Do you need a heat sink to handle your voltage regulator power dissipation? Let’s see with some simple math.

Recently, I have been looking at mounting voltage regulators on my wideband loop amplifier boards. My thinking is that this would allow me to experiment with different amplifier circuits attached to a standard 12 volt supply on the CAT cable. I realized that I had never taken the time to understand if and when a heat sink would be required for the regulator. Heat sinks can either be externally attached or built into the copper on the circuit board.

Here is what I learned. Whether or not a heat sink is required depends on voltage regulator power dissipation and its related temperature rise in the component. With the data sheets, it is straightforward to calculate the maximum power dissipation and temperature rise allowed without a heat sink.

You will find the key to everything is the thermal resistance, or effectiveness of the part in dissipating heat. You find this on the data sheet as θ_JA measured in °C/W. θ_JA tells you how much the temperature between the device junction and ambient  will increase per watt of power dissipated. Typically, junction temperature or T_J is allowed to rise to 125° or 150° before the regulator shuts down.

Power dissipation is the regulator voltage drop times load current, or PD = (V_IN – V_OUT) x I_L. Maximum allowable temperature rise is the difference between maximum junction and ambient temperatures, or T_RISEmax =T_Jmax – T_Amax.

Finally, the maximum allowed power dissipation is simply PD_max = T_RISEmax ÷ θ_JA.

Voltage Regulator Power Dissipation Example

So, do I need a heat sink if I use an LM317T to regulate 13.8 volts down to 10.0 volts for my LZ1AQ loop amplifier?

My amplifier draws 135 ma, which would be the load current on the regulator. This converts to around half a watt , calculated as (13.8 – 10.0) x 0.135, going up in heat from the regulator. If I assume a maximum ambient temperature of 40° the maximum temperature rise the LM317T can handle is 125° – 40° = 85° C.

This regulator has a θ_JA of 65° C/W. So, it can handle power dissipation of 85° ÷ 65° C/W = 1.3 watts without a heat sink. Similarly, consuming 0.5W x 65°/W gives a 32° temperature rise.

Good news. No regulator needed. The actual PD of 0.5 watt is well below the maximum allowed 1.3 watts. And the expected temperature rise of 32° is well below the allowed 85°, given the assumptions and data sheet specs.

For more information, check out this guide from Texas Instruments.