7805 OK to get hot (do I need a heatsink)

[Jack B. Pollack]

I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
Should I use a heat sink on the Vreg?
If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How would
you suggest that I make my own that might fit?


Full explanation of what I am doing for those interested:

I have a portable iPod speaker that charges the older iPods through 12V on
the FireWire pin (on the dock connector). My new iPod just charges through
the 5V USB pin. I opened the speaker and cut the trace going to the 12V
FireWire pin. This got rid of the message that the iPod doesn't support
FireWire. I then connected a voltage regulator to the 12V wall-wart power
supply (so that the iPod would charge even when the speakers were not on).
The Vreg supplies 5V to the USB power pin. For the iPod to charge through
the USB port it must also see 2.8V on data - pin and 2.0V on data + pin. I
did this with a few resistors: 33k to 5V and 47k to gnd to obtain 2.8V and
33k to 5V and 22k to gnd to obtain 2V.

This all works well and the iPod does charge correctly from the 5V. The only
question is the Vreg heat.

Thanks

 

[Michael Black]

On Sun, 26 Dec 2010, Jack B. Pollack wrote:

I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
Should I use a heat sink on the Vreg?

The heat may be coming from whatever voltage you are dropping to 5v. If
the voltage into the regulator is high enough, the regulator has to deal
with that, and get rid of it as heat.

A regulator can also get hot if it's oscillating. All the datasheets have
fairly explicit notes about the need to bypass the input, and if that
isn't done properly (ie no bypass capacitor, or the wrong type) the
regulator can oscillate badly, and thus heat up quite a bit.

Michael

If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How would
you suggest that I make my own that might fit?


Full explanation of what I am doing for those interested:

I have a portable iPod speaker that charges the older iPods through 12V on
the FireWire pin (on the dock connector). My new iPod just charges through
the 5V USB pin. I opened the speaker and cut the trace going to the 12V
FireWire pin. This got rid of the message that the iPod doesn't support
FireWire. I then connected a voltage regulator to the 12V wall-wart power
supply (so that the iPod would charge even when the speakers were not on).
The Vreg supplies 5V to the USB power pin. For the iPod to charge through
the USB port it must also see 2.8V on data - pin and 2.0V on data + pin. I
did this with a few resistors: 33k to 5V and 47k to gnd to obtain 2.8V and
33k to 5V and 22k to gnd to obtain 2V.

This all works well and the iPod does charge correctly from the 5V. The only
question is the Vreg heat.

Thanks

 

[Winston]

Jack B. Pollack wrote:

I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?

Not OK.

Your unregulated wallwart sources in the neighborhood
of ~15V into the 7805, yes?

Should I use a heat sink on the Vreg?

Always. You need to get rid of ~1.7 W

See the thermal resistance spec on page 2 of:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

It indicates that the chip will be 111 C above
ambient dissipating that amount of power.

That is above the 125 C specification for
any reasonable ambient, so you need to pull
thermal power out of that part.

If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How would
you suggest that I make my own that might fit?

First prize answer:
Have the enclosure made as a heatsink and use a switchmode
supply rather than a linear.

Full explanation of what I am doing for those interested:

I have a portable iPod speaker that charges the older iPods through 12V on
the FireWire pin (on the dock connector). My new iPod just charges through
the 5V USB pin. I opened the speaker and cut the trace going to the 12V
FireWire pin. This got rid of the message that the iPod doesn't support
FireWire. I then connected a voltage regulator to the 12V wall-wart power
supply (so that the iPod would charge even when the speakers were not on).
The Vreg supplies 5V to the USB power pin. For the iPod to charge through
the USB port it must also see 2.8V on data - pin and 2.0V on data + pin. I
did this with a few resistors: 33k to 5V and 47k to gnd to obtain 2.8V and
33k to 5V and 22k to gnd to obtain 2V.

Where did you find the spec for those voltage levels?
Just curious.

This all works well and the iPod does charge correctly from the 5V. The only
question is the Vreg heat.

--Winston

 

[ehsjr]

Jack B. Pollack wrote:

I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps?

