 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.

.... Phil

Did we all just witness a lovefest from Phil?
Mike

 

[whit3rd]

On Jun 17, 5:55 pm, "Shaun" <r...@nomail.com> wrote:

I recently bought a Blue ESR meter and assembled it and it is working well.
In the instructions it mentions making a jig with a 100 ohm 5 watt resistor
attached to some probes for discharging capacitors before testing them

Most circuitry with powered capacitors has a bleed resistor to
discharge those capacitors in a few seconds. For safety, one
is advised not to trust the bleed resistor, and various kinds of
probes
up to 'shorting chains' and down to 'connect with ammeter
setting on your VOM' can be used to ensure discharge.

For 15 kV capacitors in one-meter-cube sizes, use a steel
(but not galvanized) chain on a long insulating pole, with
eye protection.
For PC power supply filter capacitors, try that 100 ohm 5W resistor
(use a carbon resistor if you can, because the initial surge
will be more than 5W).

 

[Shaun]

"oo pere oo" <me@somewhere.net> wrote in message
news:hvfct0$k1b$1@speranza.aioe.org...

Shaun wrote:
Hello,

I recently bought a Blue ESR meter and assembled it and it is working
well. In the instructions it mentions making a jig with a 100 ohm 5 watt
resistor attached to some probes for discharging capacitors before
testing them with the ESR meter.

How does one calculate the required resistor wattage for different
capacitor values. Say I wanted to discharge a 63 volt cap at 500
microfarads, would 5 watts be enough. Right now I've made a probe jig
with four 100 ohm 3 watt resistors for a total of 100 ohms at 12 watts.

thanks in advance,

Shaun

A 1K resistor will discharge a 1000uF capacitor in 5s.
A 100R resistor achieves the same in 0.5s.
Time constant is tau=RˇC 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

Pere

Thanks for all your replies.

So I guess the important factor is it's pulse power rating for a resistor,
although I haven't seen any resistors with this spec before. If I work out
the math now that I know what to do thanks to you guys, P = 1/10 V^2/R
for 5 tau. What confuses me is that C cancels out, so it wouldn't matter
what value of C I'm discharging, the only thing that changes is the amount
of time it takes to discharge.

Does anyone know a rule of thumb for pulse power verses continuous power for
a given time frame for a resistor??

thanks again,

Shaun

 

[John Larkin]

On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

On Jun 18, 11:27 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...@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

Tau is roughly 40 seconds.

At 10 watts, this resistor gets too hot to hold (my threshold is about
55C) in around 10 seconds. That's 100 joules.

So I'd guess a 2-watt carbon comp can easily absorb 100 joules,
probably a lot more.

John

 

[John Larkin]

On Fri, 18 Jun 2010 21:12:15 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

John Larkin wrote:
On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:

On Jun 18, 11:27?am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...@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.

This graph does suggest how many joules the resistor could absorb
before it hits some scary temperature.

John

 

[George Herold]

John Larkin wrote:

On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:

On Jun 18, 11:27�am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...@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?
I bet the slope at the beginning gives the heat capacity.

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?

Perhaps we have different definitions for thermal time constant (TC).
I think of measuring TC by having a piece of stuff and a heat source.
I pulse the heat source and measure how long it takes the hunk of
stuff to come to a new temperature. If the heat pulse is twice as big
the temperature rise is twice as large, but it takes the same amount
of time. (Well as long as everything stays linear.)

At 10 watts, this resistor gets too hot to hold (my threshold is about
55C) in around 10 seconds. That's 100 joules.

So I'd guess a 2-watt carbon comp can easily absorb 100 joules,
probably a lot more.

John

I did try and measure (sometime last year) how much of the heat comes
out of the leads of a resistor and how much comes out through the
body. These were cheap Xicon 1/4 watt metal film resistors. I
thought I'd see more than 1/2 the heat leaving from the leads. But I
found that most of it was escaping though the body. A collegaue(sp)
told me my measurements were crap once the temperature got above a few
degrees cause then convection takes over. (Oh I was really trying to
answer the converse question, is the temperature of the resistor
determned by the temperature of it's leads or the temperature of the
gas that surrounds the body.)

George H.

 

[George Herold]

John Larkin wrote:

On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:

On Jun 18, 11:27�am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...@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

Tau is roughly 40 seconds.

At 10 watts, this resistor gets too hot to hold (my threshold is about
55C) in around 10 seconds. That's 100 joules.

So I'd guess a 2-watt carbon comp can easily absorb 100 joules,
probably a lot more.

OK If I'm reminded of it I'll try that on Monday. 100 Watts into a
2watt carbon comp for 1 second. The only easy way I can think to do
this is by taking a 35V 3A supply and ramping up the current knob to
max for one second...or maybe pushing the power button on and off.
(hmm I could build some power fet switch thing, but that sounds like
work.) I hope I have some 2Watt 10 ohm carbon comps. If I only find
1/2 watt ones can I scale this down by a factor of four? 25 watts
for 1 second. Hey anyone with a home lab feel like trying to blow
things up this weekend.

George H.

John

 

[George Herold]

On Jun 19, 1:01 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 18 Jun 2010 21:12:15 -0700 (PDT), George Herold





ggher...@gmail.com> wrote:

John Larkin wrote:
On Fri, 18 Jun 2010 10:05:08 -0700 (PDT), George Herold
ggher...@gmail.com> wrote:

On Jun 18, 11:27?am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Fri, 18 Jun 2010 22:22:42 +1000, "Phil Allison" <phi...@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.)

George H.

This graph does suggest how many joules the resistor could absorb
before it hits some scary temperature.

John- Hide quoted text -

- Show quoted text -