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'[EE]: PIC overheating a 7805 !'
2001\11\08@221026 by TH

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Hi, I've got this system up and running where a series of modules containing 16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 package) are all run from the same 24VDC line (long run) coming from a switching supply. Each module also has an LM7806 for driving an R/C Servo motor. All my modules work fine except that the 7805s are running rather hot even though the R/C motors are pulling much more current from the 7806 than the little PIC is from the 7805. The 7806's temperature is ok though. At first I thought that it would only be necessary to heat-sink the motor's regulator but know I think I might have to do the same with the PIC's. I can almost burn my finger from the 7805 ...

I understand that 24V isn't the optimal voltage to run the 7805 from if I intended to draw a lot of current from it but National's datasheet says that it can take up to 35V and considering the very low current consumption I don't understand why it is overheating. I suspected the length of my supply line but running a module even a few minutes on a short length gives the same result. Besides, the voltage drop is barely perceptible and I have pretty good wire gauge. Could my choice of caps for the regulators be causing the problem ? I have one 470uF 35V electrolytic before the 2 regulators and a 0.1uF ceramic cap after each of them.

thanks,

Tobie Horswill
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2001\11\08@223626 by David Duffy

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At 10:07 PM 08/11/2001 -0500, you wrote:
{Quote hidden}

Maybe the 7805 is oscillating? You should have a 10uF capacitor on the 7805
output as well.
Regards...

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2001\11\08@230441 by Kathy Quinlan

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I had a similar problem with a 7805 running from a 24V source, the load was
a 555 timer, 4020, 2716 eprom and a 74ls373  driving 3 opto triacs IIRC. not
much load, but the 7805 had to be mounted on a 1.0oC/W heatsink to keep it
working, since then I have mad a design rule, all 7805's are fed from 8 Vdc.
I know they have a wide input voltage (I have stack (not sure of PN) that
were 7805 compatible that had 120VDC input rating).

These days unless cost is an issue I either use a multi tap transformer, or
I use switching regulators.

Regards,

Kat.


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{Original Message removed}

2001\11\08@233050 by Dale Botkin

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Do you have a 'scope, or at least a decent multimeter?  A scope will tell
you if the output is funky (7805 oscillating) or just hang an ammeter on
the input & output and see what the current draw is, from that you can
figure the power dissipaton in the reguulator and see if the numbers make
sense.

Dale
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2001\11\09@014150 by myke predko
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Kathy Wrote:
> I had a similar problem with a 7805 running from a 24V source, the load
was
> a 555 timer, 4020, 2716 eprom and a 74ls373  driving 3 opto triacs IIRC.
not
> much load, but the 7805 had to be mounted on a 1.0oC/W heatsink to keep it
> working, since then I have mad a design rule, all 7805's are fed from 8
Vdc.
> I know they have a wide input voltage (I have stack (not sure of PN) that
> were 7805 compatible that had 120VDC input rating).

I see problems like this all the time and it bears going back and
understanding what is happening.

As a rough guess, I would think the circuit was drawing 300 to 350 mA
(assuming the 555 is bipolar and 2716 is NMOS) - according to the
datasheets, well within the output range of the 7805.

But, remembering that the formula for power dissipated is:

Power = Voltage Across the device * Current

In your application, the 7805 has 19 Volts across it (24 Volts in - 5 Volts
out) so the total power dissipated in the 7805 is:

Power = 19 Volts * 300 mA
     = 5.7 Watts

This power level is going to put the 7805 into thermal shutdown (crowbar)
pretty quickly.

> These days unless cost is an issue I either use a multi tap transformer,
or
> I use switching regulators.

Maxim has some nice regulator chips that are pretty inexpensive - on a
production board, a power supply using these chips would cost you about as
much as a 7805 and you don't have the heat issues (and if you are running
from batteries, the circuit will run a *lot* longer).

myke

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2001\11\09@020536 by Wagner

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> Subject: [EE]: PIC overheating a 7805 !
> From: TH (.....thorswilKILLspamspam@spam@VIDEOTRON.CA)
> Date: Thu Nov 08 2001 - 19:07:20 PST
> Hi, I've got this system up and running where a series of
> modules containing 16F628 (running @ 20MHz) and ... [snip]

Well, from 24Vdc the Vdrop will be 19V, even if your circuit is draining
only 200mA, it means 3.8Watts.  Without a proper heatsink this can be hot.
Oscillation can be an issue, but not likely. The 7806 does not heat so much,
probably because the servors could work in intermitent fashion so the
thermal average would be low.

