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I am new to the construction of DC/DC power supplies (still a university student) and have built basic supplies using simple linear voltage regulators. I have recently discovered the world of switching power supplies and their increased efficiency (in exchange for higher part counts). This is useful since I am building a project that can use 1.5A peak current at 5V, and I am using ~12V source. Linear Voltage Regulators are, from what I am reading at least, not a good selection for high current applications and heat becomes an issue.

I am wanting to use a TI TPS5420 step down switching voltage converter. I noticed the package (8-SOIC) is much smaller than many high current linear regulators, and that raises the question about heat and power dissapation. Linear regulators can require large heatsinks and larger packages at "higher currents" ( > 1A, but really counts on other factors like input voltage, output voltage, etc).

Can someone help me through how I would calculate the power dissipated through heat on this chip and if I should worry about the IC being too hot to touch? Even though the IC is more efficient than a large linear regulator, its also much smaller and doesn't have a thermal pad -- this makes me worry about how the heat is dissipated. Or am I just overthinking the issue?

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You are right in that a switcher makes a lot more sense for your application (12V in, 5V 1.5A out) than a linear regulator. A linear would waste 7V * 1.5A = 10.5W in heat, which would be challenge to get rid of. For linear regulators, current in = current out + operating current. For switchers power in = power out / efficiency.

I haven't looked up the TI part you mention (I might have if you had supplied a link). There are two broad classes of switching regulators, those with internal switches and those that drive external switches. If this regulator is the second kind, then dissipation in the part won't be a problem since it's not handling the power directly.

If it is a fully integrated solution, then you do have to look at dissipation. You can compute this dissipation from the output power and the efficiency. The output will be 5V * 1.5A = 7.5W. If the switcher is 80% efficient, for example, then the total input power will be 7.5W / 0.8 = 9.4W. The difference between the output power and the input power is the heating power, which in this case is 1.9W. That's way better than what a linear regulator would do, but is still enough heat to require some thought and planning.

80% was just a number I picked as a example. You need to look at the datasheet carefully and get a good idea what efficiency is likely to be at your operating point. Good switcher chips have lots of graphs and other information about this.

Once you know how many Watts will be heating the chip, you look at its thermal spec to see what the temperature drop from the die to the case is. The datasheet should give you a degC per Watt value. Multiply that by the Watts dissipation, and thats how much hotter the die will be than the outside of the case. Sometimes they tell you the thermal resistance from the die to ambient air. This is usually the case when the part is not intended to be used with a heat sink. Either way, you find how many deg C hotter the die will be than anything you can cool or deal with.

Now you look at the max die temp, then subtract off the above temp drop value. If that's not at least a little above your worst case ambient air temperature, then you have a problem. If so, it gets messy. You either need a heat sink, forced air, or use a different part. Higher power switchers are usually designed for external switch elements because power transistors come in cases intended to be heat sunk. Switcher chips usually don't.

I don't want to go on speculating, so come back with numbers about your particular situation, and we can continue from there.

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  • \$\begingroup\$ Thank you! You answered all my questions. I appreciate the response. I found in the datasheet all these values and the IC picked will work just fine in the worst case scenario. \$\endgroup\$ – Mike Jun 20 '11 at 20:17
  • \$\begingroup\$ excellent answer, as always. One note though: the power lost in the switcher isn't completely dissipated in the IC; there's also the external diode which takes a part of it. \$\endgroup\$ – stevenvh Jun 21 '11 at 7:40
  • \$\begingroup\$ @stevenvh - Yes, good point. I should have mentioned that. If the numbers work out assuming all losses heat the IC, then you're fine either way. \$\endgroup\$ – Olin Lathrop Jun 21 '11 at 13:08
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The datasheet has an efficiency vs output current graph on the first page. For 1.5 A peak current, it looks about 91% efficient. If it's supplying 7.5 W at 91% efficiency, it would be wasting 0.7 W from itself.

A linear regulator dropping 12 V to 5 V at 1.5 A would waste 10.5 W while supplying 7.5 W, making it 42% efficient.

Obviously, the switcher is more efficient and less wasteful. They also tend to be more expensive and more difficult to use without problems, though.

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