#Log 2: MOSFET bootstrapping

A project log for YATSS (Yet another T12 soldering station)

An arduino-based temperature controlled soldering station for the ubiquitous T12 clones.

JoelJoel 02/08/2019 at 12:580 Comments

As discussed in the previous log, I wanted to build this project with components at hand. This, combined with issues in amplification meant I would have to use a N-channel MOSFET as a high-side switch. I knew that this wasn't the common application for N-channel MOSFETS, but I never understood why. Here is what I learned: 

First, let's look at the problem. A MOSFET is controlled by the electric charge on its gate, in respect to the source. Basically, for a N-channel MOSFET, you need your gate to be at a higher potential than the source (a positive Vgs) for the transistor to turn on. In my case, the Vgs required to turn on the transistor is about 4.5V, that is; the gate needs to be 4.5V about the source. 

This is the reason that N-channel MOSFETS are generally used as a low-side switch. Any (positive) voltage at the gate will be at higher potential than the source, because the source is connected to ground. This is not the case when in the high-side configuration. Let's take a look:

If you are like me, and have a limited intuition about these things, this configuration looks like it would work fine, When the switch is closed, the gate will be at 5V, and the source is at 0V, so what's the problem? 

When the the switch is closed, the transistor does turn on, however, the voltage across the resistor cannot rise above Vgate - Vgs_threshhold (5V - 4.5V). This results in the load having 0.5V across it, and we get 0.5V / 7ohm =  71mA through our load. Far from enough to drive a soldering iron.

What if we use the supply rail to drive the MOSFET? Let's look at the following configuration: 

Here, we use a NPN BJT to drive the gate of the MOSFET. When the BJT is ON, the gate is pulled low, and the MOSFET is off. When the BJT is OFF, the gate is pulled to the supply rail through the 10k resistor.

Great! Now we have 24V at the gate, surely that solves the problem! But no, this makes our problem worse.

When the gate is pulled to 24V, the MOSFET turns on and passes current through to the load. But once again, the voltage across the load can't rise above Vgate - Vgs_threshhold (24V - 4.5V = 19.5V). This results in 2.8A through the load. Great, but wait! What about those 4.5V that are missing across the load? They don't magically disappear. Instead, they are dropped over the MOSFET. 4.5V * 2.8A is over 12W. That's way too much. Not only would the MOSFET heat up far to much, but we also lose 15% of our theoretical max power of 82W.

Here's what I did:

We can use the capacitor and diode on the left side of the diagram to "capture" the potential difference over the transistor when the transistor is off. The capacitor charges through the diode, until there is 24V across it (minus the forward voltage of the diode). The NPN BJT is initially on, which means the gate of the MOSFET is pulled low and will not conduct. When the BJT is turned of, the voltage across the capacitor is transferred to the gate of the MOSFET, turning it on. This is where the magic happens. 

As the potential across the load rises, the potential at the negative side of the capacitor rises. The cap can not discharge (bar reverse leakage current of the diode and leakage current through the gate of the MOSFET), so it will maintain 24V volts across it. This means that if there is 24V across the load, we have 24V+24V (ideally) at the MOSFET gate! Perfect, now we have produced a Vgs of 24V, plenty to keep the MOSFET fully on. (Careful though, Vgs has an upper limit, and overloading it can damage your MOSFET).

I copied this solution straight from this video by Julian Ilett and this one by 0033mer.

In conclusion: There are pros and cons with using this configuration for high-side switching.