The Microchip datasheet shows 7mA @ 20MHz, what would heats the 7805 @ less
than 150mW.  However, the maximum current into VDD could reach @ 250mA
before it blows up the chip.

Check your design, make sure no late night soldering got a short circuit at
any unused port pin or by mistake in soldering or design.  Also check the
crystal oscillator for a proper signal shape and level.  Just for testing,
change the crystal for a 4MHz and see what happens, perhaps a not so good
20MHz units lot?  Also check Microchip for 16F628 Errata.

By any analogy, what is sinking current to heats the 7805 is also heating at
least @ 5V, so it should heat 1/4 of the 7805 thermal reference.  A good wet
finger sometimes can sense what component is doing this.

In case you have an ampermeter, try and measure everything.  In case you
have a thermal noncontact measuring gun, use it, or simply point your video
camera to the unit in a dark room, you will be surprise what the camera
actually can shows you as thermal image. Most B&W video cameras are better
sensitive to IR than the color ones...

have fun.
Wagner Lipnharski
http://www.ustr.net

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2001\11\09@022257 by Gennette, Bruce

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You do have them fitted correctly don't you?

I've accidentally managed to make 78L05s 'work' from a highish voltage when
mounted back-to-front, and they got very hot.  Mine (in backwards) were
passing about 4.6V from 28V input.

Bye.

{Original Message removed}

2001\11\09@044459 by Dave King

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Sometime you gotta love this list. Sometime people go right out and let the
smoke
out of their bits long before your iron is hot. That and someone blurts out
things that
save time n money.

Just so I'm on topic slightly. We never design anything with a 7805 that
doesn't
have a head voltage in the 3-5 range. Preferably 3 if its going to pull
near rated
current. I have run one with 18.5 head and it ran red hot even with a huge
sink on
it. It was one of the least reliable designs I let out.


>Maxim has some nice regulator chips that are pretty inexpensive - on a
>production board, a power supply using these chips would cost you about as
>much as a 7805 and you don't have the heat issues (and if you are running
>from batteries, the circuit will run a *lot* longer).
>
>myke

Now this just made my day. I had a quick look at some of Maxims stuff and
found
one of these that eliminates me having to design two separate power supplies.
I can heat sink it for worst case and laugh.

You know I think I've learned two things from this list this month. That's
almost scary ;-]
As a direct result I didn't have to think nearly as hard and can now afford
to kill off
those extra brain cells with beer this weekend.

Cheers

Dave

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2001\11\09@050759 by dr. Imre Bartfai

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Hi,

try LM2574 - fine, cheap, cool!

Imre


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On Fri, 9 Nov 2001, Kathy Quinlan wrote:

{Quote hidden}

> {Original Message removed}

2001\11\09@051426 by Kathy Quinlan

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Hi hon, the 555 was a 7555 Cmos version as was the 2716 (27c16) current was
around 200mA. Even with a 20mA current draw the 7805 was hot, till we used a
separate winding (a few turns of 0.5mm2 cable over the toroidal transformer
(that's why I love them so much, cheap and can do custom winds on top :o)))

Regards,

Kat.
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{Original Message removed}

2001\11\09@052720 by solar

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Am I the only person who has just received multiple copies of this email ??

----- Original Message -----
From: "Dave King" <KingDWSspamKILLspamHOME.COM>
To: <.....PICLISTKILLspamspam.....MITVMA.MIT.EDU>
Sent: Friday, November 09, 2001 4:22 PM
Subject: Re: [EE]: PIC overheating a 7805 !


