SPICE (LT): determining dissipation by resistors, etc.?

Mike Rocket J. Squirrel E · Mar 2, 2004

Ken Smith wrote:

I obviously am having an advanced case of brain lock or too much beer.
After checking my notes here's the right answers (I hope):

In article <6ImdnVpzjvBKjNndRVn-sA@adelphia.com>,
Mike Rocket J. Squirrel Elliott <j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:
[.. I wrote ..]

The thermal resistance is in Watts per degree C. These you just call
Ohms.

So R = the inverse of thermal resistance? With thermal resistance of 2.7
C/W, then R = 1/2.7 ohm: 0.37 ohm, yes?

No I meant to say that the data sheet I'd looked at gave you Watts per
degree. The Degrees per Watt is the right form to do resistance.

The thermal mass looks like a capacitor with C = degree/J.

Degree /Joule? I'm a bit simple -- this one I don't get. Does a heatsink
data sheet provide sufficient information to calculate this number?

Another mistake, it is C=J/degree. I measured degree/J because that is
the easy way to make the measurement and then inverted. The capacitor
looks like 0.9F per gram if it is alluminum. Copper and brass are more
like 0.38F per gram.

25 degrees C ambient would be . . . 25 volts?

Yes.

Hey -- I figured one out myself!

Thanks for going back over this and clearing things up. So if a fellow
wanted to know the junction temperature of a power device mounted on a
heatsink, all he'd need to so is stick in a second R in series with the
RC network, and give it the value of the device's specified
junction-to-mounting surface thermal resistance. I'm going to play with
this a bit and see how it works. In LTspice it should be easy to create
a heatsink symbol for use on the schematic. Easy for some -- Mike,
Helmut and analogspiceman could slam-dunk it. Me, I'll slog at it.

BTW: where did the capacitor numbers for aluminum and copper/brass come
from?

--
Mike "Rocket J Squirrel" Elliott
71 VW Type 2 -- the Wonderbus (AKA the Saunabus in summer)

Ken Smith · Mar 2, 2004

In article <w62dncTrZKoqXdndRVn-vA@adelphia.com>,
Mike Rocket J. Squirrel Elliott <j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:

Ken Smith wrote:
[...]
the easy way to make the measurement and then inverted. The capacitor
looks like 0.9F per gram if it is alluminum. Copper and brass are more
like 0.38F per gram.
[...]
BTW: where did the capacitor numbers for aluminum and copper/brass come
from?

The value for C is the "specific heat" of the material times its mass.
The CRC list Al having a specific heat of 0.215 in cal/(gK) units. You
multiply cal by 4.184 to get J so 0.215 * 4.184=0.89956 or 0.9.

--
--
kensmith@rahul.net forging knowledge

Harry Dellamano · Mar 3, 2004

"Ken Smith" <kensmith@violet.rahul.net> wrote in message
news:c1t8fh$thj$1@blue.rahul.net...

In article <zyyob.438$Bf7.374091@news1.news.adelphia.net>,
Mike Rocket J. Squirrel Elliott
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:
Newbie question:

What's the quick and easy way of finding the steady-state power
dissipation of resistors and other components in a circuit with SPICE?

Here's a handy little trick I've used a couple of times. Use the arb.
voltage source to multiply I*V then feed that into an RC model of the
thermal characteristics of the heatsinking. The result is a nice plot of
the temperature rise vs time when the circuit pulses on.

--
--
kensmith@rahul.net forging knowledge

Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current source
as the heat generating driving element? The current output would be the
power dissipated. 25amps = 25watts.

Regards
Harry

Kevin Aylward · Mar 3, 2004

Harry Dellamano wrote:

"Ken Smith" <kensmith@violet.rahul.net> wrote in message
news:c1t8fh$thj$1@blue.rahul.net...
In article <zyyob.438$Bf7.374091@news1.news.adelphia.net>,
Mike Rocket J. Squirrel Elliott
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:
Newbie question:

What's the quick and easy way of finding the steady-state power
dissipation of resistors and other components in a circuit with
SPICE?

Here's a handy little trick I've used a couple of times. Use the
arb. voltage source to multiply I*V then feed that into an RC model
of the thermal characteristics of the heatsinking. The result is a
nice plot of the temperature rise vs time when the circuit pulses on.

