I am trying to get 5V,2A out of the MIC29302WU (TO-263 package) with an input voltage of 12V. The voltage drops to around 2.2V when trying to supply even 1.3A. Blowing canned air over the regulator helps so I'm assuming it's a temperature issue. This is on a 4 layer pcb without using a heat sink.

Would moving from 3A to 5A help?

Can I achieve 2A without a heat sink?

  • 1
    \$\begingroup\$ Assuming heat is the limiting factor, why are you not using a heatsink? Are there any pad down versions of your device which allow for the use of the PCB copper as a heatsink? \$\endgroup\$
    – loudnoises
    Nov 13, 2019 at 5:28
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    \$\begingroup\$ When it comes to linear regulators it is very important to understand the power burned in the package as P_loss=(Vin-Vout)*I. If you have a high input voltage you will quickly burn A LOT of power in the regulator! \$\endgroup\$ Nov 13, 2019 at 6:45
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    \$\begingroup\$ What’s your input voltage? \$\endgroup\$
    – winny
    Nov 13, 2019 at 7:21
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    \$\begingroup\$ Hi physiii - This is likely to get closed unless you are able to include a lot more design information regarding your voltage regulator. The items that are clearly missing are (1) Schematic (2) Component identification for all being used (3) Input voltage (4) Current capability of input (5) etc. \$\endgroup\$ Nov 13, 2019 at 9:45
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    \$\begingroup\$ You should to the calculation I show in my answer for the situation with a heatsink. Do realize that dropping form 12 V to 5 V means 7 V drop at 2 A that's 14 W. That's a substantial heatsink which cannot be in an unventilated (airflow is needed) box and it will be costly. I am confident that a switching regulator solution will be cheaper! \$\endgroup\$ Nov 13, 2019 at 14:19

1 Answer 1


Can I achieve 2A without a heat sink?

Probably yes, BUT only under quite strict conditions! We'll do the calculation later.

You need to realize that a linear regulator like this MIC29302 works by "burning off" the voltage just like a resistor would. That "burning off" means power is dissipated (as heat) by the chip. That heats up the chip. If the chip gets too hot it will protect itself by lowering the output voltage which should lower the current that is flowing and that should limit the amount of power that is dissipated by the chip. Basically the regulator will go in a "temperature regulation" mode where it tries to deliver as much power as it can without overheating.

So you want 2 A but no heatsink. First thing we have to figure out is what our power dissipation is allowed to be. For that we need values for the thermal resistance, this tells us how much the chip will heat up for a certain amount of dissipated power (usually 1 Watt).

You're using the MIC29302WU which uses a 5 pin TO-263 package. Some thermal resistance values are given, see the datasheet, top of page 7. Unfortunately these are \$\phi_{JC}\$ values where JC means Junction-to-Case. Junction is the chip, case is the metal part of the package. There is no \$\phi_{JA}\$ (Junction-to-Ambient) value which is what you need for use without a heatsink. As there is nothing better, let's use the \$\phi_{JA}\$ for the TO-252 package instead, that will not be accurate but it is all we have.

So the thermal resistance \$\phi_{JA}\$ will be around 56 °C/W. That means the chip will get 56 °C hotter for every Watt that is dissipated. The highest junction temperature we can operate the chip at is 125°C (also top of page 7), we should stay below that 125°C as the temperature protection will kick in at around 125°C. So let's take some margin and design for a 110°C maximum.

Let's assume an ambient temperature of 40°C that gives us:

110°C - 40°C = 70°C

of allowed temperature increase. At 56 °C/W that results in a maximum power dissipation of:

70°C / 56 °C/W = 1.25 Watt

You want 2 A to flow so that means the maximum voltage drop across the regulator can be:

P = V * I => V = P / I:

\$V_{drop}\$ = 1.25 W / 2 A = 0.625 V

So the voltage drop across the regulator should not be higher than 0.625 V or else it will get too hot. Can this regulator even work with such a low drop voltage?

The answer is again on page 7, under "Dropout Voltage". For the MIC29300 the maximum dropout voltage is 600 mV at 3 A. At 2 A this voltage will be smaller which is OK. So yes the \$V_{drop}\$ < 0.625 V can be met.

But realize that the \$V_{drop}\$ is the difference between the input and output voltage of the regulator. So if you apply a high input voltage and want a low output voltage, the regulator will need to drop much more than 0.625 V!

You want 5 V at the output, that means that your input voltage cannot be more than:

\$V_{out} = V_{in} - V_{drop}\$ = 5 V = \$V_{in}\$ - 0.625 V =>

\$V_{in}\$ < 5.625 V

So conclusion:

Yes you can operate the MIC29302 without a heatsink and draw up to 2 A IF you limit the input voltage to 5.625 V.

This is highly impractical as usually you do not have such a precise input voltage.

A possible solution is to use a heatsink!

But at 2 A you will be much better off using a switched mode regulator. You can design and build one yourself if you have enough experience but it is much easier and less risk (of it not working) to buy a ready made module. For hobby purposes an LM2596 based module might do the job.

  • \$\begingroup\$ This is what I thought as well but wanted verification. I'm dropping from 12v so, yes too much power is being dissipated triggering thermal protection mode. Looks like the package type determines \$\phi_{JC}\$ and not current output. \$\endgroup\$
    – physiii
    Nov 13, 2019 at 13:55

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