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## MOSFET Based

The following image shows a MOSFET based high-side switch:

### BJT Current Sink Driving P-Channel MOSFET Load Switch

A simple resistor divider can be used to provide the correct $$V_{GS}$$ to turn on a P-channel MOSFET based load switch, however that only works well if the $$V_{IN}$$ is known and stays at fixed voltage. If it doesn’t, then the resistor divider provides a varying $$V_{GS}$$, which could either turn the switch the MOSFET off at lower input voltages, or exceed $$V_{GS(max)}$$ at higher input voltages ($$V_{GS(max)} = \pm 20V$$ for most MOSFETs).

A better option in this case is to use a BJT current sink to set the desired $$V_{GS}$$ across a resistor, as shown in the following diagram:

We assume the BJT ($$Q_1$$) is switched with $$+3.3V$$ coming from a microcontroller or similar. The BJT is configured to be a simple current sink, with the current given by:

\begin{align} I_C &= \frac{V_B - 0.7V}{R_E} \\ &= \frac{3.3V - 0.7V}{2.7k\Omega} \\ &= 1mA \end{align}

This current goes through $$R_1$$, which provides the necessary $$V_{GS}$$ to turn the P-channel MOSFET ($$Q_2$$) on:

\begin{align} V_{GS} &= -I \cdot R_1 \\ &= -1mA \cdot 10k\Omega \\ &= -10V \end{align}

$$R_G$$ is added as good standard practise to limit gate current and gate voltages. In the above example, $$V_{IN}$$ can vary from approx. 11V right up to the maximum allowed drain-source or collector-emitter voltages (for example, $$48V$$), whilst keeping $$V_{GS} = -10V$$.

## IC Based

The following image shows an IC based high-side switch.

Some load-switches have reverse-polarity protection. More information of how they exactly implement reverse-protection with only the one MOSFET can be found in the The Substrate (Body) Connection section of the MOSFET page.

## Authors

### Geoffrey Hunter

Dude making stuff.