how to design circuit of high side MOSFET switching with bootstraping and mosfet driver ic?

Question

Hey im starting to learn about MOSFETs at my university classes, but Im having trouble trying to design a circuit for a buck boost converter for a MPPT solar charge controller. The part pushing me down is the high side switching, I kinda understand the boostrap operation with the capacitor and the diode, but dont know how to integrated with my MOSFET driver ( TC4420 ). I know that boostraping is needed because I will be using an N channel MOSFET, but I dont have a clue how the circuit might look I know is probably really simple but im kinda stuck. If someone can show me how the circuit can be I would really appreciate. Im using TC4420 mosfet ic.

Answer 1

The simplest would be to use an intergated driver with bootstrap:

http://www.farnell.com/datasheets/9094.pdf

https://www.infineon.com/dgdl/ir2110.pdf?fileId=5546d462533600a4015355c80333167e

This is a very common problem, so you will find lots of chips on the market to do this. The search terms to use are "half-bridge driver" (because a pair of FETs, one high and one low side make one half of a H-bridge). A H-bridge driver would also work, but chips intended for this application will usually include extra features like current monitoring that you probably don't need here.

If you're more interested in learning how it works, the datasheets should provide useful information.

There is a gotcha: some of these drivers are unable to start if the bootstrap cap is not charged, which means they will not start if the output voltage is present. If you want to make a buck converter to charge a battery then you need a driver which is able to turn the low side FET ON which then charges the bootstrap cap and allows it to start, even if output voltage is present.

Answer 2

Instead of bootstrap circuit you can use isolated dc dc converter for highside mosfet.You wont get 100% duty cycle if you are using bootstrap circuit.The design process will become easy if you can use DC-DC converter for high side power supply

Answer 3

A bootstrap circuit is just a capacitive voltage doubler in disguise. The current requirements are rather minor, so this type of DC-DC converter is all that is needed.

1. You start from a low-impedance AC/switching signal (in a bootstrap this would be the output node of the driver).
2. You attach a relatively small capacitor to it (no need to load the signal too much)
3. You connect a diode from the power supply to the capacitor.
4. You connect another diode from the capacitor to another similar or larger capacitor (the other node of that capacitor can be gnd or vdd). This capacitor holds the bootstrap voltage.

(Sorry, I have no access to the schematic editor right now).

Of course, the actual capacitor sizes would depend on switching frequency and current consumption of the gate drive circuitry. The charge transfer has to be enough to account for the current consumed (VppCf> Igate)

Once you see this idea, I hope you would realize that the bootstrap voltage is only needed when the NFET gate has to be turned on, so with careful design of the gate drive the second diode and capacitor can be removed altogether.