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I have designed a portable lens heater prototype. Using:

The MCU board works as expected and outputs 3.3V on the MOSFET gate when the button selects the correct mode. However, there is only 2.5V across the heating element. At full charge the 18650 should be able to output 4.2-3.6V. I chose this MOSFET because I thought RDson would be in a suitable region. Please see figure below: PMV16XN RDSon

I want to maximise the power delivered to the heater element. Is there some other MOSFET property preventing the 4.2V forming across the resistor? Can you recommend an alternative circuit/component? enter image description here

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    \$\begingroup\$ What I miss is the measured value of the voltage across the MOSFET when it is switched on. The high current loop is battery(+) - heater - MOSFET - batter(-). Measure the voltage across all of these, are they what they should be? You're suspecting the MOSFET while there is no proof that it is the issue. \$\endgroup\$
    – Bimpelrekkie
    Commented Jan 15, 2020 at 14:56
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    \$\begingroup\$ "the 18650 should be able to output 4.2-3.6V" You need to verify this in your circuit. Don't just measure the voltage across the resistor, measure the voltage across the transistor and the voltage coming from your battery. \$\endgroup\$
    – Elliot Alderson
    Commented Jan 15, 2020 at 15:40
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    \$\begingroup\$ @Delphesk no, look at the MOSFET spec and you will see that you are just "guessing"! That MOSFET is good for this application and the gate voltage should be good also. \$\endgroup\$
    – Andy aka
    Commented Jan 15, 2020 at 16:07
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    \$\begingroup\$ Your battery is sagging. Measure the voltage across it with that 5 ohm load connected and you'll see. \$\endgroup\$
    – winny
    Commented Jan 15, 2020 at 17:02
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    \$\begingroup\$ No apalogies needed as you learned something! Perhaps the battery has an internal series resistance that is higher than it should be and/or the battery contacts have more contact resistance than they should have. \$\endgroup\$
    – Bimpelrekkie
    Commented Jan 16, 2020 at 20:47

2 Answers 2

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If you add V1 and V2 you get 2.7V. Unless there's some other source of resistance in the circuit, that's the battery voltage under about 0.534A load. That's not much load for the 18650, unless it's severely aged or fairly completely discharged. Measure the voltage directly across the battery terminals while the element is active; if it reads near 2.7V try recharging it. If the same thing happens when fully charged, try a different battery.

If you see more than 2.8V across the battery, I suspect another source of resistance in your system, possibly a bad connection. If it's localized, use the voltmeter to track it down.

In any case, your MOSFET is working fine.

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This MOSFET is only specified for operation at Vgs = 4.5V.

enter image description here

The RdsOn curve in the datasheet represents "typical" values but there is quite a lot of dispersion. If you want to be sure that the FET RdsON is below, say 10-20 mOhms at Vgs=3V then you need to have this specified in the datasheet.

The voltage across the 5ohm heater is 2.67V. 30mV across the Drain and Source.

So you have about half an amp and the FET has 60 mohms RdsON or double what you'd get at Vgs=4.5V... so while it isn't exactly the right FET for the job it will probably work. Note there is a lot of dispersion in MOSFET characteristics so several FETs from the same bit of cut tape may perform differently.

If you want to select a proper FET use the DigiKey search filter "Drive Voltage (Max Rds On, Min Rds On)". For example this FET has a proper RdsON spec for Vgs=2.5V:

enter image description here

The rest of the voltage is probably lost in the wires, or a bad contact somewhere.

If the arduino can provide 3.2V gate drive, then it is receiving at least that from the battery, but you only have 2.7V on the FET and resistor.

So the battery may sag a bit down to 3.2V, or the battery holder contact has a bit of resistance... but that doesn't account for the 0.5V difference between 3.2V and 2.7V! That's probably lost in the wiring.

Are your PCB traces wide enough?

You're not running high current through a breadboard?

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  • \$\begingroup\$ Did you look at the voltage readings that were added an hour ago? There is no reason to think that the battery is providing 3.2V under these conditions, so no need to guess about voltage "lost in the wiring". \$\endgroup\$
    – Elliot Alderson
    Commented Jan 15, 2020 at 17:52
  • \$\begingroup\$ @ElliotAlderson Well he says Vds=30mV on the FET and there is 2.67V across the load, so 2.7V total. And Vgs=3.2V, so the micro must get at least 3.2V supply from the battery. So the battery probably sags from 3.8V to 3.2V but there is still 0.5V missing! Unless the measurements are wrong, of course. \$\endgroup\$
    – bobflux
    Commented Jan 15, 2020 at 18:18
  • \$\begingroup\$ I'm not sure it is safe to assume that the microcontroller is being powered directly from the battery. The OP expected the battery voltage to be at least 3.7V, so either there is some kind of dc/dc converter between the battery and the microcontroller or the microcontroller has a separate power supply. Either way, the output voltage of the microcontroller is not indicative of the battery voltage. \$\endgroup\$
    – Elliot Alderson
    Commented Jan 16, 2020 at 1:26
  • \$\begingroup\$ @MTIbio could you give more details about your circuit to clear up the above? \$\endgroup\$
    – bobflux
    Commented Jan 16, 2020 at 10:14
  • \$\begingroup\$ @ElliotAlderson yes I am powering the microcontroller from the same battery (there is a convertor onboard). The battery did sag a bit and I noticed the digital output dropped from 3.4V to an unsteady 3.2V when the transistor is on. I rewired the circuit and it works much better. Couldn't believe the voltage drop across by bad contacts! Thanks for the help. \$\endgroup\$
    – MTIbio
    Commented Jan 16, 2020 at 15:01

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