Yes - you are dissipating about 1.19 watts, and more if the
12 volt wall wart you're using is not regulated:
Power dissipated in the 7805 = (Vin - Vout) * Iout

Is it OK?

No, not if it's inside an Ipod battery compartment. The 7805
chip, without a heatsink, would probably be able to handle
the 1.19 watt dissipation if it's in free air. Putting it inside
an unventilated enclosure is likely to cause trouble.

Should I use a heat sink on the Vreg?

Yes, and/or a 25 ohm 5 watt series resistor between the wall wart
and the 7805 input pin. That series resistor would dissipate about
..7225 watts, and the 7805 would dissipate ~ .4675. If the wall wart
is unregulated, those figures will be higher.

If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How would
you suggest that I make my own that might fit?

I wouldn't put it inside the Ipod.

Ed

Full explanation of what I am doing for those interested:

I have a portable iPod speaker that charges the older iPods through 12V on
the FireWire pin (on the dock connector). My new iPod just charges through
the 5V USB pin. I opened the speaker and cut the trace going to the 12V
FireWire pin. This got rid of the message that the iPod doesn't support
FireWire. I then connected a voltage regulator to the 12V wall-wart power
supply (so that the iPod would charge even when the speakers were not on).
The Vreg supplies 5V to the USB power pin. For the iPod to charge through
the USB port it must also see 2.8V on data - pin and 2.0V on data + pin. I
did this with a few resistors: 33k to 5V and 47k to gnd to obtain 2.8V and
33k to 5V and 22k to gnd to obtain 2V.

This all works well and the iPod does charge correctly from the 5V. The only
question is the Vreg heat.

Thanks

 

[David Eather]

On 27/12/2010 3:27 PM, Winston wrote:

Jack B. Pollack wrote:
I am charging an iPod using a 7805 1amp regulator. With the iPod
screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot
enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?

Not OK.

Your unregulated wallwart sources in the neighborhood
of ~15V into the 7805, yes?

Should I use a heat sink on the Vreg?

Always. You need to get rid of ~1.7 W

See the thermal resistance spec on page 2 of:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

It indicates that the chip will be 111 C above
ambient dissipating that amount of power.

That is above the 125 C specification for
any reasonable ambient, so you need to pull
thermal power out of that part.

If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How
would
you suggest that I make my own that might fit?

First prize answer:
Have the enclosure made as a heatsink and use a switchmode
supply rather than a linear.

That is a silly answer. The guy has a working circuit and all components
required for that solution. All he needs is a heatsink which could be
the enclosure - and the heatsink/enclosure part is common to both solutions.

There is no pay-off for collecting new components, designing, and then
implementing a new circuit to produce exactly 1 item which will save a
few cents of electricity per year.

 

[Winston]

David Eather wrote:

On 27/12/2010 3:27 PM, Winston wrote:

(...)

First prize answer:
Have the enclosure made as a heatsink and use a switchmode
supply rather than a linear.

That is a silly answer.

What exactly is silly about that answer?
The switchmode approach is wise and sensible in my opinion.

The guy has a working circuit and all components
required for that solution. All he needs is a heatsink which could be
the enclosure - and the heatsink/enclosure part is common to both
solutions.

Putting the linear regulator on a heatsink exposed to
the room ambient is a 'second prize answer' because it
is inefficient electrically and a potential problem
from a thermal perspective.

You can 'get away with it', but....

That regulator is not the wisest, most sensible thing
to put in an enclosure that was never designed to
dissipate that amount of temperature rise.

There is no pay-off for collecting new components, designing, and then
implementing a new circuit to produce exactly 1 item which will save a
few cents of electricity per year.

Of course there is a payoff.

The name of the group is sci.electronics.basics,
a forum for discussion of electronics.

The payoff is in understanding a better approach
from an engineering perspective.

I am here to learn. I appreciate it when someone
points out a better way to address a problem.

Especially if it means my speaker enclosure doesn't melt.



--Winston

 

[Jon Kirwan]

On Sun, 26 Dec 2010 21:27:37 -0800, Winston
<Winston@BigBrother.net> wrote:

Jack B. Pollack wrote:
I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
snip
Should I use a heat sink on the Vreg?

