I believe that running the Panda of a 12V power source is within speck but the voltage regulator (IC3) gets very hot. As I understand it there is some sort of thermal protection so the board should be safe but will this protection kick in after hours of running?
Would it be better to bring the voltage down to about 7V with another regulator first?
This is for my boat project and I cant risk it just shutting down out on the lake.
Dropping large voltages with a linear regulator is very inefficient. For a mobile project you may find that using a DC-DC converter works better.
I just bought some 3.3V 1.0A output DC-DC converters (9-18V input) for about $15 a piece from Mouser. These will be used to power a Domino, IO40, and some actuators in a mobile application. Since both my actuators and the FEZ actually run on 3.3V this was a good choice.
I’m sure 12v input is within spec, but the question is, is your power source actually 12v?
And yet another question. How much current are you drawing through this?
Adding another regulator can avoid overloading the existing regs, but if you’re drawing a lot of current (ie have lots of connected peripherals) then you may just be asking it to do too much. Heat dissipated is a function of input voltage and current draw; if you can reduce either of those, the regs won’t be so taxed. But an additional reg also increases the inefficiency along your power-chain so you may be prematurely depleting your power (although I reckon your motor, if using the same source, will do that just a tad faster 
).
If a few minutes extra runtime (note: that’s a guess, not calculated 
) is not important, then i’d consider a second reg; or you could just get a higher current 5v reg with heatsink and power Fez with that.
Im pretty sure that its 12V as it is on a lab power supply at the moment and its pulling 250mA of current.
This is all the devices that is powered by the Panda tinkerer kit, GPS, I2C compass and XBee. Other stuff like servo and motor wont be powered through the Panda.
Inside the boat they all share the same power source (with lots of noise suppressing caps). This power source is dual 12V lead acid batteries. As these batteries hold 12Ah each the estimated run time is about 9 hours. Ill rather have 8 hours of stable run time then 9 with a potential breakdown.
So the verdict is an extra regulator?
You might want to watch out while charging the lead acid batteries.
Charged batteries mostly have a higher voltage at charged state.
To give you an idea:
I am using 2 cordless drill batteries at the moment. Charged one of them 3 days ago. Instead of 18v, the charged state is 20.2v. Which is of course a huge difference. It does not matter much, since the batteries are parallel connected, but it’s something you should keep in mind if you would like to run your fez of these batteries.
Of course I do not know which charger / batteries you are using. Having a good processor charger should stop charging with minus delta-u recognition.
Good point Robert.
I’ll add a LM317 regulator and go for about 7V and see what that does with the current consumption.
UPDATE
I added an extra voltage regulator to feed the panda with 7.2V and now the onboard regulator is only warm to the touch and not burning hot as it was. I have had it running an hour with no increase in temperature, so Im happy.
When it comes to current consumption I cant see any change at all.
I still don’t understand why the panda’s regulator gets hot.
I have never had any hot regulators. Even though I am sometimes doing things which I should not be doing :
I will overlook your specs tomorrow. I am too tired now… :-[
It is all because of how a linear regulator works. Let’s say that your circuit draws 1A at 9V for sake of simplicity. Now you decide that you want to run this circuit from a 12V battery so you pop in a 9V linear regulator in there.
The regulator is outputting 9V so your circuit is still drawing 1A(9W of power being used by circuit.) The extra 3V is being dropped across the regulator, so the regulator is dissipating 3V*1A = 3W of power.
Now let’s say you need to move this circuit to another location that has 24V only. So you say, ‘Oh, well I have a 9V regulator’ and you hook it up. Remember the regulator will still put out 9V and your circuit will still draw 1A. The difference is that now we have 24V-9V = 13V being dropped across the regulator. That means the regulator is dissipating 13V*1A=13W or power! So it will get very hot.
If a circuit is using very little current, like 100ma than a large voltage drop across the regulator is not a big deal. The more current your circuit draws the bigger deal the voltage drop is.
Right. But in his case total power consumption is around 4,25w.
Is that already too much?
Amperage is 250mA.
Total power: 12V * 0.25A = 3W
5V Regulator: (12V-5V)*0.25A = 1.75W
Thus the first regulator on the Panda, the 5V regulator, must dissipate 1.75W.
Heatsinks are rated at “temperature increase per watt”. It means that a 50C/W heatsink will be 50C above ambient if it must dissapate 1W.
The regulator doesn’t have a heatsink fitted as such, only some PCB area to act as a heatsink. The temperature rise will depend on the size of this area…
The regulator package without heatsink is spec-ed at 136C/W, so at 1.75W it will reach 238C above ambient. Not good at all.
One square inch of filled copper(no traces, solid) on PCB is 66C/W, so at 1.75W that is 115C above ambient.
It also means that your Panda is using 1.25W, and the regulator is turning 1.75 into wasted heat. Which means that your power supply has a 41% efficiency. Switching to a switching regulator will double your run time from the same battery…
Switching regulators is the answer! Very efficient and no heat at all
Take a look at ChipworkX power supply…can run at +35V and the will be no heat and no problems at 1A
lol, just edited my post to include some switching regulator info…
Errol, very well explained! ;D
Some useful tips on when to choose linear or switching regulator: