# Power loss and heating in a diode

I'm quite confused about diodes heating when conducting electricity.

I have "P1000 10A10" diode. It's supposedly rated for 10A forward current.

When I've attached my lab power supply set it to 3.5 A, I've got 0.8 V voltage drop over the diode. This gives around 3 watts of power loss inside the diode. From the datasheet, the diode should get to 30°C over ambient, but it's getting to around 100°C in a minute.

I "only" need 5 A of current through the diode, so I added another one in parallel, but at 5 A they got to above 120°C which is the max temperature of my thermal camera.

• Strange. Sometimes diodes' datasheet specifies some copper area for Rthj-a specification but this one does not. Double-check the current with a multimeter in series? Commented Oct 10, 2023 at 13:53
• The datasheet says 10°C/W junction-ambient. I say, no way, at least not without sinking heat through the leads to a heat sink. TO-220 parts (roughly similar size to this thing) typically have Rθja in the neighorhood of 60°C/W, not 10. Commented Oct 10, 2023 at 15:55
• @marcelm Maybe that's with the PCB at a fixed temp where the leads are soldered?
– Drew
Commented Oct 11, 2023 at 14:22
• @Drew That would also apply to soldered TO-220. With a large enough copper area attached to the leads, perhaps. But the linked datasheet specifies nothing of the sort, so we're left guessing how they came up with that number :( Commented Oct 11, 2023 at 15:33
• @marcelm I agree it's a bad datasheet. Surprisingly common. You'd think with the work that goes into producing a new component, some guy would at least spend a few hours on the datasheet.
– Drew
Commented Oct 11, 2023 at 15:34

The Rth(J-A) for that case is actually the Rth(J-L), the thermal resistance from the junction to the leads. A similarly-sized diode gives a similar Rth(J-L) of 11 C/W when mounted on 100 square cm of copper. This Microchip app note shows you that even with a full square inch of copper on each lead, you would only get down to a Rth(lead-ambient) of ~40 C/W which, added to the resistance from junction to lead, means that your 3 watts of heat would cause a junction temperature rise of ~150 C, which puts you over your max junction temperature.

You have some options:

• Change package to something easier to remove heat from: in the surface mount category, a DPAK only has a Rth(j-c) of 1 C/W, which removes some of the resistance (but you still need to have a pretty large land to go with it). A through-hole package like a TO-220 is very easy to heatsink, at the expense of size and cost.
• Choose a diode with lower Vf: this only gets you so far if your thermal resistance is huge to begin with, but something in the 400 mV range isn't out of the question and will definitely help.
• Use an ideal diode controller: a small IC that senses the voltage across a "backwards" MOSFET and turns it on when current is flowing through the body diode and turns it off when it tries to go the other way. Heat produced is from I2R losses in the pass element which, at the currents you're talking about, is orders of magnitude lower than a diode. You can find ones with MOSFETs integrated into the package, making it a one-chip solution.

I do not recommend paralleling diodes as the Vf drops with increasing temperature, so one diode will begin to "steal" current from the other, which starts a positive feedback cycle until it's carrying all or almost all of the current.

• The temperature coefficient is certainly an argument, but even with constant temperature you should not connect diodes in parallel. A diode, like a LED, has a low internal resistance, and will act like a voltage source. A small delta-V gives a high delta-I. Two diodes (or LEDs) will never have the same Vf, and the unbalance of both currents will be huge. Incandescant lamps are the opposite: internal resistance is high, and never identical, and series connection will most often give an unbalance in voltages, and the one getting the highest voltage will burn out quickly. Commented Oct 12, 2023 at 12:29

Lead length and PCB mounting style have effect on final temperature.

• As the effective total lead length increases so does the effective Rthja because of the increase on the effective path length of the thermal conduction. Some datasheets mention this by numbers. Try to find ones from different manufacturers to see if they provided thermal resistance vs lead length information.

• If the diode is mounted on a PCB (or a surface, generally), the temperature rise of the diode will heat the surface up, therefore the effective ambient temperature for the diode will no longer be equal to your ambient temperature.

You need to provide some heat sinking for the diode.

The datasheet doesn't say much, but I'm pretty sure the thermal resistance mentioned is when mounted on a PCB that will provide heat sinking.

Here is a datasheet for a similar diode. This diode has a similar figure of thermal resistance, but the datasheet has the following disclaimer:

Valid, if leads are kept at ambient temperature at a distance of 10 mm from case

I'm quite confused about diodes heating when conducting electricity.

Reasons for Heating:

1. Diodes dissipate power due to their forward voltage drop.
2. Inefficient heat dissipation due to thermal resistance.
3. Parallel diodes can lead to uneven current distribution and heating.
4. External factors like ambient temperature and airflow.

Solutions:

1. Attach a heat sink to the diode.
2. Improve airflow with a fan.
3. Opt for a higher-rated diode instead of paralleling multiple diodes.
4. Ensure good ventilation and ambient conditions.

Always ensure safe operation within component specifications!