Always. You need to get rid of ~1.7 W

I get (12 - 5)*(0.17+Iqreg)

See the thermal resistance spec on page 2 of:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

That link shows a max Iq of 8mA, but Figure 3 shows the
typical value around 6mA at 125C. Anyway, go with 8mA.

So this works out to about 1+1/4 watts, assuming that the 12V
the OP talked about is actually a regulated 12V. (Which,
despite being a wallwart thing, I'd expect it does include a
linear regulator whose output really is closer to 12V -- or,
in other words, there is also some unaccounted dissipation
taking place in the wall wart, too, but we don't have to
count it in for these purposes.)

I'm curious where you got the 1.7 W figure.

Was it from adding in some overage for the fact that this is
based upon a wallwart power supply and that you think the 12V
regulation might not really be regulated and instead looks
like the nasty output regulation, like 30% or worse, for the
transformer itself?

It indicates that the chip will be 111 C above
ambient dissipating that amount of power.

snip

At 65 C/W (which seems to apply to both Fairchild and OnSemi
220 style mountings), 1.25W gets to >80C over ambient. But
you were basing your figure on a different number I'm curious
about.

[Interestingly, Sipex (Exar) has a max quiescent of 0.8mA
that actually declines with increasing temperature.]

Jon

 

[Jon Kirwan]

On Mon, 27 Dec 2010 06:57:21 -0800, Winston
<Winston@BigBrother.net> wrote:

snip
Putting the linear regulator on a heatsink exposed to
the room ambient is a 'second prize answer' because it
is inefficient electrically and a potential problem
from a thermal perspective.
snip

Well, it is inefficient, though taken into the fuller context
of everything else (losses in the wallwart, possible other
dissipations in the speaker enclosure if it also includes an
amplifier system...) it is hard to know just how much worse
it makes the whole picture.

But the OP was asking a different set of questions:

"Should this be happening drawing only 0.17A?"
"Should I use a heat sink on the Vreg?"

With a tiny bit of flat metal added to distribute the heat,
the peak temperature of the regulator's IC can readily be
brought down. The issue is then the total dissipation and
the speaker case's ability to handle it, as a matter of
continuous operation.

We don't know if the roughly 1.25 W figure I get can be
handled there, but the OP does go on to conclude: "This all
works well and the iPod does charge correctly from the 5V."

The OP didn't complain about the current package temperature
or total dissipation causing problems, so we have to conclude
that if it goes still lower that things will still be okay.

The OP is just worried about the regulator IC itself over the
longer term.

I also think it is fine to bring up a switcher as a suggested
idea to also consider, consistent with providing simple and
more direct answers, first. It's a simple and direct answer
to suggest heatsinking alone.

Jon

 

[Jon Kirwan]

On Mon, 27 Dec 2010 16:21:01 +0000 (UTC), don@manx.misty.com
(Don Klipstein) wrote:

In article <38bhh6l781cing74h62qajs3mcokepa7l6@4ax.com>, Jon Kirwan wrote:
On Sun, 26 Dec 2010 21:27:37 -0800, Winston
Winston@BigBrother.net> wrote:

Jack B. Pollack wrote:
I am charging an iPod using a 7805 1amp regulator. With the iPod
screen on and charging I am drawing .17 amps. The Vreg is getting
almost hot enough so that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
snip
Should I use a heat sink on the Vreg?

Always. You need to get rid of ~1.7 W

I get (12 - 5)*(0.17+Iqreg)

See the thermal resistance spec on page 2 of:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

That link shows a max Iq of 8mA, but Figure 3 shows the
typical value around 6mA at 125C. Anyway, go with 8mA.

So this works out to about 1+1/4 watts, assuming that the 12V
the OP talked about is actually a regulated 12V. (Which,
despite being a wallwart thing, I'd expect it does include a
linear regulator whose output really is closer to 12V -- or,
in other words, there is also some unaccounted dissipation
taking place in the wall wart, too, but we don't have to
count it in for these purposes.)

I'm curious where you got the 1.7 W figure.