> Sometime you gotta love this list. Sometime people go right out and let
the
> smoke
> out of their bits long before your iron is hot. That and someone blurts
out
{Quote hidden}

as
> >much as a 7805 and you don't have the heat issues (and if you are running
> >from batteries, the circuit will run a *lot* longer).
> >
> >myke
>
> Now this just made my day. I had a quick look at some of Maxims stuff and
> found
> one of these that eliminates me having to design two separate power
supplies.
> I can heat sink it for worst case and laugh.
>
> You know I think I've learned two things from this list this month. That's
> almost scary ;-]
> As a direct result I didn't have to think nearly as hard and can now
afford
{Quote hidden}

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2001\11\09@061426 by Vasile Surducan

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On Fri, 9 Nov 2001, Dave King wrote:

> Sometime you gotta love this list. Sometime people go right out and let the
> smoke
> out of their bits long before your iron is hot. That and someone blurts out
> things that
> save time n money.
>
> Just so I'm on topic slightly. We never design anything with a 7805 that
> doesn't
> have a head voltage in the 3-5 range. Preferably 3 if its going to pull
> near rated

 3V it's not enough ! Do a simple computation for a 220V +10% -15% ( or
120V, whatever...) mains variation and see you need at least 4V in-out to
be sure with output ripple.
Better are: LM low dropout series ( 1.25...1.5 V in-out )
            KA78R05 ( 0.5...1V in-out)
and of course the standard 7805 with good design consideration up to +35V
input.
regards, Vasile

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2001\11\09@061724 by Russell McMahon

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Tobie said -

Hi, I've got this system up and running where a series of modules containing
16F628 (running @ 20MHz) and individually fed by LM7805 (TO-220 package) are
all run from the same 24VDC line (long run) coming from a switching supply.
Each module also has an LM7806 for driving an R/C Servo motor. All my
modules work fine except that the 7805s are running rather hot even though
the R/C motors are pulling much more current from the 7806 than the little
PIC is from the 7805. The 7806's temperature is ok though. At first I
thought that it would only be necessary to heat-sink the motor's regulator
but know I think I might have to do the same with the PIC's. I can almost
burn my finger from the 7805 ...

I understand that 24V isn't the optimal voltage to run the 7805 from if I
intended to draw a lot of current from it but National's datasheet says that
it can take up to 35V and considering the very low current consumption I
don't understand why it is overheating. I suspected the length of my supply
line but running a module even a few minutes on a short length gives the
same result. Besides, the voltage drop is barely perceptible and I have
pretty good wire gauge. Could my choice of caps for the regulators be
causing the problem ? I have one 470uF 35V electrolytic before the 2
regulators and a 0.1uF ceramic cap after each of them.

____________


It doesn't SOUND like it makes sense.
IF the 6v supply IS drawing more current from the same input then it DOESN'T
make sense.

- As others note, ensure the regulator is not oscillating. A scope helps a
lot for this.
Make sure caps are per the databook.

- A 7805 / LM340 in a TO220 package is good for about 1 watt (preferably
less) without a heatsink.
At 24 volts in, 5v out this gives Current = 1 watt / (24-5) = ~50 mA.
A PIC by itself will draw much less than that.

- Check that the 5 v supply is not drawing more than you think.

- A series resistor will take the load off the regulator.
eg at 100 mA and 15v drop a 15/.1 = 150 ohm resistor will dissipate V^2/R =
225/150 = 1.5 watts and allow the regulator to dissipate
(24-15-5) * 0.1 = 0.4 watt.
If you place a 150 ohm resistor in series and the regulator can't support
the 5v output then it is drawing rather more than 100 mA.
Note that you need a larger decoupling capacitor at the regulator input if
you use the series input resistor.
Using a 5 watt wire wound here will move most of the dissipation out of the
regulator.
If the 24v supply varies to below 24v you will have to design the resistor
for the lowest input voltage.

Rseries = (Vin_minimum - 5v - Vheadroom) / (Imaximum)
V headroom is the minimum voltage you want across the regulator. For a 7805
2.0v is safe.



How's the Autonomous Boat coming?





regards



           Russell McMahon

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2001\11\09@092909 by Mark Peterson

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An easy way to unoad the regulator is to put a series resistor in the
input.  Let it dissipate as much of the heat as possible.  Size it such
that you always maintain the minimum input voltage required by the
regulator.  If, for example, the 7805 needs a minimum of 7 volts at its
input and you have a 24 volt source, use a resistor that will drop 17 volts
at your load's maximum current draw.

