I try to switch a 12V non-inductive load drawing 400-500mA with a IRF3708 Logic-Level-MOSFET with the 5V from an Arduino. I am new to MOSFETs but I simply cannot find my mistake. I think I fried some MOSFETs using the 5V because Rds(on) was too high (3-4Ohm) and this resulting in too much heat which killed the MOSFETs. I hope this conclusion is correct. But the datasheet states that at 4.5V Ugs Rds(on) should be in a 9.5-13.5mOhm range. Even if it may be optimistic I don't get why it is in the Ohm range and not even near the values in the datasheet.

What am I understanding wrong here?



simulate this circuit – Schematic created using CircuitLab


Circuit Calculation of Rds(on)


simulate this circuit

With a load of 1k my Amperemeter gives me a current of 11.8mA @12V with 5V applied to the gate. This equals a series resistance of 1.0166kOhm. This results in a Rds(on) of 1.66Ohm. Ok it is lower than my wrong measuring but still a lot higher as in the datasheet. If 500mA would flow like in my application this Rds(on) gives an UDS of 0.664V which result in 265.6mW of heat at the MOSFET. Correct?

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    \$\begingroup\$ (1) Is the FET stuck open-circuit or short-circuit? (2) Did you measure the gate voltage? (3) Are your grounds connected correctly? (4) Add a photo. We might spot an error. \$\endgroup\$ – Transistor Oct 14 '17 at 22:12
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    \$\begingroup\$ "Rds(on) was ... (3-4Ohm)" - how do you know this? Got any voltage and current measurements you can share? \$\endgroup\$ – Bruce Abbott Oct 14 '17 at 22:13
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    \$\begingroup\$ Have you triple and quad checked the pin connections of the FET to make sure you have Gate, Source and Drain correct? \$\endgroup\$ – Michael Karas Oct 14 '17 at 22:14
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    \$\begingroup\$ This question looks extremely familiar... \$\endgroup\$ – user103380 Oct 14 '17 at 22:17
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    \$\begingroup\$ "I connected my multimeter at source and GND to measure the Rds(on)." - ??? To measure RDSon you should connect a load with known current draw and measure voltage between Drain and Source, then calculate the resistance. To get an accurate measurement you should probe directly onto the FET leads (breadboards are notorious for having high-resistance connections). \$\endgroup\$ – Bruce Abbott Oct 15 '17 at 6:40

Your circuit is correct. Your way of measuring the RDSON is not.

With a load of 1k my Amperemeter gives me a current of 11.8mA @12V

This is good to see that the circuit is generally working, but you don't have anywhere near the accuracy to declare the RDSON result you do. How accurate is the supply? How accurate is the resistor? What is the accuracy of the ammeter? What is the resistance of the ammeter.

Note that small errors in the first three cause large errors in the RDSON conclusion. And the resistance of the meter is likely significantly higher than RDSON. Even if the other numbers were accurate enough, all you'd really be doing is measuring the meter resistance.

For example, let's work backwards what some of your measurements would need to be to indicate 0 RDSON. The supply might only be 11.8 V. That's only 1.7% below spec. Is your supply really that accurate? Right. I didn't think so. Or the resistor might be 1.017 kΩ instead of 1.000 kΩ. That's only 1.7 % high. What range of current is really running thru the ammeter when it says 11.8 mA? A combination of errors under 1% is all it takes to make the computed RDSON actually come out negative!

Your measurements don't support your conclusions.


You can't reliably test a MOSFET that way (using a multimeter). For one, most multimeters wont put out 12V and the Vdd has to be higher than the gate voltage or you can damage the device.

Testing it in circuit is more reliable. Understand that a MOSFET is not a voltage to resistance device, its is a voltage controlled current device. The current it allows to flow from D-S is dependant on the gate to source voltage. The resistance is dynamic, i.e. it is whatever it needs to be to pass the rated current for the given voltage. If you use a multimeter though, it will have a current limited output which means you will only get so much current and the effective resistance will be the same no matter the gate voltage.

  • \$\begingroup\$ Ok I will calculate the Rds(on) with a resistive load. You say it is a "voltage controlled current device and the resistance is what it needs to pass the rated current for the given voltage". But wouldn't that mean that more/or maybe too much power gets lost at the MOSFET? \$\endgroup\$ – moessi774 Oct 15 '17 at 6:50
  • \$\begingroup\$ Not necessarily, the device is fully off if the gate voltage is 0v. The typical off resistance is in the megohm range. 9.5 megohms for that device. The electric field generated by the gate charge induces charge carriers to move from the Drain to the source. There is very little current flow from the gate to either the drain or source. The voltage on the gate is what determines the strength of the electric field and how much of the channel is passivated. The more of the channel is passivated, the more charge carriers are allowed to flow. \$\endgroup\$ – Anthony Bachler Oct 15 '17 at 8:12
  • \$\begingroup\$ Ok thats also what we learned in theory lessons. But in my understanding that means that a MOSFET is never generating waste heat which is not true. Sorry for asking so much but I am really a beginner with "real" hardware not just drawn circuits. \$\endgroup\$ – moessi774 Oct 16 '17 at 8:12
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    \$\begingroup\$ "Vdd has to be higher than the gate voltage or you can damage the device". This is just plain wrong. "a MOSFET is not a voltage to resistance device, its is a voltage controlled current device". This is true over part of its operating range, but not when used as a fully-on switch. In that case, D-S does look like a resistor. That's what the Rdson spec is all about. \$\endgroup\$ – Olin Lathrop Dec 17 '17 at 13:14

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