Was it from adding in some overage for the fact that this is
based upon a wallwart power supply and that you think the 12V
regulation might not really be regulated and instead looks
like the nasty output regulation, like 30% or worse, for the
transformer itself?
SNIP from here

My experience is that most wallwarts with iron core transformers
operating at line frequency are not regulated.

Your experience notwithstanding, we just don't know at this
point and I was leaving room for it in my comments. But
"most" is not "all," even assuming your experience is truly
comprehensive. I still take the point, of course, which is
why I brought up the whole suggestion in the first place. I'm
merely curious how Winston came up with the number, that's
all. I like to arrive at the same quantities when deducing
to specifics.

Jon

 

[Winston]

Jon Kirwan wrote:

On Sun, 26 Dec 2010 21:27:37 -0800, Winston
Winston@BigBrother.net> wrote:

Jack B. Pollack wrote:
I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
snip
Should I use a heat sink on the Vreg?

Always. You need to get rid of ~1.7 W

I get (12 - 5)*(0.17+Iqreg)

If the wallwart is regulated, sure.

See the thermal resistance spec on page 2 of:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

That link shows a max Iq of 8mA, but Figure 3 shows the
typical value around 6mA at 125C. Anyway, go with 8mA.

0.008 * 10 = 80 mW

So this works out to about 1+1/4 watts, assuming that the 12V
the OP talked about is actually a regulated 12V. (Which,
despite being a wallwart thing, I'd expect it does include a
linear regulator whose output really is closer to 12V -- or,
in other words, there is also some unaccounted dissipation
taking place in the wall wart, too, but we don't have to
count it in for these purposes.)

Jack, can you measure the voltage of your wallwart, please?

I'm curious where you got the 1.7 W figure.

I didn't assume that his wallwart was regulated.
I figured (15 - 5) * (0.17) = 1.7 W

Was it from adding in some overage for the fact that this is
based upon a wallwart power supply and that you think the 12V
regulation might not really be regulated and instead looks
like the nasty output regulation, like 30% or worse, for the
transformer itself?

Yup.

It indicates that the chip will be 111 C above
ambient dissipating that amount of power.

snip

At 65 C/W (which seems to apply to both Fairchild and OnSemi
220 style mountings), 1.25W gets to>80C over ambient. But
you were basing your figure on a different number I'm curious
about.

65 * 1.708 = 111.02

Even 80 C plus (~worst case) ambient puts it
at ~110 C which is too close to the 125 C limit
for my taste.

[Interestingly, Sipex (Exar) has a max quiescent of 0.8mA
that actually declines with increasing temperature.]

The 72 mW difference doesn't bother me.
Heating up the box with 1.71 W bothers me.

--Winston

 

[Don Klipstein]

In article <38bhh6l781cing74h62qajs3mcokepa7l6@4ax.com>, Jon Kirwan wrote:

On Sun, 26 Dec 2010 21:27:37 -0800, Winston
Winston@BigBrother.net> wrote:

Jack B. Pollack wrote:
I am charging an iPod using a 7805 1amp regulator. With the iPod
screen on and charging I am drawing .17 amps. The Vreg is getting
almost hot enough so that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
snip
Should I use a heat sink on the Vreg?

Always. You need to get rid of ~1.7 W

I get (12 - 5)*(0.17+Iqreg)

See the thermal resistance spec on page 2 of:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf

That link shows a max Iq of 8mA, but Figure 3 shows the
typical value around 6mA at 125C. Anyway, go with 8mA.

So this works out to about 1+1/4 watts, assuming that the 12V
the OP talked about is actually a regulated 12V. (Which,
despite being a wallwart thing, I'd expect it does include a
linear regulator whose output really is closer to 12V -- or,
in other words, there is also some unaccounted dissipation
taking place in the wall wart, too, but we don't have to
count it in for these purposes.)

I'm curious where you got the 1.7 W figure.

Was it from adding in some overage for the fact that this is
based upon a wallwart power supply and that you think the 12V
regulation might not really be regulated and instead looks
like the nasty output regulation, like 30% or worse, for the
transformer itself?
SNIP from here

My experience is that most wallwarts with iron core transformers
operating at line frequency are not regulated.
--
- Don Klipstein (don@misty.com)

 

[Jack B. Pollack]

"Jack B. Pollack" <N@NE.nothing> wrote in message
news:I8ydnZswuYipl4XQnZ2dnUVZ_omdnZ2d@earthlink.com...

