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'[EE] TO-92 volt regs and heat issues'
2005\10\18@164755 by Jim Robertson

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Greetings Piclisters.

I currently have a 78L05 (100mA) in the WARP-13 but with the possible
change over to USB via a 18F2550 it tends to run just a little hot.

My question  is if I change it over for a Zetec ZSR500C that is rated
for 200mA does it follow that the output drive is lower impedance and
therefore if I draw the same amount of current from it, it would not
generate as much heat as the 78L05? Right/Wrong?

I am also looking at adding a heatsink if I can find a suitable size
and shape. Any other ideas or solutions anyone. Thanks for any help.

Regards,

Jim


2005\10\18@172637 by Mike Hord

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> My question  is if I change it over for a Zetec ZSR500C that is rated
> for 200mA does it follow that the output drive is lower impedance and
> therefore if I draw the same amount of current from it, it would not
> generate as much heat as the 78L05? Right/Wrong?

Doesn't sound right to me.  After all, let's say you put 12V into the
regulator and 5V out.  As long as it's a linear regulator, does it really
matter which regulator is there?  Any linear regulator will dissipate
(Vin-Vout)*Iout watts.

Of course, some will do so much more gracefully than others for a
given current drain...

Mike H.

2005\10\18@172852 by Mike Harrison

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On Wed, 19 Oct 2005 06:47:59 +1000, you wrote:

>
>Greetings Piclisters.
>
>I currently have a 78L05 (100mA) in the WARP-13 but with the possible
>change over to USB via a 18F2550 it tends to run just a little hot.
>
>My question  is if I change it over for a Zetec ZSR500C that is rated
>for 200mA does it follow that the output drive is lower impedance and
>therefore if I draw the same amount of current from it, it would not
>generate as much heat as the 78L05? Right/Wrong?

No.
Linear regs burn off the difference between input and output voltage as heat, so the device rating
makes no difference.
Heatsinking, changing to a TO220 package, reducing input voltage or load current are the only things
you can do.
Could you run the PIC off the USB supply instead ?


2005\10\18@174554 by Spehro Pefhany

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At 06:47 AM 10/19/2005 +1000, you wrote:

>Greetings Piclisters.
>
>I currently have a 78L05 (100mA) in the WARP-13 but with the possible
>change over to USB via a 18F2550 it tends to run just a little hot.
>
>My question  is if I change it over for a Zetec f that is rated for 200mA
>does it follow that the output drive is lower impedance and therefore if I
>draw the same amount of current from it, it would not generate as much
>heat as the 78L05? Right/Wrong?

Mostly wrong. The Zetex part draws less current for its own use, so it will
run a bit cooler on account of that (600uA maximum rather than about 5mA).

Power dissipation is P = Vin* Iq + (Vin-Vout) * Iload, and usually the latter
term dominates- eg. with 16 volts in, the left term is only 80mW.

>I am also looking at adding a heatsink if I can find a suitable size and
>shape. Any other ideas or solutions anyone. Thanks for any help.

There are heatsinks for TO-92s, which might safely get you
up into the mid hundreds of mW if the ambient isn't too high.

http://rocky.digikey.com/WebLib/Aavid/Web%20Data/575200B00000.pdf

But personally, I prefer the inexpensive 78M05CDT SMT part (with a bit of
copper around it if there's no power planes) for moderate power dissipation
levels.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
spam_OUTspeffTakeThisOuTspaminterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com
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2005\10\18@175835 by olin piclist

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Jim Robertson wrote:
> My question  is if I change it over for a Zetec ZSR500C that is rated
> for 200mA does it follow that the output drive is lower impedance and
> therefore if I draw the same amount of current from it, it would not
> generate as much heat as the 78L05? Right/Wrong?

The impedence (current per droop voltage) is a function of the feedback and
other parameters inside the regulator.  This is independent of maximum
allowable current load.  In either case I expect the impedence to be
sufficiently low to not be an issue.  After all, that's what a regulator
does.

