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

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

Not OK.

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

Always. You need to get rid of ~1.7 W

See the thermal resistance spec on page 2 of:

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.

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

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

--Winston

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

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.

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.

I wouldn't put it inside the Ipod.

Ed

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.

(...)

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

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.

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

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

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?

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

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

--
 - Don Klipstein (don@misty.com)

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

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

If the wallwart is regulated, sure.

0.008 * 10 = 80 mW

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

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

Yup.

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.

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

--Winston

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

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.

Then you have potential EMI issues.

There is no "best" answer.

(...)

You grasped the major issue and addressed it properly. :)

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.

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.

Excellent! Engineering is compromise.

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

Couldn't hurt.

Cool!

--Winston

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."

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.

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

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.

2 watts! :-/

(and thanks for the adult argument)

Use a switching regulator ?

Go here and design it on line..

Jamie

(...)

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:

"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"

(...)

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. :)

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

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.

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. :)

It is in the finest tradition of USENET.

Thank you.

--Winston

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)

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

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 WEBENCH® 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