How to calculate the Temperature rise of PCB and its surrounding air from the power PCB consumes?

Question

I am having a difficulty understanding some mechanism regarding heat generation from an electronic PCB (Printed Circuit Board) and its effect on atmospheric temperature.

I don't know if this question is related to This forum or its related to Physics, but I am going to post it here as my concern is regarding Electronic PCB.

Please bear with me on this long question.

Problem:

A PCB with specifications:

• Surface mount-multi layered with few through-hole components (capacitor, LCD etc).
• The circuit designed on it consists of Capacitors, Resistors, EEPROM, 3.3 V regulators, RC Power Supply circuit, LCD, MCU etc.
• Area = 66 cm^2 Power Dissipation in the circuit on PCB = 0.44 W to 1.0 W, typically 0.5 W
• No heat sink, or specific heat generating components.
• The circuit is operated on 3.3Vdc
• Enclosed in a Box of volume 1007.8 cm^3.
• The box has holes (where wires are passed to PCB) that can accommodate air flow easily.

I want to find:

1. The temperature rise of PCB when it consumes 0.5 W power.
2. The temperature rise of air inside the box because of heat transfer from PCB and time.
3. Time to reach equilibrium when PCB surface and air inside the box are at the same temperature.

Note: I don't want to go into details or very deep into the design (as I have seen some designer do the very complex and detailed calculation for these purposes). I only need to find the general idea about the heat transfer of this topology. For that, if it's easy, you can assume that PCB is just a heat generating surface (amount of heat can be calculated from power).

Now, I came this far,(this work was done by someone else on this forum)

• The copper side with the traces is modeled as a sheet of copper rather than traces.
• The body is thin enough that thermal conductivity within the body is unimportant, and the entire device is considered to be at a uniform temperature.
• Only the two broad surfaces contribute to the heat loss, the sides are neglected.
• The surroundings, including the air and radiative syncs, are at a uniform temperature Ts
• Thermal coefficients: ϵcu=0.78, ϵpcb=0.50, hup=7.25 W/m2K, hdown=3.63 W/m2K Under these assumptions, we can estimate the temperature of the board by simply equating heat flows.

The heat coming in per unit time is from Joule heating from the current running through the copper and is given by qin=I2R = Power Dissipated

The heat flowing out has two escape mechanisms; radiative heat transfer to the surroundings which is given by

qrad=ϵσA(T^4 − T^4 * s) Here, ϵ = ϵcu +ϵpcb

and convective heat transfer to the air which is given by

qconv=hA(T−Ts) Here, h = hup + hdown

Now we just equate the heat flows

Qin=Qout

I2R=A[σ(T^4−T^4 *s)(ϵcu+ϵpcb)+(T−Ts)(hup+hdown)]

we can rearrange this to look like a quartic equation

σ(ϵcu+ϵpcb)T^4 +(hu+hd)T−[I2RA+σ(ϵcu+ϵpcb)T^4 *s+(hu+hd)Ts]=0

From here, I calculated the temperature of the PCB surface to be 28.9 degrees Celcius, assuming the temperature of the air inside the box to be 25 degree Celsius.

Questions:

1. Have I done it correctly? if not please tell me the reason (cause I am not confident about this).
2. If its ok, how do I find the temperature rise of inside the box and the time it will take to raise that temperature?
3. Will air inside the box and PCB surface ever be at equilibrium and at what temperature and what time will it take to reach that state?

I know this is a mess, but I can't find any help regarding this specific problem, I tried other methods and tricks but the results are far too much wrong that even I know they are not right (one such example; temperature cam out to be 300-degree Celsius).

Please give me the directions to how to approach this (as I am fresh electrical engineer and don't know about heating, thermodynamics of this level).

Answer 1

The thermal time constant of 1 square of PCB GROUND plane, of size 10cm (4"), is 96 second. The time constant of a square meter of foil is 9,600 seconds; of 1cm is 0.96 seconds; of 1mm square is 0.0096 seconds (9.6 milliSeconds).

If we model the heat flow as from one edge of the foil, thru the 4" of foil, and exiting the opposite edge, then that square of foil has a Thermal Resistance of 70 degree Centigrade per watt.

Thus worst case temperature rise, with 1 watt of heat generation spread along one edge of the PCB, and the heat flowing only through the foil (no heat exiting into the air), to exit into the air (or thru metal mounting posts into the case) at the opposite edge, will be 70 degree C /Watt * 1 watt = 70 Deg C.

However, if the heat has to flow through long narrow traces to reach the Ground (or power) plane so the heat can spread out, then you can have much hotter local regions, and the slowly moving air flow will easily cool regions that are 100C or 150C hotter.

SUMMARY:

reach thermal equilibrium? yes, with one TAU being 96 seconds.

assume uniform heating on the PCB? at 70 degree C per watt of heat flow? not likely. You need at least ONE plane, to really spread out the heat, or a bunch of WIDE traces exiting your hot-spots.

That 70 degree C per watt is PER SQUARE. A trace 2cm long and 2mm wide has 10 squares, thus is 700 degree C per watt. Use a plane, or wide-and-short traces. And you need to dump heat THRU the FR-4 epoxy fiberglass substrate, to move heat into the plane.

Draw some sketches of the PCB heat flow, using a bunch of resistors to constrain the heat movement.