# Choosing the correct Gate Driver for a MOSFET in a DC-to-DC Converter

I am building my first simple buck converter, stepping down 12 V to 5 V and driving a small load of 5 Ω - 10 Ω. Something like the following:

or:

The above images both show an N-channel MOSFET. I am confused as to what would be the best option to drive the N-channel MOSFET at a 10 kHz switching frequency, keeping in mind the above parameters.

I have read here, that for an N-channel MOSFET a special drive circuit may be required to turn the transistor ON, so a P-channel MOSFET may be easier to implement.

As for the gate-driver I have seen plenty, and a popular, cost-effective circuit seems to incorporate a non-inverting totem-pole driver as shown.

I was thinking whether the above circuit would be good in my case, and simply give the PWM pulse at Rb.

- What would be a suitable gate driving IC for an N-channel MOSFET, and why?

- IF a P-channel MOSFET is more suited for this application, then I would go for the least 'complex' solution and use a P-channel MOSFET. What would be a suitable P-channel drive circuit/IC?

I have never designed something like this before, so please keep this in mind :) Any tips and/or suggestions would be appreciated!

Edit: Just for clarification.

1. Cost is not an issue in this case.
2. Simpler design (fewer parts) is better.
• I would start here The frequency is rather low and a 1n4007 is a poor choice, too slow. Use a simulator to check your circuit. – lakeweb Mar 12 '18 at 16:32
• Notice that the second diagram has the MOSFET connected on the low side of the load, where your first diagram has the FET on the high side of the load. To get good efficiency with the FET on the high side of the load, you'll need to be able to apply a gate voltage higher than VIN. – The Photon Mar 12 '18 at 16:33
• Yes, you can look for a high-side gate-driver IC. I don't know exactly what is available since I mainly work with integrated controller (or integrated regulator) ICs for this kind of application. – The Photon Mar 12 '18 at 16:40
• @WarrenHill, I assume that not any 'high side' drive circuit IC would work equally well. I am not looking for any shopping suggestions, simply if the one I pointed out would be suitable, and if not what should I look for in a 'high side' drive IC? – Rrz0 Mar 12 '18 at 19:26
• I will be using a fixed load, so no need for feedback circuitry. Also, I am applying a 10 kHz signal from a signal generator. As I said above, I am not asking for any shopping suggestions, simply, if an N-channel or P-channel FET would be optimal in this case and what type of driver would be preferred. (Bootstrap driver, synchronous driver, etc) – Rrz0 Mar 16 '18 at 5:28

I'd simply use a driver like ADP3120 or one of its cousins. This one drives 2 MOSFETs in synchronous rectification mode, which will be a bit more efficient than using a FET and a diode. Considering the low price of this driver, there is really no reason to build a complex circuit for this.

If you want to keep the diode, there are also high-side driver chips.

• Many thanks for your answer and suggestion. I am failing to understand why you say that I may remove the diode when using the driver you suggested. As far as I know in a buck converter, during the charge phase, the diode prevents the capacitor from discharging, while during the discharge phase, the inductor current continues to flow into the rectification diode to charge the output. Doesn't this mean that the diode is a necessary component for the correct functioning of this converter? I may be missing your point completely, however, and going off on a tangent. – Rrz0 Mar 14 '18 at 18:08
• The chip I quoted is drives 2 FETs so the diode is replaced by a FET which has the same role as the diode but less voltage drop, hence less losses. I'm sure you could find high-side only drivers, but I cant' quote one off the top of my head. – peufeu Mar 14 '18 at 18:10
• Good! You can use the digikey search engine, "FET driver" category, select "high-side driver" for FET+diode or "high and low side" for 2 FETs. – peufeu Mar 14 '18 at 18:12
• I found the above suggestion you made extremely helpful. I have decided to go ahead with this kind of driver. I would like to confirm that an external bootstrap circuit needs to be built in conjunction with the above driver, to drive the high-side MOSFET. Can such modifications be found internally in similar drivers or would it be easier to simply a\dd these components externally? Thanks – Rrz0 Mar 23 '18 at 9:39
• It says a bit further "An external diode is required" so I presume the "internal diode" text is a mistake... when I used this driver I added the diode. – peufeu Mar 23 '18 at 11:17

There are many possible solutions. What is the best way to travel from London to Paris? By car, train or plane? They all work but the cost and performance will be different. The same applies to your problem. If you are set on a high side switch, there are a number of different options. I will outline a few common approaches:

1. Use an isolated DC-DC power supply plus a standard low side gate driver, e.g. MCP1416, whose reference point is the source of the MOSFET. This solution is expensive and has a high parts count.

2. Use a pulse transformer. Simple. The duty cycle here is limited to < 50%.

3. Use a bootstrapped high-side buck driver or a half bridge driver, e.g. LM5109. This is one of the simplest means of driving a high side switch. The following diagram illustrates this approach. You can read more here. Just be aware that you need to charge the bootstrap capacitor at startup.

For a buck converter, you can also shift the MOSFET to the ground rail. This is known as a low side buck. With a low-side buck you can use a standard low side MOSFET driver, e.g. MCP1416, which is very simple. However, measuring the output voltage becomes more complicated.

For the low power application your describing (~5W output power) and your demand for a simple design i would recommend just using a simple buck controller IC with integrated switch (some even have a integrated inductor/diode aswell). The only external components would be input and output caps, (diode and inductor) and in some cases (if output voltage is adjustable) a voltage divider in the feedback path. These ICs normally have comprehensive application notes which makes the choice of external componens more or less foolproof.