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'[OT]: 7905 getting hot!'
2003\03\09@215414 by fred jones

picon face
Hi all,
I thought I'd ask this question here as my pool of friends here can't seem
to offer any help.  I have a circuit where I am feeding -24V into a 7905
linear regulator.  It gets very hot, can't keep your finger on it.  It is on
a pretty hefty heatsink.  I took the output leg loose and measured the
current draw its supplying and it shows to be 280ma.  I would think that it
should not have any problem with this and shouldn't be getting hot.  I
thought maybe it was getting so hot because there was such a large voltage
feeding it so I then took the circuit off of the 24v supply and fed it with
12V and again it still gets too hot to touch.  I had a .1uf cap on the input
and output.  I looked at the data sheet and from that I tried changing them
out to a 2.2uf on the input and a 1uf on the output.  At the same time, I
chaged out the regulator to one of a different brand.  Still the same
results.

My questions would be should it be getting that hot?  I thought that it
shouldn't at only 280ma.  If not, can you point me toward any problems I
might be having that could cause this?  Suggestions of things to look at
would be greatly appreciated.
Thanks,
FJ

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2003\03\09@220104 by Jai Dhar

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-24 is nearing the -25 limit, which is probably why it's getting hot - but -12
is still probably a bit more than it needs. Can you drop it down to 7 for
testing purposes? As long as the rest of your circuit is right, and your sure
it's drawing 280mA... there doesn't seem to be an obvious reason why it would
heat up.

Quoting fred jones <spam_OUTboattowTakeThisOuTspamHOTMAIL.COM>:

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2003\03\09@221142 by Sean H. Breheny

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Even at 12V, there is a 7V drop across the regulator, which is about 2W of
power dissipation. A 7905 with no heatsink would definitely get hot at 2 W
dissipation. You said that it was on a pretty hefty heatsink, but did you
keep it attached to the heatsink for the 12V test?

I think that at -24 V in and 280mA, you might well be outside the maximum
power dissipation of the 7905.

Also, is the output voltage correct? Are you sure you have the pinout
correct? I'm not sure what the tab is connected to in the 7905 but are you
sure that you aren't shorting the tab to something it is not supposed to be
connected to by attaching it to the heatsink? In other words, the tab is
ground on the 78xx series, but it might not be on the 79xx ones.

Sean


At 10:01 PM 3/9/2003 -0500, you wrote:
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2003\03\09@221149 by Tom Messenger

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Hi Fred

The temperature of the 7905 will be proportional to how much power it is
dissipating. Power here is the voltage drop across it from input to output
multiplied by the current it is supplying to the load.  It's own quiescent
current figures in too, about 10mA.

So in your "It gets very hot, can't keep your finger on it" circuit, the
poor little regulator sees (24in - 5out)* 28mA = 19V * .28A = 5.32 SMOKIN'
WATTS!

With 12 volts in, it sees 7V * .28A = 1.96 WATTS.  With Iq (10mA) call it
2W.  Yes, this will get hot.

Solutions?
1. Lower the input voltage.
2. If you can't do that, put something in to drop it. A resistor between
the 24V and the regulator can take the heat off the regulator. Of course,
then the resistor will get real hot.
3. Use a switcher. Look up buck converters.
4. If your load is fixed, you can bypass the regulator with a resistor to
supply some of the current that way and the regulator supplies the rest
keeping things at 5volts. Don't do this one unless you *really* understand
how it works.
5. Use two or more regulators in parallel with their outputs separated with
"ballast" resistors to make them share the load.

Good luck! Hope the blisters heal real soon.
Tom M.

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2003\03\09@221554 by Jinx

face picon face
> to offer any help.  I have a circuit where I am feeding -24V
> into a 7905 linear regulator.  It gets very hot, can't keep
> your finger on it.  It is on a pretty hefty heatsink.  I took the
> output leg loose and measured the current draw its supplying
> and it shows to be 280ma

i/p limit for a 7905 is -35V, so you're OK there, but the power
dissipation in that 7905 is over 5W. You'd be better to drop a
lot of the voltage (16V) through a resistor. Off the top of my head
it looks like 56R 10W to give you 8V at the 7905

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2003\03\09@224810 by fred jones

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Thanks to all of you who answered my question.  I now understand the
problem.

