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I have designed a circuit that will switch on a 12V siren through a IRF540N MOSFET. The siren positive is connected to ST-3 (12V), while the negative is connected to ST-4 (SIREN-12V). A flyback diode is used for protection.

The IRF540N has a gate threshold voltage of 2V to 4V. I am using an ESP32 to turn the siren on and off via the MOSFET. The output current of IO 13 of the ESP32 is 33mA and the voltage is around 3.2V, so this should be enough to turn the MOSFET on. However, this 12V siren when activated sounds very quiet.

When I apply the 3V3 voltage from the ESP32 to the gate of the MOSFET, the siren then sounds as expected (loud).

Why is there is a difference in the loudness of the siren? The MOSFET should turn on when GPIO 13 is high but it sounds like the is only turning on the gate slightly.

enter image description here

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    \$\begingroup\$ According to the datasheet that MOSFET just barely reaches it's Miller plateau at 4V Vgs. \$\endgroup\$ Aug 1 at 16:41
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    \$\begingroup\$ schematic for connection looks right. still worth double-checking you have mosfet connected to correct leg. Another thing to be aware of is that current supplied is proportional to current through gate - ditch the 10K pulldown (possibly for a 500k) and try again. so close to the threshold, every piece counts. \$\endgroup\$
    – Abel
    Aug 1 at 16:45
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    \$\begingroup\$ @Abel - there is essentially zero o current through the gate. It is controlled by gate voltage. Your idea of increasing the pulldown value is good though. \$\endgroup\$ Aug 1 at 16:53
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    \$\begingroup\$ if 3.3V is reduced to 3.2V over drawing current through mystery load and 3.2V to ground via 10k pull (gate current neglected). If mystery load is resistor (bad assumption but rolling with it), it's 312.5 Ohms. To lose .1% (.0033V) over mystery load and 99.9% over pull, pull=3.3*.999/(.0033/312.5) =~300k. add fudge to 500k. Or as my gut said: if it loses 1/30 on current through 10k, x50 to get it below 1/1000. \$\endgroup\$
    – Abel
    Aug 1 at 18:20
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    \$\begingroup\$ in truth we don't know how much you would have to increase it. start with infinity (removal) and see if it works first. if it doesnt, no resistor increase will help you. \$\endgroup\$
    – Abel
    Aug 1 at 18:21
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The gate threshold voltage is not that at which the device is fully turned on - it means that it is just starting to conduct (typically 250uA). enter image description here

Usually there is a chart showing the drain current versus gate voltage - in this device you can see that at 3.3V a typical device is only just starting to conduct. There will also be significant differences between individual devices, one might start at 2V and another at 4V. It will also change with temperature.

In order to pass a high current you usually need 10V for many MOSFETs or 5V for those referred to as Logic Level MOSFETs.

IRF540 transfer characteristics

You don't say how much current the siren requires. If it requires more than a few hundred millAmps you will need a level shifter to create a logic signal of at least 5V and a suitable MOSFET.

For currents up to a few hundred millAmps you may be able to find a low gate threshold MOSFET that will function with 3.3V drive, or use a bipolar transistor to drive the siren.

IRF540 Datasheet

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  • \$\begingroup\$ But when I connect the 3V3 directly to the gate input from the EPS32 3V3 (pin1) the siren emits a far louder (normal) sound. Just confused why the pin1 works but not IO13 \$\endgroup\$
    – Joey
    Aug 1 at 16:49
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    \$\begingroup\$ Measure the voltage when driven from the ESP32 and when connected directly to 3.3V. The transfer curve is very steep at that point and even a few 10's of millivolts will make a difference. \$\endgroup\$ Aug 1 at 16:52
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    \$\begingroup\$ @Joey This datasheet for ESP32 lists the minimum high level output voltage as 0.8 * VCC. When the MOSFET is barely conducting small changes in Vgs can have large effect. \$\endgroup\$ Aug 1 at 16:55
  • \$\begingroup\$ I see now. I never thought these mosfets are that sensitive. Gees, this makes a lot of sense now. Thank a mil! \$\endgroup\$
    – Joey
    Aug 1 at 17:01
  • \$\begingroup\$ Current transmitted is approximately proportional to the very tiny current drawn through the gate. Because the slope is extreme, we usually look at it as an on/off. However when it's barely conducting that you can observe the behavior - small increases to the voltage produce small increases to the gate current. Gate current was tiny to begin with so it's not negligible as a percentage, and so affects the transmitted current in a big way. \$\endgroup\$
    – Abel
    Aug 1 at 18:52

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