I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough
so that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
Should I use a heat sink on the Vreg?
If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How
would you suggest that I make my own that might fit?


Full explanation of what I am doing for those interested:

I have a portable iPod speaker that charges the older iPods through 12V on
the FireWire pin (on the dock connector). My new iPod just charges through
the 5V USB pin. I opened the speaker and cut the trace going to the 12V
FireWire pin. This got rid of the message that the iPod doesn't support
FireWire. I then connected a voltage regulator to the 12V wall-wart power
supply (so that the iPod would charge even when the speakers were not on).
The Vreg supplies 5V to the USB power pin. For the iPod to charge through
the USB port it must also see 2.8V on data - pin and 2.0V on data + pin.
I did this with a few resistors: 33k to 5V and 47k to gnd to obtain 2.8V
and 33k to 5V and 22k to gnd to obtain 2V.

This all works well and the iPod does charge correctly from the 5V. The
only question is the Vreg heat.

Thanks

Thanks for everyone's replies. Some of the stuff is getting way to technical
for me.

So a quick update on what I was able to get out of these post:

I was able to find a continuous 9V trace and am now powering the Vreg off of
that.
The current draw remains the same (.17amps), but the Vreg heat is about 65%
reduced to a mild warmth.
Not sure if it is the lower voltage in (9V instead of 12V) or a reduction in
oscillation but it made a big difference.

I think I will mount a standard heat-sink to the outside back of the speaker
case.
I assume that the mild warmth is OK, and the heat-sink can only help a bit.

Should I put a .33uf cap across the input and gnd and a .1uf across the
output and gnd?

FYI: The wall-wart does look to be regulated

 

[krw@att.bizzzzzzzzzzzz]

On Mon, 27 Dec 2010 06:57:21 -0800, Winston <Winston@BigBrother.net> wrote:

David Eather wrote:
On 27/12/2010 3:27 PM, Winston wrote:

(...)

First prize answer:
Have the enclosure made as a heatsink and use a switchmode
supply rather than a linear.

That is a silly answer.

What exactly is silly about that answer?

Exposing the chip to ambient is silly for another reason; product liability.
These things get hot and you never know when some old broad will stick one
between her legs.

The switchmode approach is wise and sensible in my opinion.

Then you have potential EMI issues.

<snip>

There is no "best" answer.

 

[Winston]

Jack B. Pollack wrote:

(...)

Thanks for everyone's replies. Some of the stuff is getting way to technical
for me.

You grasped the major issue and addressed it properly.

So a quick update on what I was able to get out of these post:

I was able to find a continuous 9V trace and am now powering the Vreg off of
that.

Yes! That lowers dissipation significantly. Good move!
Please measure the 9 V rail as it powers the regulator
to make sure it doesn't sag too much. We are adding
a 688 mW load it wasn't initially designed to power.

The current draw remains the same (.17amps), but the Vreg heat is about 65%
reduced to a mild warmth.

So 4 V drop is 680 mW. Add 'chip power' and you have 688 mW
wasted as heat instead of the 1.2 W as before. This equates
to a 44 C rise or say 74 C in a hot room. Plenty of margin.

Not sure if it is the lower voltage in (9V instead of 12V) or a reduction in
oscillation but it made a big difference.

Excellent! Engineering is compromise.

I think I will mount a standard heat-sink to the outside back of the speaker
case.
I assume that the mild warmth is OK, and the heat-sink can only help a bit.

The heat sink will help a lot. Please use a dab of
silicone thermal grease when you mount the regulator
to it.

Should I put a .33uf cap across the input and gnd and a .1uf across the
output and gnd?

Couldn't hurt.

FYI: The wall-wart does look to be regulated

Cool!

--Winston

 

[David Eather]

On 28/12/2010 12:57 AM, Winston wrote:

David Eather wrote:
On 27/12/2010 3:27 PM, Winston wrote:

(...)

First prize answer:
Have the enclosure made as a heatsink and use a switchmode
supply rather than a linear.