Mark Peterson

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2001\11\09@134755 by Harold M Hallikainen

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       24V is a bit high! Calculate out the power dissipation
(24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of
your heat sink in degrees C/ watt. This should give temperature rise from
ambient.
       Another thing to watch for is oscillating regulators, which get hot.
Check the output on a scope and make sure you've got DC without a bunch
of RF.
       Finally, consider a switching regulator. The inefficiency of using a
linear regulator to go from 24V down to 5V is pretty impressive.

Harold


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2001\11\09@140722 by Harold M Hallikainen

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On Fri, 9 Nov 2001 08:26:27 -0600 Mark Peterson <EraseMEmarkpspam_OUTspamTakeThisOuTCANNONTECH.COM>
writes:
> An easy way to unoad the regulator is to put a series resistor in the
> input.  Let it dissipate as much of the heat as possible.  Size it
> such
> that you always maintain the minimum input voltage required by the
> regulator.  If, for example, the 7805 needs a minimum of 7 volts at
> its
> input and you have a 24 volt source, use a resistor that will drop
> 17 volts
> at your load's maximum current draw.
>

       Or a series zener diode so the input to the regulator does not vary as
much with load...

Harold


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2001\11\09@203446 by Brian Kraut

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Your problem is very simple.  You are dropping 19V accross the
regulator.  Multiply that by the current and you will see that you need
to dissipate considerable wattage in the regulator.

I have several devices that need to run on 24 or 12V and use a 7805 in
an enclosure with no way to use a heat sink.  I put a 12V zeener in
series with the input positive supply before the regulator.  I put a
jumper on the board in parallel with the zeener so customers can cut the
jumper if they are using a 24V supply and leave it if they have a 12V
supply.  Use a 5W zeener and try to have it as far as possible from the
regulator.  I also try to have big fat traces going to the regulator and
the zeener to conduct the hear away from it also.

Power Trends(now owned by TI) mini switching regulators are also great
if your project can justify the fourteen bucks a pop.

TH wrote:

{Quote hidden}

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2001\11\12@030140 by Mike Blakey

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The power used and dissipated by the voltage regulator is NOT calculated that way.
the power dissipated by the device is approx. = (Vin-Vout) * quiescent current and
NOT load current. Check the data sheets!





@spam@haroldhallikainenKILLspamspamJUNO.COM on 09/11/2001 18:44:56
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Subject:        Re: [EE]: PIC overheating a 7805 !

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       24V is a bit high! Calculate out the power dissipation
(24V-5V)*LoadCurrent. Take this (in watts) and multiply by the rating of
your heat sink in degrees C/ watt. This should give temperature rise from
ambient.
       Another thing to watch for is oscillating regulators, which get hot.
Check the output on a scope and make sure you've got DC without a bunch
of RF.
       Finally, consider a switching regulator. The inefficiency of using a
linear regulator to go from 24V down to 5V is pretty impressive.

Harold


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2001\11\12@032838 by SkinTech

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----- Original Message -----
From: "Mike Blakey" <TakeThisOuTmike.blakeyEraseMEspamspam_OUTBAESYSTEMS.COM>
To: <RemoveMEPICLISTspamTakeThisOuTMITVMA.MIT.EDU>
Sent: Monday, November 12, 2001 8:58 AM
Subject: Re: [EE]: PIC overheating a 7805 !


> The power used and dissipated by the voltage regulator is NOT calculated
that way.
> the power dissipated by the device is approx. = (Vin-Vout) * quiescent
current and
> NOT load current. Check the data sheets!
>
>I beg to differ. I think you mix up power consumption and power
diddipation. Power consumption is what the regulator uses to do the
regulation, so to speak. That is very little indeed. But the dissipation is
(Vin-Vout)*load current (plus the power consumption of course, but that is
normally insignificant). Look at it as a (variable) resistor with a voltage
difference of Vin -Vout, and a current = load current.
If it was us you seem to think, no regulator would ever become hot and Ohms
law would be violated...