--
--
kensmith@rahul.net forging knowledge

Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current
source as the heat generating driving element? The current output
would be the power dissipated. 25amps = 25watts.

One use a Thevenin or Norton equivalent circuit. Putting a current
source into a series RC circuit will result in an infinite temperature
(voltage) in steady state!!!

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

http://www.anasoft.co.uk/NewBeginning.mp3

"quotes with no meaning, are meaningless" - Kevin Aylward.

Harry Dellamano · Mar 3, 2004

"Kevin Aylward" <kevindotaylwardEXTRACT@anasoft.co.uk> wrote in message
news:6sq1c.150$2F6.140@newsfep3-gui.server.ntli.net...

Harry Dellamano wrote:
"Ken Smith" <kensmith@violet.rahul.net> wrote in message
news:c1t8fh$thj$1@blue.rahul.net...
In article <zyyob.438$Bf7.374091@news1.news.adelphia.net>,
Mike Rocket J. Squirrel Elliott
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:
Newbie question:

What's the quick and easy way of finding the steady-state power
dissipation of resistors and other components in a circuit with
SPICE?

Here's a handy little trick I've used a couple of times. Use the
arb. voltage source to multiply I*V then feed that into an RC model
of the thermal characteristics of the heatsinking. The result is a
nice plot of the temperature rise vs time when the circuit pulses on.

--
--
kensmith@rahul.net forging knowledge

Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current
source as the heat generating driving element? The current output
would be the power dissipated. 25amps = 25watts.

One use a Thevenin or Norton equivalent circuit. Putting a current
source into a series RC circuit will result in an infinite temperature
(voltage) in steady state!!!

Kevin Aylward

But, but doesn't this current flow thru the thermal resistances and
finally end at a voltage source which is the ambient temperature or infinite
heat sink? It may get slowed down along the way by shunt capacity (thermal
mass).
Regards
Harry

Ken Smith · Mar 3, 2004

In article <Pzp1c.37675$C65.18320@nwrddc01.gnilink.net>,
Harry Dellamano <harryd@tdsystems.org> wrote:

"Ken Smith" <kensmith@violet.rahul.net> wrote in message
news:c1t8fh$thj$1@blue.rahul.net...
[...]
Here's a handy little trick I've used a couple of times. Use the arb.
voltage source to multiply I*V then feed that into an RC model of the
thermal characteristics of the heatsinking. The result is a nice plot of
the temperature rise vs time when the circuit pulses on.
[...]
Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current source
as the heat generating driving element? The current output would be the
power dissipated. 25amps = 25watts.

Sorry about the slight muddle on that. Yes a current source is the right
answer if the model you use is a more or less direct model of the real
heat sink. If you use resistance to model thermal resistance and
capacitance to model thermal mass, you use a current source.

If you have just a pre-calculated thermal time constant, you can just use
a voltage source and the right RC value to make the right time constant.

--
--
kensmith@rahul.net forging knowledge

Ken Smith · Mar 3, 2004

In article <KZr1c.38424$C65.29904@nwrddc01.gnilink.net>,
Harry Dellamano <harryd@tdsystems.org> wrote:
[.. RC model of thermal ..]

But, but doesn't this current flow thru the thermal resistances and
finally end at a voltage source which is the ambient temperature or infinite
heat sink? It may get slowed down along the way by shunt capacity (thermal
mass).

Yes, the current (heat) flows through the resistance (thermal r) to go to
a voltage (outside temperature). Capacitance does the thermal mass.

--
--
kensmith@rahul.net forging knowledge

Helmut Sennewald · Mar 3, 2004

"Mike Rocket J. Squirrel Elliott"
<j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> schrieb im Newsbeitrag
news:w62dncTrZKoqXdndRVn-vA@adelphia.com...

Ken Smith wrote:
I obviously am having an advanced case of brain lock or too much beer.
After checking my notes here's the right answers (I hope):

In article <6ImdnVpzjvBKjNndRVn-sA@adelphia.com>,
Mike Rocket J. Squirrel Elliott
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:
[.. I wrote ..]

The thermal resistance is in Watts per degree C. These you just call
Ohms.

So R = the inverse of thermal resistance? With thermal resistance of 2.7
C/W, then R = 1/2.7 ohm: 0.37 ohm, yes?

