IRF530 getting very hot

Question

I have a IRF530 connected to a 5V micro. The trace is on a PCB and very short, with only a pull down resistor on the gate:

^{simulate this circuit – Schematic created using CircuitLab}

The MOSFET is switching a long length of 24V LED strips, which draws 6A. There is a 5m 2 core cable from the MOSFET to the strip.

When I set the pin to high, the mosfet gets very hot. On shorter lengths of LEDs, it only gets marginally warm. When I PWM the mosfet (f=500Hz), it still gets hot but takes a little longer to warm up.

How come this is happening? I am above the threshold voltage of 4V, so I assume the FET is not in a linear state, and the power dissipation according to Rds(on) is only 0.96W, which the un-heatsinked (un-heatsunk?) package should handle.

Answer 1

How come this is happening? I am above the threshold voltage of 4V, so I assume the FET is not in a linear state

The FET datasheet specifies RdsON for Vgs=10V. It is not specified for 5V. Therefore it will not conduct fully and act as a resistor. Threshold voltage is the Vgs at which is begins to conduct just a little bit, you need way more to have good switching.

Solution: use a FET with RdsON specified for Vgs=5V, and suitably low RdsON. For 6A and without heatsink, aim for 15 mOhms or lower. There are plenty of such FETs.

Answer 2

Your problem is insufficient Vgs for the device chosen. Hence not the rated RdsOn.

MOSFET's come a many different part numbers, with distributor search filters to help selection.

How can you choose one to keep cool?
How much does a heatsink cost vs a better FET? Copper area? Alum. sub.?
How many do you need? 1? 1k? +?

Generally for slow PWM, high current, low voltage;

• consider devices rated for > 5x the current you plan on using. e.g. >=30A
• Ensure RdsOn is rated at your available Vgs
  • or in other words Vgs >> 2x Vgs(th) pref. 3x.
• prefer SMD over THT for better heat transfer to board if suitable
• always do a thermal resistance analysis of temp rise including case and ambient cooling or lack of.

This change will:

• reduce heat loss, reduce cost of thermal design and only slightly increase FET cost but multiply input capacitance to gate

Example:

IRFH5301 $1.25 (1pc) 1.85 mOhm @ 50A, 10V

Alternatives

Use 12V to drive the gate 1k pullup, 1k down divider from 24V and shutoff with an additional NPN inverter to shut off gate.

^{simulate this circuit – Schematic created using CircuitLab}

R values may be increased slightly to reduce Pd. with Ciss affecting risetime.

Answer 3

While others have adequately answered the question, I thought I'd approach it from another perspective, in order to give you some insight as to how to read data sheets.

Go back to your IRF530 data sheet, and look at the Vgs(th) rating of 2 to 4 volts. Now look closely. Notice that at the rated voltage, the drain current will only be 250 uA. Even at 5 volts gate drive, you are not guaranteed to get a whole lot more than this, and this explains your high Rds. Now, it's true that 4 volts is worst-case, and you are surely doing better than this, but it's the heart of the problem.

Now look at figure 3. This deals with "normal" behavior, rather than worst-case. At 5 volts gate voltage, you can only "expect" 3 to 4 amps of current, and you want 6, so this also should give you warning.

Finally, look at figure 4, Rds vs temperature, and notice how it rises. This allows the temperature to stabilize, rather than running away. It's very useful for connecting FETs in parallel, since if one is hogging the current, its resistance will rise and it will draw less current, allowing equal current sharing without too much fuss on your part. This is in contrast with BJTs (bipolar junction transistors) whose effective resistance drops with rising temperature, which can lead to runaway thermal failure.