Regards, jan Didden
{Quote hidden}

hot.
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2001\11\12@034116 by Vasile Surducan

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Sorry Mike, but is this case your datasheet is wrong.
Any  device working as linear device ( not switching ) is dissipating
delta_V * I where delta_v is in-out difference and I is the current
flow through device. ( Ic and Uce at transistors, Uin_out and Iload at
7805 )
Regards, Vasile


On Mon, 12 Nov 2001, Mike Blakey wrote:

{Quote hidden}

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2001\11\12@072925 by Kathy Quinlan

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Hi Mike,

I have searched data sheets from natsemi and TI, and cannot see how to
calculate power disapation.

Could you please provide a reference.

I disagree with your comment, to me a linear device working in a linear
region has to dissipate the power not consumed by the load ie

If I have 24 volts in, 5volt out to the load, load consumes say 1 amp.

Power into regulator is 24W, power out is 5W, so the regulator must consume
the missing 19W.

With your idea, the max power consumption would be 152mW ( vin - vout * Qc)
= (24 -5) *0.008A (from the TI data sheet maximum value)

Regards,

Kat.

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2001\11\12@074837 by artstar

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The datasheet is right, the interpretation of it wasn't, that's all 3#-)

Adios,
LarZ

---------------  TAMA - The Strongest Name in Drums  ---------------

{Original Message removed}

2001\11\12@081148 by Gerhard Fiedler

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At 20:32 11/12/2001 +0800, Kathy Quinlan wrote:
>I disagree with your comment, to me a linear device working in a linear
>region has to dissipate the power not consumed by the load ie
>
>If I have 24 volts in, 5volt out to the load, load consumes say 1 amp.
>
>Power into regulator is 24W, power out is 5W, so the regulator must consume
>the missing 19W.

Of course :)  Just one comment to add: the device doesn't have to work in a
linear region for this to be true.

Ah, yes: the difference between current in and current out has to do with
the power consumption of the device (which is probably what Mike confused
with the dissipation under load).

ge

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2001\11\12@084302 by Vasile Surducan

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On Mon, 12 Nov 2001, Gerhard Fiedler wrote:

> Of course :)  Just one comment to add: the device doesn't have to work in a
> linear region for this to be true.
>
 But also the relationship is different then.
 In switching mode for example, device dissipating power is proportional
with voltage drop across device and flowing current only in t_on and t_off
time. ( where t_on is comutation time between low and hi state and t_off
is the reverse ) So in that expression we have also an integral. Which is
missing under linear usage of device. This is the only difference.
Regards, Vasile

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2001\11\12@092044 by Gerhard Fiedler

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At 15:38 11/12/2001 +0200, Vasile Surducan wrote:
>On Mon, 12 Nov 2001, Gerhard Fiedler wrote:
>
> > Of course :)  Just one comment to add: the device doesn't have to work in a
> > linear region for this to be true.
> >
>   But also the relationship is different then.
>   In switching mode for example,

Of course. The complete phrase I was referring to was "to me a linear
device working in a linear region ..."  A switching device is not
considered a linear device, mostly, probably.

OTOH, a DC current through a transistor may well be in a region where the
transistor is not considered linear, but the simple calculation of
dissipated power is still valid enough for most purposes.

I guess "linear" is not defined well enough to be really useful here...
:)  There are few if any fully linear devices in real world circuits, and
what you consider linear depends on your specific point of view. "Linear"
power regulators have some quite unlinear regions, and we're dealing with
them every now and then :)

ge

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2001\11\12@113219 by Mike Blakey

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Hi Kat,

I Think I misunderstood the original point of the thread and only interperated what
I thought.

According to National Semiconductors data sheets the TOTAL Power is

Pd = ((Vin -Vout) * Load Current) + (Vin * Quesient Current)

MAximum allowable temperature is Tr(max) = Tj(max) - Ta(max)

Tj(max) is 125deg C    and Ta(max) is the maximum ambiant for your application.

Thermal Resistance(ja) = Tr(max) / Pd

Use this to calculate the TR of the heatsink.

I hope that helps?






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Hi Mike,

I have searched data sheets from natsemi and TI, and cannot see how to
calculate power disapation.

