I have designed a small custom pcb which uses a 5v regulator . I have gone through 2 designs of this perfecting it. In the first design i everything ran at the temperatures expected and worked fine on the second i changed the voltage regulator to another type and it is hitting over 100 degrees celcius. Now i understand calculating the heat disapation can be quite complex but in this case the initial design the reg its self is only about 10 degrees above ambient in the second deign the very similar pcb is 60 degrees above ambient 30mm from the regulator.
The formula i can find is p=v*i=(Vin-Vout)*I . I have checked and both boards are drawing 240 mA with a supply of 12v so i would asume the temperature should be similar for both? both regulators are rated at 1 A and well witin there voltage rating.
Does any one have a clue whats going on here or is there alot more to calculating the heat generated than that formula ?
Let's start of by saying you're a bit spars on the details. For example, which regulators EXACTLY.
simon100:
so i would asume the temperature should be similar for both?
That is a false assumption 
That formula only gives you the dissipated power. Which will indeed be the same if both board ahve the same Vin, Vout and current. But it tells you nothing about the actual temperature. That depends not only on the power but also on the ambient temperature and more important, the thermal resistance of the part 
If that's high it will give you a greater rise in temperature per watt of power. The thermal resistance not only depends on the part but also on how you mount it.
septillion:
Let's start of by saying you're a bit spars on the details. For example, which regulators EXACTLY.
That is a false assumption
Thanks for the reply , I went from https://www.onsemi.com/pub/Collateral/MC78M00-D.PDF to a http://www.ti.com/lit/ds/symlink/lm2940-n.pdf , yes i have gone from a dpak package to a tiny wson but i had throught the 9cm2 of pcb that is at 82degrees would get rid of far more heat than the dpak package alone which is generating so little heat that the surounding board hardly warms.
So is a pcb just far worse at getting rid of heat than i had estimated? I had never really looke dinto voltage regulators previously and had thought they would operate like a mosfet switching on and off fast so could have different efficiencies but it appears that a linear regualtor is just basically a voltage divider ?
DPak is a huge package, of course it will dissipate a lot more heat both to the air and to the PCB. The
temperature of the PCB close to the package will always be higher for a smaller package.
When you mount stuff to a PCB and rely on that for cooling it all gets a bit complicated... 9cm2 might sound like much but can it really put the heat into that? And that's a bit of a difficult question because you don't get a nice rating for that...
I had never really looke dinto voltage regulators previously and had thought they would operate like a mosfet switching on and off fast so could have different efficiencies
Switching regulators work that way. A switching regulator can be nearly 100% efficient which means very little heat is generated and you can get more current out than you put-in (with lower voltage-out). Energy is stored in an inductor so power can be delivered continuously to the load while the transistor/MOSFET switches on & off. Ideally/theoretically, the MOSFET doesn't dissipate/consume any power.
Of course, a simple MOSFET switching circuit isn't a regulator. You could get 5V average PWM for a motor or LED but you can't power an Arduino that way.
but it appears that a linear regualtor is just basically a voltage divider ?
That's right. ...Like a "smart-variable" voltage divider. The voltage gets divided between the regulator and the load with the same current flowing through both. With 12V regulated to 5V, the regulator consumes/dissipates more power than the load.
I have checked and both boards are drawing 240 mA with a supply of 12v so i would asume the temperature should be similar for both?
Your logic is correct. The total heat-energy is the same. But, the temperature depends on the area/volume that the heat is dissipated into. (It will run cooler with a heatsink because the heatsink spreads-out the heat.)
both regulators are rated at 1 A and well witin there voltage rating.
The specs can be tricky... You have to consider the power dissipation and the heat sinking... 1A with 1V dropped across the regulator is 1W. 1A with 10V dropped across the regulator is 10W. Even with heatsinking, the part might not work with maximum-rated current and maximum-rated voltage at the same time.
simon100:
So is a pcb just far worse at getting rid of heat than i had estimated? ?
Do not estimate , calculate from the devices specification sheet.
if there are other devices nearby with significant dissipation this will affect the dissipation of just pcb copper.
You may need a heatsink.
simon100:
Now i understand calculating the heat disapation can be quite complex but in this case the initial design the reg its self is only about 10 degrees above ambient in the formula ?
Not once you understand it .