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I'm working on a battery powered circuit that is using a 1N4001 diode to bring the 6V from 4xAA closer to 5V as well as for protection in case the batteries are put in backwards.

The total circuit draws around 0.6A when driving some LEDs and, when doing so, the diode gets pretty toasty. Definitely uncomfortable for prolonged contact. Probably in the high 100's (F).

I'm really more of a software guy so a lot of this is new to me. I tried looking at the diode datasheet but it was mostly a bunch of stuff I didn't quite understand. Since it's dropping 0.7V @ 0.6A I realize that it's dissipating 0.42W of power, but it doesn't seem that that would be all that much.

Should I be worried about it being too hot or is this sound like it would be in normal operating conditions.

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  • \$\begingroup\$ Sounds like you need a Schottky diode. \$\endgroup\$ – Connor Wolf Jan 31 '13 at 4:16
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From the datasheet:

enter image description here

This is saying the junction temperature will rise 100 K per watt dissipated. The case temperature will be somewhat less than this, but this datasheet doesn't break out the junction-to-case thermal resistance separately, only the junction-to-ambient.

A change of one K is the same as one degree C, so you will see both units used for this spec.

When you are dissipating that 0.42 W at room temperature, the junction is about 25 + (100 * 0.42) = 67 C. If we guess that the case temperature is halfway between junction and ambient, that would be about 46 C, or 115 degrees Fahrenheit. Does that sound about right?

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  • \$\begingroup\$ Sounds reasonable... so, I guess that I should be good then :P \$\endgroup\$ – Adam Haile Jan 31 '13 at 3:12
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The thermal resistance for plastic diode is 50 C/W. So you may get about 20-40C over room temperature. That is, say 45-65C. But 0.6A current is RMS current, with diode being non-linear, with voltage drop increasing with increased current. I'd estimate power to be about 1W, so the worst temperature can be about 75C.

This is at the rated limit, meaning, that diode can easiliy exceed it and unsolder itself. I'd consider currents higher than 0.1 A a high current. And for 1A use a bit more bulky diodes, like some shottki with short and thick leads.

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  • \$\begingroup\$ The current is DC from a battery. And the voltage drop doesn't change (much) with current. \$\endgroup\$ – markrages Jan 31 '13 at 3:16
  • \$\begingroup\$ I can confirm that... I've measured the voltage drop without and with load and it's effectively the same. \$\endgroup\$ – Adam Haile Jan 31 '13 at 3:22

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