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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: Rather than using a simple resistor divider to provide the necessary gate-source voltage to turn on a P-channel load switch, you can use a BJT current sink, which has the added benefit of providing a constant Vgs over a wide range of input voltages.

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.

As highlighted in the below screenshot of it’s datasheet, during start-up it only supplies an average of 28mA to the load, even though the part is designed to pass up to 1.5A during normal operation1. If $$V_{OUT}$$ doesn’t climb to equal $$V_{IN}$$ within 250ms during start-up, it times out. So any significant resistive load on the output that drew more than 28mA but less than 1.5A would always cause this IC time out during start-up. This suggests that it is designed to work in tandem with an external high-side switch that is placed between this circuit breaker IC and the load. If the high-side switch is kept off whilst the circuit breaker IC start-up, it will only have capacitance to charge up on it’s output, hence the 28mA will be ok (up to a max. capacitance, and explain how to calculate this in the datasheet).