heatsink Rth calculation
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Tjunction = P * R_therm + Tambient
Tj = 5*4 + 5 * R_therm_heat_sink + T_ambient
T_case = 5 * R_therm + T_ambient
Peter H Anderson, http://www.phanderson.com
--- In basicstamps@y..., "peter verkaik" <peterverkaik@b...> wrote:
> Hi,
>
> Does anyone know of a formula to calculate the
> required thermal resistance Rth of a heatsink for a 7805
> in which the junction-to-case and case-to-ambient thermal
> resistance appear? All formula I find use some compound
>
Tj = 5*4 + 5 * R_therm_heat_sink + T_ambient
T_case = 5 * R_therm + T_ambient
Peter H Anderson, http://www.phanderson.com
--- In basicstamps@y..., "peter verkaik" <peterverkaik@b...> wrote:
> Hi,
>
> Does anyone know of a formula to calculate the
> required thermal resistance Rth of a heatsink for a 7805
> in which the junction-to-case and case-to-ambient thermal
> resistance appear? All formula I find use some compound
>
Comments
> Does anyone know of a formula to calculate the
> required thermal resistance Rth of a heatsink for a 7805
> in which the junction-to-case and case-to-ambient thermal
> resistance appear? All formula I find use some compound
> thermal resistance that I cannot directly relate to the given
> parameters from the datasheet.
**************
Peter:
Here is a copy of an old post I sent regarding a 7805. I hope it will help
you.
Ray McArthur
**************************************
Some power calculations:
For max power calculations, you must know max allowed junction temp, and
thermal resistances, which are degree C rise per watt. My 7805 data book
for TO220 case lists max Tj (junction temp) =125 C, and TRjc (thermal
resistance-junction to case)=4 deg C/watt. Also, TRca (thermal res case to
ambient, no heat sink)=50 deg C/watt. Now we can do some calculations:
*No heat sink*, 14 volts in, 0.5 amp load, 25 deg C ambient;
14-5=9 volts across regulator. 0.5amp*9V=4.5 watts reg dissipation.
Total TR (junction to ambient)=4+50=54 deg C/watt.
Temp rise, junction to ambient=4.5W*54 deg C/W=243 deg C.
Add ambient temp, Tjunction=243+25=268 deg C!
....spec is 125 C max so you need a heat sink.
*Infinite heat sink* This is a heat sink so large that the case is same temp
as ambient air.
The case = 25 C, so the junction rises TRjc above case.
4.5watt*4deg C/W=18 deg rise. Add case temp=25,
Tjunction=18+25=43 deg junction...no sweat.
This shows that a big enough heat sink will allow operation without a series
resistor.
*Now, try a series resistor*:
You should keep at least 2.5 volts across the 7805 for good regulation.
So you have 5V out + 2.5V=7.5V into 7805.
The resistor will drop 14V-7.5=6.5volts.
For 0.5 amp, R=6.5/0.5=13 ohms.
resistor pwr=0.5*6.5=3.25 watts.
Regulator pwr=2.5V*0.5amps=1.25 watts.
(notice we still dissipate 1.25+3.25=4.5 watts total).
Now check Tjunction, assuming no heat sink.
Total TR junction to ambient is 54 deg C/W as above.
Since the 7805 now only dissipates 1.25W,
1.25*54 deg C/W=67.5 deg C rise.
Add ambient of 25C,
Tjunction=67.5+25=92.5 deg C...no sweat.
This tells you that a series resistor would let you get away without a heat
sink. (For 25 deg air). You still have to dissipate the resistor heat, but
sometimes you can mount the hot resistor away from the electronics.
Hope this helps,
Ray McArthur
Original Message
From: Miguel Puchol <mpuchol@w...>
To: <basicstamps@egroups.com>
Sent: Tuesday, July 25, 2000 3:38 AM
Subject: RE: [noparse][[/noparse]basicstamps] 7805 regulator running hot!!
> The input voltage is 12 to 14 volts, current consumption should be no
> greater that 0.5A at max load.
> I will do some test, and will try your idea of placing a resistor at the
> input. I suppose that being an energy problem, dissipated heat will be the
> same, just distributed over two components.