I am using a 7912 in the circuit as well.  I guess i could feed the 7905 off
of it and improve things.  Should I do as Jinx suggests and put a resistor
on the output of the 7912 to drop it and then feed the 7905?  I believe the
right answer here is to use a switching regulator but this circuit can't
tolerate any extra noise it would generate.

I knew you guys would be able to help.
Thanks so much,
FJ





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2003\03\10@050008 by Eric T. Brewer

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You don't say what you are trying to do with power other than you would
like it to be low noise. Does all of the -5 need to be low noise or just some?

Some options:
1. You can use a buck regulator to drop -24 down to -7 or so. Then use your
7905 to drop to -5.
2. Low pass filter (L C) the output of the buck regulator before it
goes to the the
7905.
3. Use an opamp such as a LM324 to give a very low noise -5 for your low noise
section.
4. Use an opamp with a hefty transistor to do the linear regulation
from -7 down
to -5. This eliminates the 7905 all together.

cheers,
eric

But, if you need a low noise -5. Use an OPAMP such as LM324
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2003\03\10@050803 by Alan B. Pearce

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Dop you have it correctly connected? The centre leg is NOT the ground leg.

Don't ask how I know - but it was learnt many years ago. :)))))

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2003\03\10@052247 by Russell McMahon

face
flavicon
face
re 7905 connections -

> Do you have it correctly connected? The centre leg is NOT the ground leg.
> Don't ask how I know - but it was learnt many years ago. :)))))

Knowing WHY these things are so can be helpful:

(Almost) all 3 terminal regulators have the most negative part of the
circuit connected to the tag and the centre electrode. This is due to the
fact that the most negative part of the die is the substrate and this is
electrically and mechanically and thermally connected to the mounting frame.

Therefore, on positive regulators the middle tag is ground as this is most
negative in that case.

AND

on negative regulators the middle leg is Vin as this is the most negative
potential in that case.


(It is possible to build regulators differently but in most cases this is
the easiest and cheapest and most effective way to do it so that's how it's
usually done.)



           Russell McMahon

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2003\03\10@070722 by Rick C.

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face
Just a thought, it is connected up correctly isn't it? The negative
regulators have a different pinout than the positive regulators. IGO for
positive and GIO for negative. If connected up wrong, and damaged, can
sometimes put out a strange voltage and gets hot. Keep the bypass caps
at 0.1uf.
Rick

>From: fred jones <RemoveMEboattowTakeThisOuTspamHOTMAIL.COM>

>Hi all,
>I thought I'd ask this question here as my pool of friends here can't
seem
>to offer any help.  I have a circuit where I am feeding -24V into a
7905
>linear regulator.  It gets very hot, can't keep your finger on it.  It
is
>on
>a pretty hefty heatsink.  I took the output leg loose and measured the
>current draw its supplying and it shows to be 280ma.  I would think
that it
>should not have any problem with this and shouldn't be getting hot.  I
>thought maybe it was getting so hot because there was such a large
voltage
>feeding it so I then took the circuit off of the 24v supply and fed it
with
>12V and again it still gets too hot to touch.  I had a .1uf cap on the
>input
>and output.  I looked at the data sheet and from that I tried changing
them
>out to a 2.2uf on the input and a 1uf on the output.  At the same time,
I
>chaged out the regulator to one of a different brand.  Still the same
>results.
>
>My questions would be should it be getting that hot?  I thought that it

>shouldn't at only 280ma.  If not, can you point me toward any problems
I
>might be having that could cause this?  Suggestions of things to look
at
>would be greatly appreciated.
>Thanks,
>FJ

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2003\03\10@081813 by fred jones

picon face
Hi Alan,
Yes I'm sure its connected correctly, it puts out a nice clean -5V.  Its
dropping 19V which equates to 5W so now that I understand this, I understand
why I can cook dinner on it.
Thanks,
FJ






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2003\03\10@175739 by Wagner Lipnharski

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face
Jai Dhar wrote:
> -24 is nearing the -25 limit, which is probably why it's getting hot
> - but -12 is still probably a bit more than it needs. Can you drop it
> down to 7 for testing purposes? As long as the rest of your circuit
> is right, and your sure it's drawing 280mA... there doesn't seem to
> be an obvious reason why it would heat up.