That is a silly answer.

What exactly is silly about that answer?

The switchmode approach is wise and sensible in my opinion.

I gave my answer just below your paragraph and also:

">> There is no pay-off for collecting new components, designing, and then

implementing a new circuit to produce exactly 1 item which will save a
few cents of electricity per year."


The guy has a working circuit and all components
required for that solution. All he needs is a heatsink which could be
the enclosure - and the heatsink/enclosure part is common to both
solutions.

Putting the linear regulator on a heatsink exposed to
the room ambient is a 'second prize answer' because it
is inefficient electrically and a potential problem
from a thermal perspective.

Its a maximum thermal problem of 1.7 watts! It's nothing. If you need to
talk "efficiency" lets talk "total" efficiency. To be viable your
solution has to be(including power consumption), in the long run, not
just cheaper than the 7805 solution but also cover the cost of the
working 7805 solution which already exists and works which you propose
to abandon. At a maximum of a few cents per year savings in power that
is *never* going to happen with this type of consumer product.

You can 'get away with it', but....

That regulator is not the wisest, most sensible thing
to put in an enclosure that was never designed to
dissipate that amount of temperature rise.

But you propose to use the enclosure as a heatsink for the switch mode
design, so this is a fallacious argument especially in light of a less
than 2 watt dissipation

There is no pay-off for collecting new components, designing, and then
implementing a new circuit to produce exactly 1 item which will save a
few cents of electricity per year.

Of course there is a payoff.

The name of the group is sci.electronics.basics,
a forum for discussion of electronics.

The payoff is in understanding a better approach
from an engineering perspective.

I am here to learn. I appreciate it when someone
points out a better way to address a problem.

In that case post your own thread. The OP wanted to know if the heat
dissipated from a linear regulator was a problem. In this case a yes/no
answer. Yes it might be a bit of a problem, but a small heatsink will
fix that.

Especially if it means my speaker enclosure doesn't melt.

2 watts! :-/

--Winston

(and thanks for the adult argument)

 

[Jamie]

Jack B. Pollack wrote:

I am charging an iPod using a 7805 1amp regulator. With the iPod screen on
and charging I am drawing .17 amps. The Vreg is getting almost hot enough so
that you cant touch it.

Should this be happening drawing only .17amps? Is it OK?
Should I use a heat sink on the Vreg?
If yes, I am planning on putting the Vreg in the battery compartment that
holds 2 AAA batteries. I doubt that a standard heat sink will fit. How would
you suggest that I make my own that might fit?


Full explanation of what I am doing for those interested:

I have a portable iPod speaker that charges the older iPods through 12V on
the FireWire pin (on the dock connector). My new iPod just charges through
the 5V USB pin. I opened the speaker and cut the trace going to the 12V
FireWire pin. This got rid of the message that the iPod doesn't support
FireWire. I then connected a voltage regulator to the 12V wall-wart power
supply (so that the iPod would charge even when the speakers were not on).
The Vreg supplies 5V to the USB power pin. For the iPod to charge through
the USB port it must also see 2.8V on data - pin and 2.0V on data + pin. I
did this with a few resistors: 33k to 5V and 47k to gnd to obtain 2.8V and
33k to 5V and 22k to gnd to obtain 2V.

This all works well and the iPod does charge correctly from the 5V. The only
question is the Vreg heat.

Thanks


Use a switching regulator ?

http://www.dimensionengineering.com/de-sw050.htm

http://www.simplecircuitdiagram.com/2009/05/20/5v-switching-regulator-very-simple-circuit/

Go here and design it on line..

http://www.national.com/analog/power/simple_switcher?gclid=CJHYoOmwjaYCFUdN4AodR297oQ

Jamie

 

[Winston]

David Eather wrote:

(...)

At a maximum of a few cents per year savings in power that
is *never* going to happen with this type of consumer product.

Never is a long time.

Perhaps you've seen the evolution of the wall wart supply
from the old linear transformer type to the new switch mode
type over the last few years? I sure have.

I suppose that approach is the best answer for manufacturers
considering shipping cost and 'energy star' regulatory compliance.

