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I am working on adjustable voltage controlled current source (2 to 3 A) for my project.

I tried different NMOS for the projects, but they all burned.

In my application, the resistance of my load is very low, like 0.5 ohm to 1 ohm.

I want to control the current on the load from 2 to 3 A (maximum 5 A also fine) using the DAC function of a microcontroller. In my opinion, MOSFET is burns due to limitations in Pds.

enter image description here

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  • \$\begingroup\$ You suspect power limit issues. What does the datasheet say the expected power dissipation limit is? (What is the datasheet?) Does your layout match the test setup used for the power rating? \$\endgroup\$ Commented Aug 12, 2024 at 3:55
  • \$\begingroup\$ R1 resistor is also a "big" too high ... 100 ohm should be ok. \$\endgroup\$ Commented Aug 12, 2024 at 6:41
  • \$\begingroup\$ Consider using a MOSFET driver with built-in overcurrent protection. Also, is it some manner of inductive load like a solenoid? In that case reverse EMF might as well be the culprit. \$\endgroup\$ Commented Aug 12, 2024 at 11:04
  • \$\begingroup\$ In datasheet, at Vds 5V, Ids can be 4A in dc. And actually, through datasheet I dont see any wrong. \$\endgroup\$ Commented Aug 13, 2024 at 6:32

3 Answers 3

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In my opinion, my mosfet is burnt due to limitation in Pds.

You are probably right.

With a 0.5 ohm load and 3 A output, the output FET is consuming $$3.5\ {\rm V}\times 3\ {\rm A}=10.5\ {\rm W}$$

Even if your FET is designed for this kind of power consumption it will only work with an adequate heat sink. In the case of SMT FET's, that heatsink will be provided by the copper of the circuit board it is mounted to.

Your options include:

  • Design your board to provide adequate heat sinking

  • Choose a FET in a package like TO-220 that can fit an air-fin heat sink.

  • Redesign your circuit entirely to use a switching design that doesn't require using a transistor that consumes much more power than the actual load.

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  • \$\begingroup\$ What about the 5A and 0.5 Ohm ? \$\endgroup\$ Commented Aug 12, 2024 at 5:33
  • \$\begingroup\$ @D.A.S., 12.5 W. Doesn't change the general gist of the answer: OP needs to do something different. \$\endgroup\$ Commented Aug 12, 2024 at 15:54
  • \$\begingroup\$ OK but this ignores the variable current to a load which shares 50% power at half load, \$\endgroup\$ Commented Aug 12, 2024 at 17:25
  • \$\begingroup\$ @D.A.S., no, it doesn't. At 5 A, 0.5 ohms, the load drops 2.5 V. That leaves 2.5 V across the control FET. 2.5 V x 5 A is exactly how I calculated the 12.5 W. The load would also be consuming 12.5 W, which means 50% of the total power provided by the 5 V supply goes to the load and 50% to the FET, just as you said. \$\endgroup\$ Commented Aug 12, 2024 at 17:27
  • \$\begingroup\$ OK I agree but I read " 2 to 3 A (maximum 5 A"" as 2~5A @ 0.5ohm ,so I am just commenting on your assumptions not the results. \$\endgroup\$ Commented Aug 12, 2024 at 17:35
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You are operating a part designed to be used as a switch, as a linear pass element. The surface-mount package can dissipate very little heat. The datasheet is very clear about the thermal limitations of this component.

With R3 = 0.5 ohm and 3 A, the FET package is trying to dissipate 10.5 W. At that power level, even a TO-220 part would burn up without a heatsink. Note that the FET would be preforming as a 1.167 ohm resistor, nowhere near its rated Rdson. In this type of application, the Rdson spec rarely applies.

Choose a FET designed for this power level and add an appropriate heatsink.

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You need an active cooler sink & fan, the size of a CPU heatsink (est.) from an old PC tower.

Active Load Design Specs

  1. I = 5A max
  2. R = 0.5 min
  3. P = I^2*R = 12.5W load
  4. Load = V(R)=IR=2.5V
  5. Vdd= 5V
  6. FET Vds also = 2.5V = Vdd-V(R) must deliver 5A*5V=25W
  7. Case Temp rise = 25’C max.(pref.)
  8. Load thermal Rth = 2 ‘C/W = 25/12.5 ‘C/W .

Notes

If dummy load R is a 25W 0.5 Ohm resistor with sink mount flange then both FET and fixed R load can share heatsink with a rating of 1’C/W which is like a standard modern tower CPU heatsink & fan 5V with silver oxide heat grease & FET drain TO-220 insulator.

This way the desired 25W max. power is shared by 2 parts to increase surface area to the sink.

To compute final thermal resistance, and temp rise, the equivalent R network adds all series and parallel parts to the shared heating Rca (case to ambient).

Added

Also if you want 5A always choose a 50 mV max drop or less.

Even if you use a 2W resistor in this case for R3 that is excessive.

R3 temp rise

I^2R = 450 mW for 3A and 1.25W for 5A. To estimate R temp rise using Pd/Max *125’C. So for a 2W part dumping only 1.25W heat rises 80’C which will sizzle water.

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  • \$\begingroup\$ Yes, sure. It will be great if I could add heat sink like ordinary electronics. However, due to size limitation, it is not possible... Thererfore, I am finding a better MOSFET that can manage 2A to 3A. \$\endgroup\$ Commented Aug 13, 2024 at 11:13
  • \$\begingroup\$ This is a thermodynamics , power resistance per degree 'C problem, not a linear RdsOn choice. Sorry, that won't work for linear loads, only PWM nonlinear load types. \$\endgroup\$ Commented Aug 13, 2024 at 12:14
  • \$\begingroup\$ ..unless you intended to switch fixed resistors rated at high power \$\endgroup\$ Commented Aug 13, 2024 at 16:11

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