Impedence is also not related to power dissipation.  Dissipation is simply
(Vin - Vout) * Iin.  Note that Iin can be considered the same as Iout for
this purpose.  The difference is the regulator's ground pin current, which
will be negligeable in the realm where you have to worry about power
dissipation.

So the short answer is that replacing the regulator with a higher current
version will have no effect at all on the power dissipated in that
regulator.

> I am also looking at adding a heatsink if I can find a suitable size
> and shape. Any other ideas or solutions anyone. Thanks for any help.

Assuming the regulated output voltage is a fixed requirement and that the
circuit will draw what it will draw, the only way to decrease power
dissipation in the regulator is to reduce its input voltage.  Of course you
must still maintain the regulator's minimum required drop voltage at your
worst case current.  This can be only a few 100mV for a good LDO, or over 2V
for a brute force 7805.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

2005\10\18@183231 by Jinx

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> Any other ideas or solutions anyone

A series resistor to drop some of the source voltage before
it gets to the regulator. Get i/p down to the minimum you can
to keep a stable o/p voltage

I've got a circuit here with 3 PICs, 2 Scenix's (boy, do they
chomp through the power) and assorted bits and pieces.
Although the 7805 was on a fairly big h/s it still got red hot.
Adding a 22R 10W in series with the 12V source takes a
lot of the sting out and the 7805 could probably run without
a h/s now

2005\10\18@204716 by Chen Xiao Fan

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If you change to USB, maybe you can do without the external
power and generated the target Vpp with a step-up boost
converter or charge pump.

At least you can supply the 18F2550 with USB power
and supply the rest of the +5V from the 7805. In that case,
the current consumption of the 7805 should be lower
than the original Warp13a.

Regards,
Xiaofan

{Original Message removed}

2005\10\18@233544 by Russell McMahon

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1.    The Zetex E-line package (which looks similar to TO-92) has
slightly more dissipation capability than TO-92 - aboy 500 - 600mZW
AFAIR depending on actual product.

BUT

2.    A TO-92 7805 is seldom much cheaper and somtoimes dearrer than a
TO-220 which can dissipate far more unheatsunk power. A better
solution as long as you can tolerate the larger size.

3.    All 7805's should be destroyed on sight :-). While in their day
they were marvellous their day has long gone. Low price is their only
good feature.



       RM




2005\10\18@235150 by Chen Xiao Fan

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But low price is one of the most important features. Besides
they are cheaper they are also more robust and easier to deal
with than some LDOs. Microchip LDOs, for example, are not really
very useful since they can not withstand slightly higher voltage.

Regards,
Xiaofan

-----Original Message-----
From: Russell McMahon
Sent: Wednesday, October 19, 2005 11:36 AM
To: Microcontroller discussion list - Public.
Subject: Re: [EE] TO-92 volt regs and heat issues

3.    All 7805's should be destroyed on sight :-). While in their day
they were marvellous their day has long gone. Low price is their only
good feature.

       RM

2005\10\19@053525 by Michael Rigby-Jones

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>-----Original Message-----
>From: Russell McMahon [.....piclist-bouncesKILLspamspam@spam@mit.edu]
>Sent: 19 October 2005 04:36
>To: Microcontroller discussion list - Public.
>Subject: Re: [EE] TO-92 volt regs and heat issues
>
>
>3.    All 7805's should be destroyed on sight :-). While in their day
>they were marvellous their day has long gone. Low price is their only
>good feature.

And a feature that is of critical importance in many designs!  For a low current regulator, the 78LM05's are pretty dire in terms of quiescent current, but the 1 Amp TO220 devices have a very reasonable price/performance ratio IME.

