Introduction:

As we know, MOSFETs are voltage-driven devices. They only conduct when the gate voltage (G) is at least Vth higher than the source voltage (S). Let's look at the circuit below:

Here, the gate voltage (G) is 12V, but due to the voltage drop across resistor R7, the gate is raised.

In general, for non-low-voltage MOSFETs, the datasheet's driving voltage is set to 10V or 12V. In the circuit above, we set the driving voltage to G-S = 12-8.42 = 3.58V. 3.5V can also achieve conduction, but the conduction resistance will be high, leading to MOSFET heating.

This is where the bootstrap capacitor circuit comes in.

First, let's explain the bootstrap capacitor circuit briefly.

Bootstrap refers to using a switch power MOSFET (here referring to the upper MOSFET) and a capacitor to form a boost circuit. The capacitor is usually charged by the power supply to make its voltage higher than Vin.

The simplest bootstrap circuit consists of a capacitor. To prevent the boosted voltage from backflowing to the original input voltage, a diode is usually added.

Its advantage lies in using the characteristic that the voltage at both ends of the capacitor cannot change suddenly to increase the voltage.

In the problem of the circuit mentioned above, we can use the bootstrap capacitor method to solve it.

Let's take a look at this bootstrap circuit:

-The left end of the capacitor is VB, which is Vboost, and the right end of the capacitor is VS floating ground; -C3 is the bootstrap capacitor; -M is the inductive load, and the current flows to the right.

MOSFET Q turns on:

Assuming that the bootstrap capacitor C3 is fully charged to 14V at this time.

When PWM is 1, Q1 conducts, and the voltage at the C terminal is low, then the voltage at the B terminal of Q2 is also low, and Q2 conducts; At this time, the voltage at the E terminal of Q2 is 14V, and after passing through Q2, D2, R4, the G terminal of MOSFET is approximately 12V, and the MOSFET conducts. Here we can know that the voltage of the bootstrap power supply needs to be about 2V higher than the MOSFET driving voltage.

After the MOSFET conducts, VM (the motor M is an inductive load) is directly applied to the S terminal of Q. Because the S terminal is connected to the right end of the capacitor, the right end of the bootstrap capacitor C3 is raised to approximately 24V.

At this time, the voltage at both ends of the capacitor cannot change suddenly. The voltage at the left side of the capacitor is also raised. At this time, 14V + 24V = 38V.

Subsequently, the 38V voltage continues to supply power to the G terminal of the MOSFET through Q2, D2, R4.

Finally, the S and G terminals of the MOSFET are both raised to 24V, and Vgs = 12V.

Next, let's talk about the case when MOSFET Q is turned off:

When PWM becomes 0, Q1 is turned off, and there is no current path for the BE of Q2, so Q2 will be turned off. At this time, the discharge voltage of the bootstrap capacitor is 0, and the MOSFET is turned off.

The current of the motor M (inductive load) flows to the right, and the current flows through the body diode of the MOSFET for conduction. At this time, the voltage at the right end of C3 capacitor is -0.7V, which cannot boost. Diode D1 conducts, and the 14V power supply charges C3 capacitor, and the charging is completed.

Then, PWM switches from 0 to 1 to continue the cycle.

Conclusion: The above is all the content for today. Thank you for your attention and support!