Is the following N-FET circuit sufficient to act as a power switch for an IC?

Question

I'm trying to build a circuit which will turn an IC on and off using a MCU signal voltage of 3.3V. This is for a low-power circuit, so every coulomb is critical.

I selected the BSS816NW N-type MOSFET for the job. One can see that Vgs minimum is 1.8V for this FET, which worries me with the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

One can see that as the chip saturates in voltage, the FET will increase in resistance, which is problematic, as there will always be a voltage drop between the gate and source, which may cause brown-outs for U1.

How can I use simple FETS or BJTs to improve this low-power toggle for U1? There are no other voltage rails besides ground and 3.3V pos.

Answer 1

One way to do this (with a PMOS):

^{simulate this circuit – Schematic created using CircuitLab}

You have to make sure that the PMOS can be properly turned on with a Vsg of 3.3V (i.e look for a logic level PMOS and check the datasheet). This one could work: DMP2035U-7—rated down to 2.5V for turn-on. Since you care about low power, you can increase the value of the 10k pullup, to something like 47k or 100k.

Notice that if the control signal is either 3.3V or left floating, the PMOS will be in the cutoff region and therefore acts as an open switch. If the control signal goes low (0V), then the PMOS will conduct the current required by the load.

Using an NMOS, as you show in your post, would make things a lot more difficult since you'd need to provide a voltage higher than 3.3V at the gate of the transistor—which you probably don't have (google high-side switching with an NMOS or NPN). You can avoid all that by using a P-channel MOSFET.

Answer 2

Try a PMOS FET and pull the gate to gnd.