# N-MOSFET switching problem in LTspice

In this circuit as per theory it should switch. But it is not switching. Why it is happening.

• You won't see any change in drain voltage, because the drain is directly connected to a voltage source. What do you expect to see? To see something you need to add a load resistance.
– G36
Commented Aug 30 at 4:42

When you switched the MOSFET M1 on, the circuit will turn into this:

simulate this circuit – Schematic created using CircuitLab

Why it is happening.

Assuming V1 has zero output impedance (i.e. ideal source), it's normal you see 48V at the Drain_Voltage node.

If you add a load (e.g. resistor) between the Drain_Voltage node and the V1 + node then the Drain_Voltage node will be the output of a voltage divider:

simulate this circuit

So if the on-resistance of M1 is much lower than RL then you'll see a very low voltage (close to zero) at the Drain_Voltage node when the NMOS M1 is switched on.

I don't know what you are trying to achieve in your simulation but if you are trying to simulate shorting a voltage source then you do need to change the output resistance value of the V1 to a non-zero value because real sources have non-zero output impedance.

You can't change the potential of the ground node. In the circuit below $$\V_A = 0\$$, and there's nothing you can do to change that:

simulate this circuit – Schematic created using CircuitLab

You can't change the potential difference between two ends of a voltage source. So if you hold one end of voltage source V1 at 0V, the other end must be $$\V_B = 0V + V_1 = +12V\$$, and you can't change $$\V_B\$$ either, whatever you do.

For this reason potential $$\V_C\$$ of node C is undefined because you have two voltages sources, V2 and V3, fighting to set potential there. The simulator will fail with an error, because this situation can't be resolved. It's like trying to join both ends of two rigid rods of different lengths together; the rods are too "stiff" to permit that.

By this same reasoning, $$\V_D=+12V\$$ and $$\V_F=+24V\$$, unconditionally, regardless of what else is connected between them, but if you do connect something between them it must have non-zero resistance. If there were only a switch between D and F, then this condition would be violated when switch SW1 is closed.

R1 provides some "elasticity", so that nothing bad happens when SW1 is closed, and so that potential $$\V_E\$$ can have some value in the range $$\V_D\$$ to $$\V_F\$$. With SW1 open, $$\V_E=V_D\$$, and when it's closed $$\V_E=V_F\$$.

Think of the transistor in your circuit as a switch, open or closed (or somewhere in between those extremes). In your circuit that MOSFET is connected directly between 0V and $$\V_{DRAIN\_VOLTAGE}=+48V\$$. Neither of those nodes can have any other potential than 0V and +48V respectively, as I explained above, so it is unreasonable for you to expect $$\V_{DRAIN\_VOLTAGE}\$$ to ever change. If you want it to change, then provide "elasticity" with something that has resistance, like a motor, a lamp, or a resistor, R6 here:

simulate this circuit