Could you please provide a reference.

I disagree with your comment, to me a linear device working in a linear
region has to dissipate the power not consumed by the load ie

If I have 24 volts in, 5volt out to the load, load consumes say 1 amp.

Power into regulator is 24W, power out is 5W, so the regulator must consume
the missing 19W.

With your idea, the max power consumption would be 152mW ( vin - vout * Qc)
= (24 -5) *0.008A (from the TI data sheet maximum value)

Regards,

Kat.

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{Original Message removed}

2001\11\12@130024 by Harold M Hallikainen

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       I'll respectfully disagree.

Harold


On Mon, 12 Nov 2001 07:58:13 +0000 Mike Blakey
<.....mike.blakeyspamRemoveMEBAESYSTEMS.COM> writes:
{Quote hidden}

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2001\11\12@132152 by Kirk Lovewell

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I'll believe the datasheets right up to the point where they violate the
laws of physics!
A simple application of the principle of conservation of energy should tell
you that that spec is bogus!

Kirk

{Original Message removed}

2001\11\13@172710 by Peter L. Peres

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For regulators with a large voltage drop across them the power drain of
the regulator proper becomes a significant part of Pd at low loads. With
devices used without a heatsink this can be the straw that breaks the
camel's back and cause strange faults in cased devices (regulator enters
thermal shutdown under hard to duplicate conditions - i.e. at the
customer's location, over the bridge, down the lane, etc, on hot Sundays
exclusively).

Pdreg = U1 * Iddreg + (U1 - U0) * Iload

7805 series regulators use little extra Iddreg to drive the output
transistors because they are darlingtons and the base currents also flow
to the load. With other regulators this is not so.

The self-heating problem is especially visible with TO92 regulators. The
usual fix is to add a R in series with the voltage input such that at max.
load Uin on the regulator is higher than the required min. input voltage.
A decoupling capacitor is required after the resistor.

R = (U1 - Uinmin) / Iloadmax

It is possible to split an arbitrary part of the power to be dissipated
into the resistor (i.e. 50%, 80% etc) and it need not be just one
resistor, it can be a series combo. Me, when I do this, I arrange for the
power dissipation in each device to be equal (including in the regulator),
so I have no hot spots on the board (with similar sized devices - e.g.
TO220 7805 and 2W resistors). The formula for that is:

Rtot = (U1 - Uload) / (2 * Iload)

This is inefficient but nobody cares whether a couple of resistors
dissipate 2W in a unit that draws 100 or 3000W. You can put other things
in series if the current is reasonably constant, like dial lights, pilot
lights, anti-dew heaters, LCD backlights, even a small fan with
decoupling. The fan trick is neat because as the circuit draws more
current the fan accelerates ... (assuming DC brushless fan).

A 7805 TO220 is not to dissipate more than about 1.5W without a heatsink
(at 50C ambient) and it will run very hot to the touch at that. Up to
about 1W into a device of that size will keep it 'touchably' warm. At
24Vin and 5 Vout that is less than 50mA total, leaving only 40mA for the
load (acc. NS data book).

With an 'equal dissipation' R the current can be doubled under the same
conditions (R = 180 Ohms, 2W). Decouple the regulator input after the
resistor with 10 uF 16V to GND. The resistor will run about as hot as the
TO220 case (not as hot as the TO220 case without the R). This will allow
loads up to 80mA and may save your design as is now (cut trace, connect
resistor with wires - after packing it in heatshrink tubing and add
decoupling capacitor on bottom of board - if you have room). Remember that
heat does not go away, you will have to put the resistor in a place where
it does not heat something else - like making ugly dents in the plastic
case after a while).

Peter

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2001\11\14@203933 by Friedel Bruening

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As fixed regulators tend to be cheaper than power resistors,
I prefer to use an other regulator in front with even better results, let´s say you got 24V
and need 5V with like 100mA, so take a 12V regulator first (24V-12V=12V*100mA=1.2W)
and than the 5V regulator (12V-5V=7V*100mA=0.7W) and you get off without needing
the extra cap, which results even cheaper.
Friedel

At 09:30 p.m. 13/11/01 +0200, you wrote:
{Quote hidden}

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