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enter image description here

I would like to do something like this application note: http://www.onsemi.com/pub/Collateral/AND9093-D.PDF

And I'm using this transistor: https://www.fairchildsemi.com/datasheets/FD/FDS5670.pdf

For me Vin = 20 V, and I apply 5 V on the gate. Load = 22k

But it's like the MOS is always open. That's I don't understand, Vgs > Vth, so why is it always open? And what I should do for make it work?

Thanks edit: Add schematic, sorry I forgot it

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  • \$\begingroup\$ Where is the schematic of your circuit? \$\endgroup\$ – Andy aka May 4 '18 at 12:41
  • \$\begingroup\$ I edited it. Sorry I forgot it \$\endgroup\$ – Tack May 4 '18 at 12:45
  • \$\begingroup\$ That needs to be a low side switch not a source follower. \$\endgroup\$ – Brian Drummond May 4 '18 at 12:50
  • \$\begingroup\$ You're using an NMOS, in the "high side" configuration as in your schematic, the NMOS will only will act as a low value resistor when it is in triode mode. For that the gate voltage needs to be higher than the drain voltage. Make V1 25 V and the NMOS fully conduct. R2 of 1 milli ohm is rather pointless, you should remove it or give it a proper value. \$\endgroup\$ – Bimpelrekkie May 4 '18 at 13:00
  • \$\begingroup\$ I'm new in the design, what is the difference between low side switch and source follower? They look like similar. I don't think I use in high side, because mu load is on the source. Am I wrong? The 1 mOhms is a "sensor" resistor. I wanted a switch, something with the smallest Vds. \$\endgroup\$ – Tack May 4 '18 at 13:21
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In order to adequately turn the MOSFET on there needs to be "several" volts between gate and source. If you apply 5 volts to the gate and the gate-source threshold voltage is exceeded you will begin to see some voltage at the source but you won't see 20 volts because this is a source follower circuit.

A better circuit is to put your 22 kohm resistor in series with the drain up to 20 volts and have the source connected to ground/0 volts. This is called a common-source configuration and you will be able to switch virtually the full rail across the load.

Alternatively, if you need a ground connected load you would use a PMOS FET: -

enter image description here

R3 stops the full 20 volts being applied across GS of the PMOS FET because that might otherwise be close to the limit for a typical device.

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  • \$\begingroup\$ Thank you for your verry good answer. If get it, every time I want to use a MOSFET as control (vd = vs if vg is ON), I should give a Vg close from Vd. Looking at Id(Vgs), I understand that. So I couldn't control a MOSFET with 3.3 V (if Vd is higher than 3.3V)? What is the need of Q1? \$\endgroup\$ – Tack May 4 '18 at 13:26
  • \$\begingroup\$ Q1 is to allow a low voltage from typically an MCU to switch Q1 and that does the proper level shifting to control the MOSFET. BTW I've added an extra resistor... \$\endgroup\$ – Andy aka May 4 '18 at 13:28
  • \$\begingroup\$ Ah, yes, Q1 has the 20 V on collector, it can do 0V to 20 V. Thanks \$\endgroup\$ – Tack May 4 '18 at 15:35

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