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I have a project I am trying to put together. My understanding is lacking though. I would really appreciate some help in helping me to understand what I am missing!

The circuit would be a a logic level mosfet that would be controlled by a microcontroller. So 3.3v at the gate. It will be set in between the battery and the ignition wiring harness. Essentially, i would be breaking one wire in the harness. Which connect to the steering column, and inserting the mosfet/circuit. so when the car is started if the correct pin is entered, then the mosfet would complete the wiring harness circuit and allow the 12v to pass through.

To my understanding so far, I would need a logic level mosfet that has a low level rds(on) that is applied when the VGS is 3.3 or lower.

If the above were true, and all I had to do was apply a small 3.3v from the MCU and that would allow 12v to pass through the drain/source??

It seems to simple! I feel like there is more and like I am missing something. Could and would someone either point me in the right direction or help break this down, correct me, and or explain to give me a better understanding. I would very much appreciate it!

EDIT: I am just trying to get a better understanding of how these things work! If I apply the MCU's limited 3.3v to the gate, then that would allow 12v to pass from drain to source?? Do I need to accommodate for anything else? Anything like leakage voltage, kickback voltage, etc?

Schematic:

https://goo.gl/photos/sKPDi2RFykXcvL8X7

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    \$\begingroup\$ Hint: use more than one MOSFET - a nice power MOSFET as the main switch, and a second logic level one to drive its gate. \$\endgroup\$ – Tom Carpenter Dec 29 '16 at 19:28
  • \$\begingroup\$ Couldn't i just use a BJT as the driver or would another mosfet be more suitable in this application? Also, how far off is my understanding? Only the proper voltage needs to be applied to the gate in order for the mosfet to be able to allow current to pass from drain to source? We want a low RDS(on)!? The lower the better. Is it really as simple as that? I know BJTs are much more complicated than that! You need to saturate the base, accomodate for b/e leakage, integrate a spdt relay with a flywheel diode, etc. \$\endgroup\$ – Aguevara Dec 29 '16 at 19:41
  • \$\begingroup\$ Please provide a schematic. That will give everyone something to build from. \$\endgroup\$ – Daniel Dec 29 '16 at 19:44
  • \$\begingroup\$ You can use anything to drive the MOSFET gate. A BJT would work fine. \$\endgroup\$ – Tom Carpenter Dec 29 '16 at 19:48
  • \$\begingroup\$ @DanielGiesbrecht goo.gl/photos/sKPDi2RFykXcvL8X7 \$\endgroup\$ – Aguevara Dec 29 '16 at 20:16
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You're on the right track, and a MOSFET could be used.

The gate-to-source voltage (Vgs) must be greater than a threshold specified by the datasheet to conduct from Drain-to-Source (assuming you are using a N-Channel MOSFET). This threshold will be listed as Vgs(th).

Below, the N-Channel MOSFET is shown to be controlling the high-side of R (your load, in this case the 'steering wheel'). This is based on your drawing, which shows a N-Channel MOSFET, with current flowing to the steering column, Drain-to-Source.

schematic

simulate this circuit – Schematic created using CircuitLab

This could work, but the voltage applied to the gate would have to exceed the threshold voltage added to the voltage drop across R:

Vg > Vgs(th) + I(R)

In your case, Vg is supplied by the Arduino. Having a N-Channel MOSFET on the high side will likely not work.

Solution:

While you could use a P-Channel MOSFET on the high side, it is much easier to use a N-Channel MOSFET on the low side, shown below:

schematic

simulate this circuit

Your required gate voltage formula now becomes:

Vg > Vgs(th)

Alternatively, you could also use a solid state relay.

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  • \$\begingroup\$ the only difference I see in this circuit is the resistor. Is this what you call a pulldown resistor? Why would changing the position of the resistor change anything? One placement uses high-sdie switching while the other, low-side? \$\endgroup\$ – Aguevara Dec 29 '16 at 21:15
  • \$\begingroup\$ No, this is not a pull-down resistor. The resistor symbol represents the steering wheel which you are powering. It is not a physical resistor. When I say 'high-side', I mean that you are using the MOSFET to control the POSITIVE connection to your steering wheel. 'Low-side' means the NEGATIVE side of your steering wheel. The choice of high or low position has to due with the voltage drop described above. \$\endgroup\$ – Daniel Dec 29 '16 at 21:18
  • \$\begingroup\$ Oh OK. Thank you! Should I use a smaller logic MOSFet to drive a larger one or can I just use a regular logic level MOSFet? Also that IRF530 has a RDS(on) of 10V. That doesn't seem like a logic level MOSFet. \$\endgroup\$ – Aguevara Dec 29 '16 at 21:24
  • \$\begingroup\$ It's like putting a switch to interrupt the positive connection or negative connection to your steering wheel. They both achieve the same thing, but because of the way MOSFETs work, you end up with different voltage requirements for the gate based on which place you choose. \$\endgroup\$ – Daniel Dec 29 '16 at 21:24
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    \$\begingroup\$ @Aguevara - "yeah I know what RDS(on) is" - not if you think it can be 10 volts. However, your idea (that the IRF530 gate-source voltage is too high for the 530 to be a logic-level FET) is entirely correct. The built-in schematic editor which Daniel used to create the schematics defaults to the IRF530, and Daniel mistakenly thought that you would realize that his circuit is not intended as a "real" circuit, but that you would realize that he was just making a general point about circuit topology. \$\endgroup\$ – WhatRoughBeast Dec 30 '16 at 0:12

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