If you think that 280mA x 19V (more than 5W) is not hot enough to burn your
finger, than I wonder your finger is dead already... :)

I would suggest you to use some voltage drop device to reduce this 24V to
something around 9 to 12V, so the 7905 heat dissipation would be less than
1 Watt, what would be nice, isn't it?

Suggestions would go from power resistors to lamps in series with the input
of the 7905. Lets see, a regular 50 ohms 5 Watts resistor would drop the
input voltage to the 7905 to around 10V.  A truck (24V x 10W) would reduce
the voltage to around 9V.  All of this if your circuit continuous to drain
around 290mA.

Of course, there are nice dc-dc conversions that will produce 5V with much
less power dissipation loss.

Wagner.

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2003\03\12@152656 by Barry Gershenfeld

face picon face
>I am using a 7912 in the circuit as well.  I guess i could feed the 7905 off
>of it and improve things.  Should I do as Jinx suggests and put a resistor
>on the output of the 7912 to drop it and then feed the 7905?

Taking 280 mA from 24v to 5v will generate 5 watts of heat no matter
how you do it (linear, not switching).  So think of it as a budget--
you can dissipate it in the 7095 or partly in the 7912 or even add a
resistor and shed part of it there.  It all adds up.

Also not a bad idea to measure the current going INTO the regulator;
you would catch any surprises such as oscillation, big leakage
current, shorted tab, and all that.

Barry

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2003\03\12@184610 by fred jones

picon face
Hi,
By this, what exactly do you mean.  What I did was cut the output pin of the
7905 and put a current meter across this and measured 280ma.  I am still
trying to figure out how I want to handle this problem.  I want to try using
a negative buck regulator to go from -24 to -5 but haven't had much success
at finding a suitable circuit.  In the past when I've played with switchers
I've had too much noise and this particular circuit can't tolerate more
noise.  My inexperience with these probably doesn't help :0)
Thanks,
FJ


>Also not a bad idea to measure the current going INTO the regulator;
>you would catch any surprises such as oscillation, big leakage
>current, shorted tab, and all that.
>
>Barry


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2003\03\12@185710 by hard Prosser
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Take a look at Roman Black's regulator - just use PNP transistors instread
of NPN & reverse the diodes etc.
Should work well for this application - except you may need to look
carefully at the output noise - add more filtering etc. You'll get noise
issues with any switcher regardless!
One possibility might be to use the switcher to drop the voltage to 8V or
so & then use the linear regulator (7905) from here.  This would cut your
dissapation to about a watt. and should help the noise problem.

Richard P




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Hi,
By this, what exactly do you mean.  What I did was cut the output pin of
the
7905 and put a current meter across this and measured 280ma.  I am still
trying to figure out how I want to handle this problem.  I want to try
using
a negative buck regulator to go from -24 to -5 but haven't had much success
at finding a suitable circuit.  In the past when I've played with switchers
I've had too much noise and this particular circuit can't tolerate more
noise.  My inexperience with these probably doesn't help :0)
Thanks,
FJ


>Also not a bad idea to measure the current going INTO the regulator;
>you would catch any surprises such as oscillation, big leakage
>current, shorted tab, and all that.
>
>Barry


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2003\03\12@192215 by Kyrre Aalerud

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My 7805 has a nominal operating temp of about 50C, it actually sais so in
the docs...
If your 79 is anything similar I'd check the docs and look at max temp
values etc...

It may be as simple as you running a little too much power through it
without anything but the basic blade to cool it.  A small heatsink may be
required...  Docs should tell you this.

   KreAture

{Original Message removed}

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