I am new to electronics could any one explain to me what is the Bootstrap capacitor and what is used for I am doing research about buck convertor using MP1482 chip and I do not understand bootstrap section

  • \$\begingroup\$ It is difficult to understand for the MP1482 without realising that this chip contains a half-H bridge. If you look at some datasheets (or questions on this site) about H-bridges using N-channel MOSFETs, all will become clear. A good example is (58849). \$\endgroup\$ – Oleksandr R. Feb 17 '16 at 12:24

Your question is a good one, even if it is short.

Linguistically the word bootstrap comes from "Pulling yourself up on your bootstraps", which is of course in practise nonsense, but it does give a hint to the purpose, once it's more clear to you how capacitors work, when connected to a square wave signal.

What happens is, there is a diode in the system that forces the "top" of the bootstrap capacitor to be at least VCC, because when it is lower the diode will conduct. (Technically, the top of the capacitor will be (VCC - Vf(diode)), but in these systems they will try to make the diode forward voltage at 0.2V or below in all practical situations, so in most cases it's negligible).

From here it becomes a bit theoretical, since I don't want to keep saying "In practice, of course, with leakage and imperfect behaviour...", so it's not all as black and white, but the principles hold.

Now, when the top of that capacitor goes up above VCC, the diode will block current (it becomes reverse biased) and allow that part of the capacitor to stay at this higher voltage.

A capacitor is a voltage storing device, to put it in the simplest words, so it first tries to keep the voltage difference between its pins the same. In the same way an inductor is a current storing device, it will want the current through it to stay the same, if it can help it. But that's not so important now.

So, if you flip the bottom pin up to a higher voltage than it is at, the top will jump up by the same amount, as long as nothing in the system prevents that, of course. In the same way, if you toggle the bottom to a lower voltage, the top will jump down by the same amount.

So, now the chip connects the bottom of the capacitor to a signal that toggles between 0 and VCC, or between two voltages very quickly.

If we assume we start with the top of the capacitor at VCC and the bottom at 0V, and only the diode connected to the top, nothing else, then if the first toggle comes switching the bottom from 0V to VCC, the top will jump the same amount. Because the diode will not conduct the extra energy away, now the top jumps from VCC to 2*VCC.

Now, you suddenly have a doubled voltage to supply internal circuitry. Then there's some stuff inside of MP1482, which take that voltage and puts it into a small energy buffer, probably through another MP1482's internal diode into another capacitor, or some such system.

Then, if the bottom switches back, the top of the capacitor drops down again too. If a bit of energy was taken away, the top will be below VCC, and the diode will just fill it up again. Then at the next toggle: Boom double VCC again.

Since this pulsing 2*VCC signal is rectified in a small buffer, you have a tiny, but steady supply inside the chip that is quite a bit above VCC.

Now the N-MOSFETs come in: An N-MOSFET is easier to get to a lower Rds(on) than a similar sized P-MOSFET (to put it extremely simply!), so for high-power stuff people really prefer using N-MOSFETs.

But, to turn on an N-MOSFET, you need to drive the gate with a voltage above the voltage that's on its Source. So if the top MOSFET is an N-MOSFET, driving the load, its source will go to VCC whenever it is turned on, or at least, that's what you want to happen, because you want no losses in that transistor. To be able to have a turned on N-MOSFET, while its Source is at VCC, you need to connect the Gate to a voltage higher than VCC, so that's why the chip needs that supply of a higher voltage.

  • \$\begingroup\$ "Bootstrapping" simply means to subtract the same voltage from the input voltage (by connecting an equivalent opposing voltage source in series). As a result, there is no current flowing and the input voltage source "sees" infinite impedance. \$\endgroup\$ – Circuit fantasist Feb 18 at 21:13

From the datasheet of the step-down converter MP1482, you need a bootstrap capacitor, between SW pin and the N-Mosfet gate pin BS. This bootstarp capacitor, allows to keep the high side MOSFET gate voltage greater than the input voltage. As you can see in the figure 1 from the datasheet, a diode is connected to the BS pin allowing to charge the boost capacitor when the SW is low.


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