>
>Does anyone know of a formula to calculate the
>required thermal resistance Rth of a heatsink for a 7805
>in which the junction-to-case and case-to-ambient thermal
>resistance appear? All formula I find use some compound
>thermal resistance that I cannot directly relate to the given
>parameters from the datasheet.
>
>7805:
>maximum junction temperature 150C
>junction-to-case Rth 4C/W
>case-to-ambient Rth 50C/W (TO-220)
>
>I need a formula that gives me the Rth for a
>heatsink that reaches a steady temperature T
>for any given power dissipation P.
>Example: 10V input, 5V output, 1A load -> P=5W
>wanted temperature for heatsink: 65C
>
>Thanks.
>Peter
Maybe I misunderstand what you are looking for, but at 25 degrees
ambient, the required thermal resistance of the heat sink would be
about,
R = (65 - 25) / 5 = 40/5 = 8 degrees C per watt.
In which case the junction temperature will get to about:
65 + 4*5 = 85 degree C.
assuming good thermal contact to the heat sink.
-- Tracy
Does anyone know of a formula to calculate the
required thermal resistance Rth of a heatsink for a 7805
in which the junction-to-case and case-to-ambient thermal
resistance appear? All formula I find use some compound
thermal resistance that I cannot directly relate to the given
parameters from the datasheet.
7805:
maximum junction temperature 150C
junction-to-case Rth 4C/W
case-to-ambient Rth 50C/W (TO-220)
I need a formula that gives me the Rth for a
heatsink that reaches a steady temperature T
for any given power dissipation P.
Example: 10V input, 5V output, 1A load -> P=5W
wanted temperature for heatsink: 65C
Thanks.
Peter
Richard
Original Message
From: "Ray McArthur" <rjmca@u...>
To: <basicstamps@yahoogroups.com>
Sent: Thursday, January 10, 2002 11:41 PM
Subject: Re: [noparse][[/noparse]basicstamps] heatsink Rth calculation
> >
> > Does anyone know of a formula to calculate the
> > required thermal resistance Rth of a heatsink for a 7805
> > in which the junction-to-case and case-to-ambient thermal
> > resistance appear? All formula I find use some compound
> > thermal resistance that I cannot directly relate to the given
> > parameters from the datasheet.
> **************
> Peter:
> Here is a copy of an old post I sent regarding a 7805. I hope it will
help
> you.
>
> Ray McArthur
>
> **************************************
> Some power calculations:
>
> For max power calculations, you must know max allowed junction temp, and
> thermal resistances, which are degree C rise per watt. My 7805 data book
> for TO220 case lists max Tj (junction temp) =125 C, and TRjc (thermal
> resistance-junction to case)=4 deg C/watt. Also, TRca (thermal res case
to
> ambient, no heat sink)=50 deg C/watt. Now we can do some calculations:
>
> *No heat sink*, 14 volts in, 0.5 amp load, 25 deg C ambient;
> 14-5=9 volts across regulator. 0.5amp*9V=4.5 watts reg dissipation.
> Total TR (junction to ambient)=4+50=54 deg C/watt.
> Temp rise, junction to ambient=4.5W*54 deg C/W=243 deg C.
> Add ambient temp, Tjunction=243+25=268 deg C!
> ....spec is 125 C max so you need a heat sink.
>
> *Infinite heat sink* This is a heat sink so large that the case is same
temp
> as ambient air.
> The case = 25 C, so the junction rises TRjc above case.
> 4.5watt*4deg C/W=18 deg rise. Add case temp=25,
> Tjunction=18+25=43 deg junction...no sweat.
>
> This shows that a big enough heat sink will allow operation without a
series
> resistor.
>
> *Now, try a series resistor*:
> You should keep at least 2.5 volts across the 7805 for good regulation.
> So you have 5V out + 2.5V=7.5V into 7805.
> The resistor will drop 14V-7.5=6.5volts.
> For 0.5 amp, R=6.5/0.5=13 ohms.
> resistor pwr=0.5*6.5=3.25 watts.
