# Why won't this circuit work when the load resistor is connected to the source of the MOSFET?

I want to use a MOSFET as a switch but wanted to put the the load to the source terminal of the MOSFET. I tried putting the load between the voltage source and drain terminal which works perfectly, but this doesn't work. Why won't this work?

• If it does not work so why do you want to do that? Or more specifically, what do you mean by "it does not work", what do you expect to happen (your expectation is likely false), and what actually happens instead (which is a pure observation)? Commented Apr 6, 2023 at 10:01

That configuration is called a "source follower", or "common drain". To switch on a MOSFET, the difference between its gate potential and its source (called $$\V_{GS}\$$) must exceed some threshold, called $$\V_{GS(TH)}\$$, commonly around 3V. Note that drain potential has no effect on whether the MOSFET is "on" or "off", and in this common-drain setup, it is fixed.

In the circuit you have drawn, when you place +5V at the gate, $$\V_G=+5V\$$, the source can rise to 5V−3V=2V, but if it rises any further, $$\V_{GS}\$$ falls below that threshold, and the MOSFET begins to switch off again. What happens is that the circuit finds an equilibrium where the MOSFET is switched on just enough to raise the source potential equal to $$\V_S=V_G-V_{GS(TH)}=+2V\$$.

In other words, $$\V_S\$$ follows $$\V_G\$$ as it varies, but always $$\V_{GS}\$$ lower. Of course the source can't ever go outside the power supply potentials, so it gets "clipped" for gate potentials beyond a certain range.

By contrast, when the source is held at a fixed potential (like 0V), a configuration called "common source", it can no longer rise or fall in potential with respect to the gate, and the state "on" or "off" is more emphatic, more "digital", switching over as the $$\V_{GS}\$$ crosses that threshold.

Here are those two configurations:

simulate this circuit – Schematic created using CircuitLab

Here I sweep the gate up and down in potential (blue), and the very different behaviour of the output in each configuration (orange is common drain, tan is common source) is clear:

There are some noteworthy points:

• The output is always constrained to stay within the power supplies potentials, 0V and 12V in this case.

• The common source setup inverts. That is, when input is high, output is low, and vice versa.

• The vertical distance between blue (input) and orange (common drain output) traces is $$\V_{GS(TH)}\$$, 4V in this case. It depends on the MOSFET model.

• The common source configuration switches output state (tan) as gate potential passes through the point where it is $$\V_{GS(TH)}\$$ above the source (0V+4V=+4V here).

• Voltage gain is very high (hundreds) for the common source setup, giving rise to the sudden rise and fall of the output.

• Common drain voltage gain is 1.

It's called a source follower: the output follows the input, minus an offset, and less a little bit of gain.

A MOSFET switch requires Vgs(on), not "Vg(on)". Voltage is always a difference between two points. The MOSFET only knows what's between its terminals. Vgs means the voltage between gate and source. In this diagram, 5V is applied to the gate, but the source voltage is allowed to float (equal to the load voltage).

To make it switch, either increase the drive voltage to V2 + Vgs(on), or use a common-source PMOS switch, with the drive referenced to +12V (because again, look at where the source is connected).

For a moment let's assume the MOSFET is fully on (Obviously it's not but don't ruin this beautiful moment). So,

• the load resistor would see full 12V
• and the VGS (the voltage across gate and source, not gate and ground) of the MOSFET would be higher than its threshold (VGS-th)

And, for a moment, let's assume the latter is 0.001 VDC but still the MOSFET is fully on.

So, what would the required VGS be? From a simple Kirchhoff voltage law run, V3 should be at least 12.001 VDC:

$$\mathrm{V_3=V_{GS}+V_{1k}=0.001+12=12.001 \ VDC}$$

What voltage do you have? 5 VDC.

This means that it's impossible to have full 12V across the load with a gate voltage of 5V.

It's a source follower and, it cannot have a voltage gain of more than unity. In fact, for a MOSFET, the voltage gain will usually (with some hand-waving) be around 0.5 so, if the gate-source voltage is 5 volts, then you might see 2.5 volts at the source.

When the load is connected in the drain circuit, the gain is much, much higher.

Think about what voltage is needed between gate and source to activate a MOSFET; it might be around 2.5 volts so, with the load in the source circuit and, the gate at 5 volts, you'll only see 2.5 volts across the load because the other 2.5 volts is between gate and source.

With the load in the drain the gate-source voltage will be 5 volts and, there's a fair chance that the MOSFET will fully turn on.