No I meant to say that the data sheet I'd looked at gave you Watts per
degree. The Degrees per Watt is the right form to do resistance.

The thermal mass looks like a capacitor with C = degree/J.

Degree /Joule? I'm a bit simple -- this one I don't get. Does a heatsink
data sheet provide sufficient information to calculate this number?

Another mistake, it is C=J/degree. I measured degree/J because that is
the easy way to make the measurement and then inverted. The capacitor
looks like 0.9F per gram if it is alluminum. Copper and brass are more
like 0.38F per gram.

25 degrees C ambient would be . . . 25 volts?

Yes.

Hey -- I figured one out myself!

Thanks for going back over this and clearing things up. So if a fellow
wanted to know the junction temperature of a power device mounted on a
heatsink, all he'd need to so is stick in a second R in series with the
RC network, and give it the value of the device's specified
junction-to-mounting surface thermal resistance. I'm going to play with
this a bit and see how it works. In LTspice it should be easy to create
a heatsink symbol for use on the schematic. Easy for some -- Mike,
Helmut and analogspiceman could slam-dunk it. Me, I'll slog at it.

BTW: where did the capacitor numbers for aluminum and copper/brass come
from?

--

Hello Mike,
I have looked for some literature with Google about thermal
transistor and heatsink models.

http://www.iisb.fraunhofer.de/en/arb_geb/powersys_thermmod_gb_fhg.pdf
http://www.infineon.com/cmc_upload/migrated_files/document_files/Appli
cation_Notes/mmpn_eng.pdf

http://www.fairchildsemi.com/an/AN/AN-7533.pdf
http://www.irf.com/technical-info/designtp/temp002.pdf
http://www.irf.com/product-info/datasheets/data/100bgq045.pdf

They are all based on the equation Tfinal=Pv*Rth
with an exponential temperature rise from Tamb to Tfinal.

The most important question now is, how can we influence the
transistor(NPN, MOS) during a simulation run.
My understanding about(LT,P-)SPICE is, that there is no chance to do
that with the basic transistor models.
What I have seen in the above articles is that they build more or
less complex circuits around the basic transistor models.

Is that the only way to get temperature dependent transistor
behaviour with SPICE transistor models in a .TRAN simulation?

Best Regards,
Helmut

Mike Rocket J. Squirrel E · Mar 4, 2004

Harry Dellamano wrote:

"Ken Smith" <kensmith@violet.rahul.net> wrote in message
news:c1t8fh$thj$1@blue.rahul.net...

In article <zyyob.438$Bf7.374091@news1.news.adelphia.net>,
Mike Rocket J. Squirrel Elliott

j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:

Newbie question:

What's the quick and easy way of finding the steady-state power
dissipation of resistors and other components in a circuit with SPICE?

Here's a handy little trick I've used a couple of times. Use the arb.
voltage source to multiply I*V then feed that into an RC model of the
thermal characteristics of the heatsinking. The result is a nice plot of
the temperature rise vs time when the circuit pulses on.

--
--
kensmith@rahul.net forging knowledge

Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current source
as the heat generating driving element? The current output would be the
power dissipated. 25amps = 25watts.

Harry, I'm glad you saw that. I thought that a voltage source was
incorrect, too, but knowing that I rank in the bottom 10 percentile of
brainpower for sci.electronics.cad, I doubted myself. Thanks for
bringing it up.

--
Mike "Rocket J Squirrel" Elliott
71 VW Type 2 -- the Wonderbus (AKA the Saunabus in summer)

Jim Thompson · Mar 4, 2004

On Wed, 3 Mar 2004 23:52:38 +0100, "Helmut Sennewald"
<HelmutSennewald@t-online.de> wrote:

"Mike Rocket J. Squirrel Elliott"
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> schrieb im Newsbeitrag
news:w62dncTrZKoqXdndRVn-vA@adelphia.com...
Ken Smith wrote:
I obviously am having an advanced case of brain lock or too much beer.
After checking my notes here's the right answers (I hope):

In article <6ImdnVpzjvBKjNndRVn-sA@adelphia.com>,
Mike Rocket J. Squirrel Elliott
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote:
[.. I wrote ..]

The thermal resistance is in Watts per degree C. These you just call
Ohms.

