I am trying to understand why someone would want to use a high-side MOSFET to switch on and off a load if they could simply use a low-side MOSFET. Using a low-side MOSFET requires a much lower VG compared to a high-side device.

In the context of a half-bridge driver, I can see how you would need both low- and high-side devices with a charge pump integrated to the circuitry to charge the gate of the high-side MOSFET.

But my question is: How and why would you use a high-side device for various circuits other than half-bridge?

  • 3
    \$\begingroup\$ There are lots of times when a load has to be physically connected to ground e.g a starter motor. Then a high side switch needs to be used. \$\endgroup\$ Commented Dec 14, 2016 at 21:43
  • 2
    \$\begingroup\$ Sometimes there are good reasons not to disconnect the ground connection between that device and other circuitry. \$\endgroup\$
    – user16324
    Commented Dec 14, 2016 at 21:43
  • \$\begingroup\$ same reason breakers are used on line instead of Neutral \$\endgroup\$ Commented Dec 14, 2016 at 21:57

2 Answers 2


If your load is referenced to a system ground, putting a switch in series with the low side can result in a ground potential difference proportional to load current. (Load * RDSon)

This can erode noise margin on digital signals and inject noise and error into analog circuits.

To get around this, high side switching is often used. You can use a P-Channel FET for this, or develop a supply that's high enough above the rail you are switching to fully enhance an N-Channel FET.

For the same RDSon, the N-Channel FET will be cheaper than the P.

You can use a charge pump, develop an auxiliary supply to drive the FET gate. Some integrated load switch drivers have a charge pump built in and provide other useful features.

There are also integrated high side load switches that can be useful for certain load ranges. Here's an example from Micrel:

Micrel Driver

  • 3
    \$\begingroup\$ Switching the low side can also be dangerous in high-energy systems because the live conductor is always in circuit even when the unit is switched off. Switching in the high side ensures power is removed when the unit is switched off. \$\endgroup\$
    – vofa
    Commented Dec 14, 2016 at 22:14

Many reasons, really:

  1. If your load is exposed, it may be a shock hazard even if the controller is supposedly off, if the rail is on. This may also may be considered when analyzing safety features and risks of the circuit, specially in case of circuit failure.

  2. If the load is grounded, then you won't be able to control it with a low-side driver, because the current will flow independent of the switch state.

  3. Even with low RGSon, there will be a voltage in the MOSFET, which my invalidate zero-potential-in-the-load (for safety?) requirements, if you have one.

  4. If you have a resistor as current sensor, even if a parasitic resistance one, it is easier to read it if shares ground with the controller. If you have a low-side switch, if you place it below the sensor, the sensor will need a differential circuit to read. If you place the switch above the sensor, the load current will affect the MOSFET Vgs.

  5. You don't need a charge pump if you use a P-Channel MOSFET in the high side. PMOS devices are usually more expensive than their NMOS equivalents, but if your power requirements are low enough (and a few amps are "low enough"), the cost of a PMOS plus its driver (which may be a single, smaller transistor and one or two resistors) are less than the cost of a NMOS plus its driver (which might include a charge pump). Consider that the cost also includes complexity and board space, which will most probably be higher for the NMOS in the high-side case.

  6. Besides the half-bridge, the full-bridge (or H-bridge) also requires a high-side device.

  7. If your load is not self-contained (like a motor, LED, etc), but a more complex one (like any 3-port or higher device), disconnecting the low side will not only not disconnect the other circuits, but will cause current to flow in unexpected situations. In the worst case, it may destroy your load. For example, disconnecting a microcontroller ground will cause current to flow from its Vcc to its I/O ports, and the ground will float, which may cause complementary digital ports to activate both sides at once, overloading the circuits.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.