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I am supposed to build a boost converter that steps-up a 1.5 V input to 5 V (I only have 1.5 V as a power source). I have designed the circuit but the only problem I face right now is that I can't turn the MOSFET on with 1.5 V peak-max. I am using a B012N03LX3 MOSFET as my switch.

This is my schematic connection (assume the diode is ideal and assume that the NE555 can operate with 1.5 V which the simulation shows it can)

enter image description here

I checked the datasheet and I know that the Vgs(th) must be at least 1 V to turn the MOSFET on.

enter image description here

The NE555 has the output of this 10 kHz with 0.7 duty cycle.

enter image description here

But the voltage output at Vload is:

enter image description here

So from this I am pretty sure that the MOSFET did not get sufficient voltage at Vgs(th) to turn on. But due to the fact that I only have 1.5 V as voltage supply I am not sure what correction I need to make to my circuit to make sure that the MOSFET is able to turn on properly (with only 1.5 V as a voltage source).

BTW I think the output should look like this if the MOSFET is able to turn on properly:

enter image description here

So any idea on what I should change in my circuit to make it work?

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    \$\begingroup\$ @Andyaka ya its the waveform when I connect to the MOSFET gate, I think all I have to the to fix it is to fix the waveform for the current (not sure if I am right,correct me if I wrong) but I have no idea how to solve it tbh \$\endgroup\$
    – tw.C4444
    Jun 14, 2020 at 19:54
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    \$\begingroup\$ VCC=1.5V will not power the 555 correctly. Also as you can see, it's not enough to turn on the MOSFET properly. You need a higher voltage (at relatively low current) to run the 555 AND to drive the MOSFET gate properly. Powerthe 555 off at least 5V and come back to this when the rest is working. \$\endgroup\$
    – user16324
    Jun 15, 2020 at 11:49
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    \$\begingroup\$ No, 1V is where the MOSFET barely begins to turn on. Look at its datasheet : the Rds((on) is specified at Vgs = 4.5V (and at 10V) and that is what you should aim to drive it with. And a 555 won't work at 1.5V (check its datasheet) - though your simulation model might. \$\endgroup\$
    – user16324
    Jun 15, 2020 at 13:52
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    \$\begingroup\$ You say your duty-cycle is "70%" but your waveform is about 30%. As Brian Drummand said, according to your very own datasheet the turn-on is MINIMAL 1V (and only 250 uA at that) but is MAXIMAL 2.2V i.e. the manufacturer is telling you "in black and white" that your MOSFET might remain 100% OFF until as high as 2.2V so your design is basically D.O.A the way you've conceived it. \$\endgroup\$
    – Atomique
    Jun 15, 2020 at 14:16
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    \$\begingroup\$ You could rig up 2 transistors as an astable multivibrator. But replacing the 555 won't solve the gate drive voltage problem. \$\endgroup\$
    – user16324
    Jun 15, 2020 at 15:10

4 Answers 4

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Also I check the datasheet and I know that the Vgs(th) must be at least 1V to turn the MOSFET on

No, you are misreading the data sheet. What it says is that, in order to get 250 uA drain current, some FETs will only require 1 volt, but some will need 2.2 volts.

You are using 1.5 volts as a supply voltage, so you would have to get quite lucky in picking your FET to get one which will work in your circuit. It's even worse than you think since you want much more than 250 uA, which means you need an even higher gate voltage.

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Vgs(th) it's just to start driving, to really serve as a switch you need VGS = 4.5 V (see rds(on))

You must try other circuits as follow

boost converter 1.5 volts input 5 volts output 200 mA, boost converter 1.5 volts input 5 volts output 10 mA, boost converter 1.5 volts input 5 volts output 500 mA, boost converter 1.5 volts input 5 volts output 300 mA

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Question: "MOSFET did not get sufficient voltage at Vgs(th) to turn on..what correction"
Answer: Since there is an oscillator, make a charge pump (voltage double or triple). Then switch that voltage on the FET gate.

One thing you may try, just to see it on the simulator is:
Shift the operating point, by adding two resistors and a capacitor
1.5V--Ra--(X)--Rb--GND, (X)--FET gate, (X)--Cx--555 output.

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If you ever manage to get it started, add oring diodes from 1.5vdc/output to vcc. Keep in mind this will make your start-up challenges more difficult but your gate drive will be much stronger once its operational. Remember to use shottky diodes.

Edit: you could use a transiator (add a series resistor with the base). You only need .7v to drive a 2222a.

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