> Regulator pwr=2.5V*0.5amps=1.25 watts.
> (notice we still dissipate 1.25+3.25=4.5 watts total).
>
> Now check Tjunction, assuming no heat sink.
> Total TR junction to ambient is 54 deg C/W as above.
> Since the 7805 now only dissipates 1.25W,
> 1.25*54 deg C/W=67.5 deg C rise.
> Add ambient of 25C,
> Tjunction=67.5+25=92.5 deg C...no sweat.
>
> This tells you that a series resistor would let you get away without a
heat
> sink. (For 25 deg air). You still have to dissipate the resistor heat,
but
> sometimes you can mount the hot resistor away from the electronics.
>
> Hope this helps,
> Ray McArthur
>
>
Original Message
> From: Miguel Puchol <mpuchol@w...>
> To: <basicstamps@egroups.com>
> Sent: Tuesday, July 25, 2000 3:38 AM
> Subject: RE: [noparse][[/noparse]basicstamps] 7805 regulator running hot!!
>
>
> > The input voltage is 12 to 14 volts, current consumption should be no
> > greater that 0.5A at max load.
> > I will do some test, and will try your idea of placing a resistor at the
> > input. I suppose that being an energy problem, dissipated heat will be
the
> > same, just distributed over two components.
>
>
>
>
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>
>
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>
>
If I understand you correctly, the case-to-ambient parameter is only used
when no heatsink,
otherwise the Rth of the heatsink is used.
So
T_junction = (P * R_therm_junction_to_case) + (P * R_therm_heatsink) +
T_ambient <= T_junction_max (1)
T_case = T_heatsink = (P * R_therm_heatsink) + T_ambient <= T_heatsink_max
(2)
in which P is my dissipated power and T_heatsink_max is my wanted heatsink
temperature.
Rewriting (1)
R_therm_heatsink <= ((T_junction_max - T_ambient) / P) -
R_therm_junction_to_case (3) (must be >= 0)
Rewriting (2)
R_therm_heatsink <= (T_heatsink_max - T_ambient) / P (4) (must be >= 0)
The smallest value from (3) or (4) is the value to use.
Please correct me if I am wrong.
Regards peter
Oorspronkelijk bericht
Van: pha555 [noparse]/noparse]mailto:[url=http://forums.parallaxinc.com/group/basicstamps/post?postID=9M66QueJZ6FoWSE_Na7uw5qdNpKSnaBc_VsB6qPjfeJF3Tc6uN2EcBzdR53NjxCzYWspUhGprXkGdt0]pha@p...[/url
Verzonden: donderdag 10 januari 2002 21:09
Aan: basicstamps@yahoogroups.com
Onderwerp: [noparse][[/noparse]basicstamps] Re: heatsink Rth calculation
Tjunction = P * R_therm + Tambient
Tj = 5*4 + 5 * R_therm_heat_sink + T_ambient
T_case = 5 * R_therm + T_ambient
Peter H Anderson, http://www.phanderson.com
--- In basicstamps@y..., "peter verkaik" <peterverkaik@b...> wrote:
> Hi,
>
> Does anyone know of a formula to calculate the
> required thermal resistance Rth of a heatsink for a 7805
> in which the junction-to-case and case-to-ambient thermal
> resistance appear? All formula I find use some compound
>
To UNSUBSCRIBE, just send mail to:
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from the same email address that you subscribed. Text in the Subject and
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ambient air temp would result in a low enough junction temp.
Add up rth junction-to-case plus (about 0.1 to 1) for the rth case-to-sink
plus the heatsinks rated rth sink-to-ambient, then multiply this total known
as rth Junction-to-Ambient times P and then add in the ambient air temp.
This will be the temperature of the junction at these operating conditions
which is really what is important.