Regards

Mike

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2005\10\19@053925 by Michael Rigby-Jones

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>-----Original Message-----
>From: piclist-bouncesspamKILLspammit.edu [.....piclist-bouncesKILLspamspam.....mit.edu]
>Sent: 19 October 2005 04:52
>To: 'Microcontroller discussion list - Public.'
>Subject: RE: [EE] TO-92 volt regs and heat issues
>
>
>But low price is one of the most important features. Besides
>they are cheaper they are also more robust and easier to deal
>with than some LDOs. Microchip LDOs, for example, are not really
>very useful since they can not withstand slightly higher voltage.
>
>Regards,
>Xiaofan

Yes, I did wonder about the usefulness of a 5 volt regulators with maximum input of 6 volts!

Regards

Mike

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This e-mail is intended for the person it is addressed to only. The
information contained in it may be confidential and/or protected by
law. If you are not the intended recipient of this message, you must
not make any use of this information, or copy or show it to any
person. Please contact us immediately to tell us that you have
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No part of this message can be considered a request for goods or
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2005\10\19@055205 by Vasile Surducan

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On 10/19/05, Jinx <EraseMEjoecolquittspam_OUTspamTakeThisOuTclear.net.nz> wrote:
> > Any other ideas or solutions anyone
>
> A series resistor to drop some of the source voltage before
> it gets to the regulator. Get i/p down to the minimum you can
> to keep a stable o/p voltage

Right. But also induce high instability in the input ripple because of
equivalent impedance which grows. So it must be
accomplished with a capacitor between stab input and ground.
Unfortunately the heat is heat no matter if is generated by a passive
or an active component. For a closed box it means problems.


Vasile

>
> I've got a circuit here with 3 PICs, 2 Scenix's (boy, do they
> chomp through the power) and assorted bits and pieces.
> Although the 7805 was on a fairly big h/s it still got red hot.
> Adding a 22R 10W in series with the 12V source takes a
> lot of the sting out and the 7805 could probably run without
> a h/s now
>
> -

2005\10\19@075150 by olin piclist

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Russell McMahon wrote:
> 3.    All 7805's should be destroyed on sight :-). While in their day
> they were marvellous their day has long gone. Low price is their only
> good feature.

They are great when you've got the input voltage headroom.  In that case
it's hard to beat the brick outhouse performance.  When the asteroid hits,
only cockroaches and 7805s will be left.


*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

2005\10\19@090629 by Russell McMahon

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>> 3.    All 7805's should be destroyed on sight :-). While in their
>> day
>> they were marvellous their day has long gone. Low price is their
>> only
>> good feature.

> They are great when you've got the input voltage headroom.

and quiescent current doesn't matter.

> In that case it's hard to beat the brick outhouse performance.  When
> the asteroid hits,
> only cockroaches and 7805s will be left.

I agree that they still have their place (despite my above comment) -
I even bought some recently :-). But they are not as rugged as many
more modern devices. In particular they can be destroyed by suddenly
dropping Vin when the output has substantial capacitance on it, so
that the cap reverse discharges through the regulator. Death happens.
This can be protected against by adding a reverse diode from Vout to
Vin - but it would have been nice to have it internally. To their
credit they (NatSemi anyway) do cover this in the application notes
section of the data sheet.


       RM


2005\10\20@171038 by Jim Robertson

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Thanks to all those who replied to my post.

I have some good suggestions to work with and to forward to those who
may do a WARP-13 USB retrofit.

Regards,

Jim


2005\10\21@144254 by Wouter van Ooijen

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> My question  is if I change it over for a Zetec ZSR500C that is rated
> for 200mA does it follow that the output drive is lower impedance and
> therefore if I draw the same amount of current from it, it would not
> generate as much heat as the 78L05? Right/Wrong?

It would of course generate the same amount of heat, unless something
weird is happening (like an oscillating 78L05).

> I am also looking at adding a heatsink if I can find a suitable size
> and shape. Any other ideas or solutions anyone. Thanks for any help.

I don't think heatsinking an 78L05 (TO92) makes sense. Take an 7805
(TO220).

Wouter van Ooijen

-- -------------------------------------------
Van Ooijen Technische Informatica: http://www.voti.nl
consultancy, development, PICmicro products
docent Hogeschool van Utrecht: http://www.voti.nl/hvu


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