I'm trying to get my head around the circuit theory behind the high side NMOS switch, here is the shematics:

enter image description here

My question is not about whether this is a good or a bad way to make a high side switch (I know it is bad), but really how to compute the value of the voltage across the resistor for a given transistor and resistor value based on voltage on the gate ? What data from the transistor's datasheet is used there? What are the equation that allow to make the calculations (like in the figure for example)?

Thanks :)

  • \$\begingroup\$ You need a high side gate driver. The gate must be high enough above the source pin to turn on, not ground. The MOSFET doesn't care or know about the voltage at ground. If the MOSFET is on, it is usually neglible resistance compared to the load so you just assume 0 ohms. If you must know use the Vds vs I vs Vgs curves in the datasheet \$\endgroup\$ – DKNguyen Apr 18 '19 at 15:14
  • 2
    \$\begingroup\$ "(which I know it is)" -- Actually, no. If your intent is to switch power to a load of some kind with minimal power loss in the transistor, this is a very poor circuit. For most MOSFETs, the gate wants to be at least 5 V above the source (not GND) for low Rdson. Putting the load in the source circuit makes this more of a constant-current circuit, the opposite of a voltage switch. \$\endgroup\$ – AnalogKid Apr 18 '19 at 15:31
  • \$\begingroup\$ @AnalogKid *putting a resistor in the source circuit and running off a ground referenced gate voltage makes it more of a current source \$\endgroup\$ – DKNguyen Apr 18 '19 at 16:09
  • \$\begingroup\$ @AnalogKid my sentence meant "I know it is a bad way to make a switch", sorry for confusion \$\endgroup\$ – JeanMi Apr 19 '19 at 8:01
  • \$\begingroup\$ @Toor The hard part is actually how to compute Vgs based on R, the transistor curves and Vgs+Vresistor (the voltage of 3.5V), because you cannot just use 3.5V as Vgs. \$\endgroup\$ – JeanMi Apr 25 '19 at 9:04

You call this a switch, so the first thing to know is that to turn on a FET, Vgs should be greater than the threshold voltage.

If it's a good switch the source will be close to 5V when on. Therefore the gate voltage should be >Vth ABOVE 5V. In your picture the gate is at 3.5V, so you DO NOT have a good switch.

There are a couple of ways to determine the voltage across the resistor. You can use the FET equation with your resistor load (Info from here): enter image description here

Or you can do it graphically with a load line on the output characteristic of your FET. Your VDS is $$VDS = 5-Id*R$$ so plot that on the characteristic curve and see where the intersection point is.

Here's an example from Wikipedia for a bipolar transistor but the technique applies to FETs as well:

enter image description here

You could also simulate or use numerical methods or a MATHCAD or MATLAB model to figure it out.

If you are actually using the FET as a switch, then you can get a very good approximation by looking for the RDSon on the datasheet for the Vgs you are applying, and just considering the voltage divider formed by the on resistance and the load resistance.

  • \$\begingroup\$ Yeah, plotting the characteristic is ok, point is, when you do have it, how do you place the point ? You don't have access to Vgs directly. Here the 3.5V are the voltage Vgs+Vresistor, and Vresistor depends on Id, so ? \$\endgroup\$ – JeanMi Apr 25 '19 at 8:51
  • \$\begingroup\$ You know the solution has to be at the intersection of the load line and the output characteristic, so if you have to you can make a guess and iterate until you approach the solution. \$\endgroup\$ – John D Apr 25 '19 at 19:57

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