As an example we will first try one radio shack carries "THM6106" rth
sink-to-ambient 16 degrees Celsius per watt
The TO-220 78xx series of regulators are rated typically 3.0 max 5.0 degrees
Celsius per watt
so assuming worst case conditions
5.0+1+16=22 degrees C/W
22*5w=110 plus an ambient air temp of 25C results in a junction temp of 135C
this will work for the 78xx which is rated for a max junction temp of 150C
but not for the Commercial version 78xxC which is rated for 125C max
possible solutions include improving the case-to-sink by using the newer
formulation of heatsink compound such as Radio Shacks 276-1372A (better than
the old silicone zinc oxide), reducing input voltage, increasing heatsink
capacity ( I often use a THM6106 plus radio shacks newer 276-1368)
Also keep in mind that the 78xx series will shut down rather than burn up if
the heatsink is not adequate.
KF4HAZ Lonnie
Original Message
From: "peter verkaik" <peterverkaik@
> Hi,
>
> Does anyone know of a formula to calculate the
> required thermal resistance Rth of a heatsink for a 7805
> in which the junction-to-case and case-to-ambient thermal
> resistance appear? All formula I find use some compound
> thermal resistance that I cannot directly relate to the given
> parameters from the datasheet.
>
> 7805:
> maximum junction temperature 150C
> junction-to-case Rth 4C/W
> case-to-ambient Rth 50C/W (TO-220)
>
> I need a formula that gives me the Rth for a
> heatsink that reaches a steady temperature T
> for any given power dissipation P.
> Example: 10V input, 5V output, 1A load -> P=5W
> wanted temperature for heatsink: 65C
>
> Thanks.
> Peter
Note that there is a parallel to electrical circuits;
V = I * R T = P * Rtherm
When using no heat sink there are two resistances
V_total = I * (R1 + R2) + V_offset
Tj = P * (Rth_jc + Rth_ca) + T_ambient
When using a heat sink;
Tj = P * (Rth_jc + Rth_cs + Rth_heatsink_sa) + T_ambient
where Rth_jc is the conductive properties of the IC - usually about
4.0, Rth_cs is the bonding of the sink to the case - about 1.0 and
Rth_heat_sink is the property of the heat sink.
Best wishes.
Peter H Anderson, http://www.phanderson.com
--- In basicstamps@y..., "peter verkaik" <peterverkaik@b...> wrote:
> Thanks Ray and Phil,
>
> If I understand you correctly, the case-to-ambient parameter is
only used
> when no heatsink,
Ray McArthur
> I liked reading it then and I liked reading it again
>
>
> Richard
>
>
>
>
Original Message
> From: "Ray McArthur" <rjmca@u...>
> To: <basicstamps@yahoogroups.com>
> Sent: Thursday, January 10, 2002 11:41 PM
> Subject: Re: [noparse][[/noparse]basicstamps] heatsink Rth calculation
>
> > **************
> > Peter:
> > Here is a copy of an old post I sent regarding a 7805. I hope it will
thermal drop between the regulator case and heat sink. It usually varies
from a few tenths to as high as 1 deg C/watt, depending on flatness and if
grease is used. I use thermal grease because it beautifully compensates for
poor surfaces. You may want to add approx 0.5 deg c/W to either Rj-c or
Rheatsink for this additional temp rise.
Ray McArthur
***************
From: peter verkaik <peterverkaik@b...>
Subject: RE: [noparse][[/noparse]basicstamps] Re: heatsink Rth calculation
> Thanks Ray and Phil,
>
> If I understand you correctly, the case-to-ambient parameter is only used
> when no heatsink,
> otherwise the Rth of the heatsink is used.
> So
>
> T_junction = (P * R_therm_junction_to_case) + (P * R_therm_heatsink) +
> T_ambient <= T_junction_max (1)
>
> T_case = T_heatsink = (P * R_therm_heatsink) + T_ambient <= T_heatsink_max
> (2)
>
> in which P is my dissipated power and T_heatsink_max is my wanted heatsink
> temperature.
>
> Rewriting (1)
>
> R_therm_heatsink <= ((T_junction_max - T_ambient) / P) -
> R_therm_junction_to_case (3) (must be >= 0)
>
> Rewriting (2)
>
> R_therm_heatsink <= (T_heatsink_max - T_ambient) / P (4) (must be >= 0)
>
> The smallest value from (3) or (4) is the value to use.
>
>
> Please correct me if I am wrong.
> Regards peter