# How to choose heatsink for power devices (transistor, LED, regulator)?

In many projects I need to choose small, but big enough heatsink for transistor or other power devices (LED, amplifiers, voltage regulators).

I know that thermal resistance of many devices is known (from datasheet), but almost all heatsinks available on the market where I live have no datasheets.

My primitive (?) method for choosing heatsinks is:

1. Put device on "probably big enough heatsink" (use thermally conductive paste, insulation etc. if needed)
2. Put device with heatsink in conditions where it's supposed to work (casing, ambient temperature etc)
3. Turn on device (transistor, LED) with 20% power
4. Wait some time (30s or more for bigger heatsinks or casings with big thermal capacity)
5. Measure heatsink temperature
6. Increase power by 20% or more if temperature is low and repeat step 3

I'm observing maximum heatsink temperature, I'm changing heatsinks and when I think some heatsink is probably good enough - I'm estimating junction temperature from power applied to device and heatsink temperature (close to device).

This method was good enough for me until now - I need to build something really small and I need smallest possible heatsink.

I can protect device (measure heatsink temperature and turn device off when temperature is too high).

Is there any better method to choose heatsink?

Edit:

Im using thermocouple and multimeter or homemade thermometer with LM35 sensor glued to small flat copper heatsink, insulated thermally from air on "air side" and covered with some thermal grease on measured radiator side.

• What are you using to measure temperatures? For example, a non-contact IR thermometer or a thermocouple such as would come with a digital multimeter? – JYelton May 22 '14 at 22:39
• Since you don't have data sheets for the heatsinks you're trying, I'd say that what you are doing is about what you need - with one BIG exception. You MUST measure the device temperature, not the heatsink. I'd recommend using a thermistor, and glue it to the device with thermally conductive epoxy (as little as possible). – WhatRoughBeast May 23 '14 at 0:54

It is possible to determine the thermal resistance of a heatsink that is at hand, but for which there is no data sheet. This can be done relatively simply and without iteration. First, weigh the heatsink, then heat it to some uniform steady temperature in an oven, finally remove from the oven and allow to cool. Cool down time will be related to the overall thermal resistance and mass of the heatsink.

To see how heatsink temperature change is related to mass and thermal resistance to ambient, use an analogous RC electric circuit. The least complicated circuit that's useful is a parallel RC with initial voltage condition ($V_o$) on the capacitor. In the thermal analog of the RC circuit, resistance becomes thermal resistance between the sink and ambient ($\Theta _{\text{SA}}$) in $\frac{\text{{}^{\circ}C}}{W}$. Heat stored in the heatsink can be mapped into the capacitance as $m$ $C_p$, where $m$ is heatsink mass and $C_p$ is specific heat capacity of the material (~0.9 $J$/$g$/$\text{{}^{\circ}C}$ for aluminum). An equation for heatsink temperature ($T_{\text {hs}}$) can be written for the thermal circuit as:

$T_{\text {hs}}$ = $\left(T_{\text{hso}}-T_{\text{amb}}\right) e^{-\frac{t}{m C_p \Theta _{\text{SA}}}}+T_{\text{amb}}$

Rearranging, thermal resistance is:

$\Theta _{\text{SA}}$ = $\frac{t}{m C_p \text{Ln}\left(\frac{T_{\text{amb}}-T_{\text{hso}}}{T_{\text{amb}}-T_{\text{hsf}}}\right)}$

Thermal time constant for the heatsink is:

$\tau$ = $m C_p \Theta _{\text{SA}}$

It is convenient to use $\tau$ to set the target heatsink temperature ($T_{\text {hsf}}$) to terminate the measurement, because with the measured time the only remaining unknown is $\Theta _{\text{SA}}$ which can now be calculated.

Method with more detail and example numbers

• Weigh the heatsink. Let's just say you get 100g.
• Install thermal probe. Attach probe where a device would be mounted.
• Put heatsink, on thermal insulator (like a piece of wood), in oven and heat to elevated temperature. While waiting for readings to stabilize calculate target cool down temperature $T_{\text {hsf}}$ by setting $t$ to $\tau$ in $T_{\text {hs}}$ equation. For example, using $T_{\text {hso}}$ = 100$\text{{}^{\circ}C}$ and $T_{\text {amb}}$ = 25$\text{{}^{\circ}C}$, $T_{\text {hsf}}$ will be 52.6$\text{{}^{\circ}C}$.
• When heatsink temperature stabilizes, remove insulator and heatsink from oven (don't burn yourself) and place in ambient environment. Record time when heatsink reaches target temperature. For this example that's 52.6$\text{{}^{\circ}C}$ and $\tau$ would be 225 Sec.
• Use recorded time and equation for $\Theta _{\text{SA}}$ (or even $\tau$) to calculate heatsink thermal resistance. For this example $\Theta _{\text{SA}}$ = 2.5$\frac{\text{{}^{\circ}C}}{W}$.

When deciding how high a temperature to use as heatsink initial condition, use a temperature that makes sense for the application. 100$\text{{}^{\circ}C}$ is probably as high as you should ever go (it would mean that the junction of whatever device was mounted to the heatsink would be at 110$\text{{}^{\circ}C}$ or more, and that's hot).

• Could you please give more detais on how you get Thsf 52,6 degrC? – GR Tech Jul 17 '15 at 4:31
• @GRTech The method relies on use of thermal time constant $\tau$, and exponential decay of temperature after removal of heatsink from oven . In the first equation for Ths, set t=$\tau$ so Ths=((100-25)*e^-1)+25=52.6. Measure time it takes to get to 52.6 for a value for $\tau$. – gsills Jul 25 '15 at 3:43

The process of selecting an appropriate heat sink is pretty straightforward.

1. Determine the maximum power dissipation, $P_{MAX}$, of the transistor or IC.

2. Determine the maximum junction (or die) temperature of the transistor/IC. This is usually given in the data sheet as a maximum $T_J$. This value can range from 70 C to over 125 C.

3. Determine the maximum ambient temperature, $T_A$, where the device will be used.

4. Calculate the maximum temperature increase of the device from ambient, $T_J - T_A$

5. Divide the maximum temperature change by the maximum power consumption. This will be the maximum allowable thermal resistance, $\Theta_{MAX}$, from the junction to ambient, in C/W (or the equivalent K/W). In other words, $\Theta_{MAX} = (T_J - T_A)/P_{MAX}$

6. From the device data sheet, find the value of the device's thermal resistance from the junction to the outside surface of the device case (i.e. the package). This is usually called $\Theta_{JC}$

7. Subtract $\Theta_{JC}$ from $\Theta_{MAX}$. The result is the maximum allowable thermal resistance $\Theta_{SA}$ from the sink to ambient for the selected heatsink. You may also need to subtract the thermal resistance between the device case and the heatsink, $\Theta_{CS}$, for high power devices. You can greatly reduce $\Theta_{CS}$ by using thermally conductive grease or a thermally conductive pad between the device and the sink, but 1.0 C/W is a reasonable estimate for $\Theta_{CS}$.

• The OP specifically stated that data sheets for possible heatsinks are not available. – WhatRoughBeast May 23 '14 at 0:49