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In common applications of MOSFET driving methods, there are non-isolated direct drive, bootstrap drive, and isolated transformer drive with optocoupler isolation. In commonly used switching power supply topologies such as Buck, full bridge, and half bridge, NMOS is often used as the switching element. Some of the source terminals (S) of these tubes are not grounded and their potential changes with the conduction state of the MOSFET. The gate signal for driving the MOSFET is referenced to the potential of the S terminal. The turn-on voltage of the MOSFET is around 4.5V, and to ensure that the MOSFET is fully turned on, the driving voltage is generally greater than 10V, ideally around 12-15V. Thus, to ensure normal operation of the MOSFET, the gate-source voltage (G-S) should reach a stable 12V.
If direct drive mode is used, the level is referenced to the ground of the circuit. When the potential of the S terminal increases, the G terminal potential remains unchanged, preventing the MOSFET from operating properly. In this case, a capacitor can be selected, with its negative terminal connected to the S terminal and the positive terminal kept at a stable 12V power supply. Charging the G terminal through this capacitor allows the GS terminal to stabilize at 12V. This is akin to a water level analogy: the capacitor acts like a water tank floating on a river. The potential of the S terminal is like the river's water level, while the circuit ground serves as sea level, typically used as a reference. The river's water level changes with the weather, but the water tank maintains a constant water level, ensuring a stable difference between the river and the tank. This driving method is known as bootstrap driving, with the capacitor storing energy referred to as the bootstrap capacitor. The potential of the S terminal can be termed "floating ground." As shown in Figure 1 below, C5 is the bootstrap capacitor, usually ranging from a few uF to 100uF.
However, the circuit shown in Figure 1 is incomplete. While the bootstrap capacitor powers the front drive, the question arises: who is charging the bootstrap capacitor? Assuming the circuit operates normally, only two positions are relatively stable: one at 30V input voltage and the other at 12V output voltage. If the input power supply is used for a long time, there is an 18V voltage difference (30V - 12V). If the current-limiting resistor is too large, the 30V would pass through the resistor slowly to charge the capacitor, resulting in insufficient current to start the driving section. On the other hand, selecting a smaller current-limiting resistor results in high losses, but since the output voltage of 12V matches the voltage requirement of the bootstrap capacitor, during normal operation, it can provide sufficient charging current to maintain a stable bootstrap capacitor voltage.
Can the output voltage be directly connected to the bootstrap capacitor? The answer is no. During normal circuit operation, when the MOSFET is turned off, the potential of the S terminal is clamped at -0.7V due to the diode, making the G terminal potential less than or equal to 11.3V, allowing the output to charge the bootstrap capacitor. However, when the MOSFET is turned on, the S terminal potential rises to approximately 30V. At this point, the G terminal potential exceeds the output voltage. Direct connection would result in the bootstrap capacitor charging the output, quickly discharging and failing to start the drive in the next cycle. Hence, the charging has a direction, requiring a fast-recovery diode to be connected.
Next, let’s analyze the working process of the circuit. When the power is turned on, it will be found that the circuit cannot start because the bootstrap capacitor has no voltage. Without driving, there is no output. Therefore, when charging, the power supply needs to be manually connected to charge the bootstrap capacitor first. Once charged to 12V, the front drive circuit can be powered. Given the 18V voltage difference, to minimize losses, a current-limiting resistor of about 20kΩ can be chosen, providing a small charging current of about 1mA for the bootstrap capacitor. Additionally, this approach achieves soft-start of the circuit, reducing stress. However, the charging current is only 1mA. Components such as a triangular wave generator, level-adjustable circuit, and driving circuit require currents of at least tens of mA. Direct charging finds that the bootstrap capacitor remains uncharged due to insufficient current. Here, a switch can be designed. During initial power-on, the 30V input power supply charges the bootstrap capacitor with a current of 1mA. At this stage, the switch is off, not supplying power to the rear circuit. Once the bootstrap capacitor voltage reaches 12V, the switch turns on, allowing the bootstrap capacitor to supply power to the drive, enabling the circuit to have a certain output. The output voltage then adds energy to the bootstrap capacitor in a timely manner, ensuring normal operation. The purpose of this switch is to charge the capacitor to 12V initially and remain open thereafter.
Following this logic, the switch tube is selected first. Since it is connected to the power supply terminal, a P-channel MOSFET is chosen here. To charge to about 12V, an 11V zener diode can be connected in series between the base and ground. A current-limiting resistor is added to limit the current. However, due to the presence of this zener diode, when the bootstrap capacitor voltage drops slightly, for instance to 11V, the switch turns off, and the output voltage rises to 12V, taking some time. This doesn’t reach the potential of the bootstrap capacitor, preventing further charging and circuit operation. The key issue lies with the 11V zener diode. If the diode is short-circuited when the switch is turned on, this problem can be resolved. Therefore, a switch is needed here, shorted to ground, and an N-channel MOSFET is selected. As shown in Figure 2, Q5 is a PNP transistor, Q4 is an NPN transistor. When powered on, Q5 is off, allowing 30V to charge C5 through R5 until it reaches 12V. When Q5 turns on, its collector potential reaches 11.3V, using this potential to drive Q4 on, allowing Q4 to short-circuit to D4. Even if the bootstrap capacitor reduces to 6V, Q5 can still turn on, maintaining the output current to keep the driving circuit running, thereby driving the entire circuit.
At this point, the bootstrap driver circuit for the Buck circuit has been designed. Bootstrap drivers are widely used due to their simple circuitry, particularly in Buck and bridge circuits. However, in practical design, from the perspectives of circuit simplicity and reliability, choosing an integrated chip is more important. There are many integrated chips available, such as IR's IR21XX series, commonly used IR2104, IR2110, and TI's. As shown in Figure 3 below:
According to the datasheet, select the appropriate integrated chip to complete the circuit design.
After studying, I feel that my knowledge is reaching new heights.
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