Phase one was the movement of the regulator outside of the
main enclosure. Phase two is boosting the efficiency of
the regulator so less power is converted to heat.

I'm all for it.

From: http://en.wikipedia.org/wiki/AC_adapter

"The report concluded that about 32 billion kilowatt-hours (kWh) per
year, about 1% of total electrical energy consumption, could be
saved in the United States by: replacing all linear power supplies"

(...)

But you propose to use the enclosure as a heatsink for the switch mode
design, so this is a fallacious argument especially in light of a less
than 2 watt dissipation

Yes, essentially make the old 'battery door' an aluminum heatsink
and sink thermal power through it. Way cooler than trying to
push that power through a little PVC door.

So it's design overkill and the switchmode supply would only need
to lose ~1 watt with a junction rise of ~5 C.

I've been accused of worse.

There is no pay-off for collecting new components, designing, and then
implementing a new circuit to produce exactly 1 item which will save a
few cents of electricity per year.

Of course there is a payoff.

The name of the group is sci.electronics.basics,
a forum for discussion of electronics.

The payoff is in understanding a better approach
from an engineering perspective.

I am here to learn. I appreciate it when someone
points out a better way to address a problem.

In that case post your own thread.

I was placing myself in the position of the OP.
Lots of times I *am* the OP.

The OP wanted to know if the heat
dissipated from a linear regulator was a problem. In this case a yes/no
answer. Yes it might be a bit of a problem, but a small heatsink will
fix that.

I answered both in my original reply.

* Yes, it is a problem.
* Yes you can use a linear if you pull heat out of the regulator.

* There is a better approach.

Especially if it means my speaker enclosure doesn't melt.

2 watts! :-/

Recall the original approach was to place the regulator
completely within the 2 'AAA' battery compartment.

What is the temperature rise of a resistor dissipating
two watts into a plastic enclosure that has only one
path to ambient measuring 10 mm x 50 mm through a plastic
door that has a thermal resistance of 1300 C per watt?

Answer: Too Hot.

(and thanks for the adult argument)

It is in the finest tradition of USENET.

Thank you.

--Winston

 

[Don Klipstein]

In article <RC7So.36024$uS7.18304@newsfe23.iad>, Jamie wrote:

Jack B. Pollack wrote:

<The story of a linear regulator IC producing enough heat to invite
asking about heatsinking it>

Use a switching regulator ?

http://www.dimensionengineering.com/de-sw050.htm

http://www.simplecircuitdiagram.com/2009/05/20/5v-switching-regulator-
very-simple-circuit/

Go here and design it on line..

http://www.national.com/analog/power/simple_switcher?
gclid=CJHYoOmwjaYCFUdN4AodR297oQ

Jamie

As said elsewhere in this thread, this is sci.electronics.basics, and
the fellow starting this thread got something working here except for
maybe to maybe-likely need or at least wanting a heatsink.

And, I would not ask a newbie to trash something that the newbie already
put effort into making work, with exception of maybe-likely need for a
heatsink.

The heatsink will cost the newbie an order of magnitude less than
rebuilding as switchmode the linear regulator solution that the newbie
appears to have about 90% of the way worked out.

So, for bringing up switchers as opposed to linear regulators, I would
look for much more diplomatic ways. I would put such ways in a class
dependent on the newbie wanting to build more regulators,

or, in terms of offering advice to next-in-line newbies that want to
make a voltage regulator circuit.
(Should newbies be good targets for making their electronics learning
experiences taking the plunge into switchmode regulators on the very day
they could be purchasing the 1st IC exposed to a soldering iron at their
hands. Meanwhile, we have a newbie that got a linear one largely working
and then posted here.)

When a newbie builds something and it works, or the newbie asks for help
in getting something to work, I would want to encourage the newbie. It
appears to me that suggestion to trash something 90% of the way working
and then go back to Square One is discouragement.

Something I see - I would let people actually building something to
either take pride in what they built, or to take pride in how they would
be ashamed to repeat what they did in their younger 1st-project days.

So, I would say, let them follow through, guide them but don't turn them
back unless they're on some outright collision course, give them advice
when they ask for it, and *let them complete something 80-90% done* when
feasible!