So R = the inverse of thermal resistance? With thermal resistance of 2.7
C/W, then R = 1/2.7 ohm: 0.37 ohm, yes?

No I meant to say that the data sheet I'd looked at gave you Watts per
degree. The Degrees per Watt is the right form to do resistance.

The thermal mass looks like a capacitor with C = degree/J.

Degree /Joule? I'm a bit simple -- this one I don't get. Does a heatsink
data sheet provide sufficient information to calculate this number?

Another mistake, it is C=J/degree. I measured degree/J because that is
the easy way to make the measurement and then inverted. The capacitor
looks like 0.9F per gram if it is alluminum. Copper and brass are more
like 0.38F per gram.

25 degrees C ambient would be . . . 25 volts?

Yes.

Hey -- I figured one out myself!

Thanks for going back over this and clearing things up. So if a fellow
wanted to know the junction temperature of a power device mounted on a
heatsink, all he'd need to so is stick in a second R in series with the
RC network, and give it the value of the device's specified
junction-to-mounting surface thermal resistance. I'm going to play with
this a bit and see how it works. In LTspice it should be easy to create
a heatsink symbol for use on the schematic. Easy for some -- Mike,
Helmut and analogspiceman could slam-dunk it. Me, I'll slog at it.

BTW: where did the capacitor numbers for aluminum and copper/brass come
from?

--

Hello Mike,
I have looked for some literature with Google about thermal
transistor and heatsink models.

http://www.iisb.fraunhofer.de/en/arb_geb/powersys_thermmod_gb_fhg.pdf
http://www.infineon.com/cmc_upload/migrated_files/document_files/Appli
cation_Notes/mmpn_eng.pdf

http://www.fairchildsemi.com/an/AN/AN-7533.pdf
http://www.irf.com/technical-info/designtp/temp002.pdf
http://www.irf.com/product-info/datasheets/data/100bgq045.pdf

They are all based on the equation Tfinal=Pv*Rth
with an exponential temperature rise from Tamb to Tfinal.

The most important question now is, how can we influence the
transistor(NPN, MOS) during a simulation run.
My understanding about(LT,P-)SPICE is, that there is no chance to do
that with the basic transistor models.
What I have seen in the above articles is that they build more or
less complex circuits around the basic transistor models.

Is that the only way to get temperature dependent transistor
behaviour with SPICE transistor models in a .TRAN simulation?

Best Regards,
Helmut

480)460-2350 | |
| E-mail Address at Website Fax

Harry Dellamano · Mar 4, 2004

"Mike Rocket J. Squirrel Elliott"
<j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote in message
news:b-> >>kensmith@rahul.net forging knowledge

Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current
source
as the heat generating driving element? The current output would be the
power dissipated. 25amps = 25watts.

Harry, I'm glad you saw that. I thought that a voltage source was
incorrect, too, but knowing that I rank in the bottom 10 percentile of
brainpower for sci.electronics.cad, I doubted myself. Thanks for
bringing it up.

--
Mike "Rocket J Squirrel" Elliott

Hey Rocket, even a blind squirrel will find some nuts.

Harry

Mike Rocket J. Squirrel E · Mar 4, 2004

Harry Dellamano wrote:

"Mike Rocket J. Squirrel Elliott"
j.michael.elliottAT@REMOVETHEOBVIOUSadelphiaDOT.net> wrote in message
news:b-> >>kensmith@rahul.net forging knowledge

Hi Ken, very interesting stuff but I am confused. Above you state, "a
voltage source to drive the RC model". Should not that be a current

source

as the heat generating driving element? The current output would be the
power dissipated. 25amps = 25watts.

Harry, I'm glad you saw that. I thought that a voltage source was
incorrect, too, but knowing that I rank in the bottom 10 percentile of
brainpower for sci.electronics.cad, I doubted myself. Thanks for
bringing it up.

--
Mike "Rocket J Squirrel" Elliott

Hey Rocket, even a blind squirrel will find some nuts.

This is probably more akin to chimps and typewriters.

--
Mike "Rocket J Squirrel" Elliott
71 VW Type 2 -- the Wonderbus (AKA the Saunabus in summer)

SPICE (LT): determining dissipation by resistors, etc.?

Mike Rocket J. Squirrel E

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