Should the newbie go on to build more, especially related, electronics
projects, then fair chance the newbie will be interested in how to make
them better than Model A.

If I knew what I know now back when I first made my own Model A
electronic project, constructed outside the 60-or-whatever-in-1 "kit"
from Radio Shack (in their better days) on some birthday or Christmas in
the mid 1970's, I would have saved my 1st "free bird" electronic
contraption in a trophy display case. (IIRC, that was a Hartley audio
oscillator by intent, but ended up being a "ringing choke" one.)
--
- Don Klipstein (don@misty.com)

 

[Jamie]

Don Klipstein wrote:

In article <RC7So.36024$uS7.18304@newsfe23.iad>, Jamie wrote:

Jack B. Pollack wrote:


The story of a linear regulator IC producing enough heat to invite
asking about heatsinking it

Use a switching regulator ?

http://www.dimensionengineering.com/de-sw050.htm

http://www.simplecircuitdiagram.com/2009/05/20/5v-switching-regulator-
very-simple-circuit/

Go here and design it on line..

http://www.national.com/analog/power/simple_switcher?
gclid=CJHYoOmwjaYCFUdN4AodR297oQ

Jamie


As said elsewhere in this thread, this is sci.electronics.basics, and
the fellow starting this thread got something working here except for
maybe to maybe-likely need or at least wanting a heatsink.

And, I would not ask a newbie to trash something that the newbie already
put effort into making work, with exception of maybe-likely need for a
heatsink.

The heatsink will cost the newbie an order of magnitude less than
rebuilding as switchmode the linear regulator solution that the newbie
appears to have about 90% of the way worked out.

So, for bringing up switchers as opposed to linear regulators, I would
look for much more diplomatic ways. I would put such ways in a class
dependent on the newbie wanting to build more regulators,

or, in terms of offering advice to next-in-line newbies that want to
make a voltage regulator circuit.
(Should newbies be good targets for making their electronics learning
experiences taking the plunge into switchmode regulators on the very day
they could be purchasing the 1st IC exposed to a soldering iron at their
hands. Meanwhile, we have a newbie that got a linear one largely working
and then posted here.)

When a newbie builds something and it works, or the newbie asks for help
in getting something to work, I would want to encourage the newbie. It
appears to me that suggestion to trash something 90% of the way working
and then go back to Square One is discouragement.

Something I see - I would let people actually building something to
either take pride in what they built, or to take pride in how they would
be ashamed to repeat what they did in their younger 1st-project days.

So, I would say, let them follow through, guide them but don't turn them
back unless they're on some outright collision course, give them advice
when they ask for it, and *let them complete something 80-90% done* when
feasible!

Should the newbie go on to build more, especially related, electronics
projects, then fair chance the newbie will be interested in how to make
them better than Model A.

If I knew what I know now back when I first made my own Model A
electronic project, constructed outside the 60-or-whatever-in-1 "kit"
from Radio Shack (in their better days) on some birthday or Christmas in
the mid 1970's, I would have saved my 1st "free bird" electronic
contraption in a trophy display case. (IIRC, that was a Hartley audio
oscillator by intent, but ended up being a "ringing choke" one.)

Ok, understood how ever, the first link I gave was a direct drop in
replacement for the 7805 with out changing nothing to the remainder of
the design..

It's obvious by now that the OP was only looking for a solution and
not wanting to expand their horizons. If you looked at the first link, I
think you would agree that it'd be the choice for those just wanting to
getting it working better and not do a total overhaul.

Additional options were supplied in the event that wasn't the case.

Jamie

 

[Jon Kirwan]

On Mon, 27 Dec 2010 16:47:34 -0500, Jamie
<jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:

http://www.national.com/analog/power/simple_switcher?gclid=CJHYoOmwjaYCFUdN4AodR297oQ

Well, damn! Impressive start. But then it eventually moves
you towards, "A Sign-On or Personal Workspace IS required for
Technical Support, Sample Orders, and WEBENCHŽ Designer Tools
use." So to actually get access to the tool, I need to do a
little sign-on jig. Fancy bait for such a simple hook.

Jon