I just learned analog circuits and designed a buck circuit. For some reason, the MOS tube burned out as soon as it was powered on (I am an amateur)

1123445

I just learned analog circuits and designed a buck circuit. For some reason, the MOS tube burned out as soon as it was powered on (I am an amateur) [Copy link]

 

I just learned analog circuits and designed a buck circuit. For some reason, the MOS tube burned out as soon as it was powered on (I am an amateur)

火辣西米秀

I guess it's hard for me to understand the buck circuit.

What are the input and output parameters for designing this circuit?

wangerxian

The wiring is a bit messy. Is it designed that the MOS tube burns out when the power is turned on? Which textbook is the reference for the circuit?

maychang

[For some reason, the MOS tube burned out as soon as it was powered on]

If it burns, it must be the actual circuit. Since it is an actual circuit, you should explain clearly what load your circuit carried during the experiment, right?

maychang

The main circuit (Q1, D1, L1, C2, C5, etc.) in the first post looks like a Buck circuit. But the driving MOS tube part in the figure is completely different. Even the power supply of U1A, U1B and two complementary transistors is a big problem.

maychang

The original poster seems to want to solve the problem of MOS tube burning. But whether this circuit can work is still unknown. Judging from the power supply of U1A and U1B, this circuit cannot work properly.

maychang

Try to see what the host thinks first.

Since it is a Buck circuit, the gate of the N-channel MOS tube needs to be supplied with a PWM signal with sufficient amplitude for the source of the MOS tube. It is estimated that the original poster uses U1A and U1B to generate this PWM signal.

maychang

U1A and U1B are common integrated operational amplifiers LM358. From the figure, U1A is a multivibrator. The voltage across the capacitor C4 at its inverting input is an approximate triangular wave. This triangular wave is sent to the inverting input of U1B, and the resistors R6 and R7 at the non-inverting input of U1B divide the voltage across the capacitor C6 and compare it with the triangular wave at the inverting input. The output of U1B is a rectangular wave after the voltages at the two input terminals are compared. This rectangular wave is then output to the gate of the N-channel MOS tube by the emitter follower composed of the two complementary transistors on the right.

maychang

This idea is not wrong, but the problem lies in the power supply of the driving circuit composed of U1A, U1B and complementary transistors. In this part of the circuit, the positive and negative ends of the power supply are connected to both ends of capacitor C6, and the upper end of C6 is connected to the input voltage (+18V) through diode D2, and the lower end of C6 is connected to the negative electrode of freewheeling diode D1.

maychang

The original poster may have referred to the bootstrap circuit for driving the half-bridge, and wanted to use the voltage across capacitor C6 to power U1A, U1B, and complementary transistors. However, there is a prerequisite for the bootstrap circuit for driving the half-bridge to power the driving circuit of the upper tube of the half-bridge, that is, the lower tube of the half-bridge can be fully turned on to ground the lower end of the bootstrap capacitor. In the first circuit, the socket J1 is powered, and the current flows to the upper end of capacitor C6 through diode D2, and then flows to the lower end of C6 through U1A, U1B, and complementary transistors, and returns to J1 through capacitor C5.

maychang

The problem lies with capacitor C5.

When capacitor C5 is charged, the voltage at both ends will increase, and it will no longer be the zero voltage when J1 is powered. In a normal Buck circuit, an appropriate load will be connected to J2, and C5 will discharge through the load, so that the output end will get an appropriate voltage. However, if the host does not connect a load during the experiment, or the load is too light (the load resistance is too large), then C6 will have nowhere to discharge, so that U1A, U1B and the complementary tube will not get the proper power supply, the MOS tube cannot be fully turned on, and the first pulse of the Buck circuit will start, and the entire Buck circuit will not start working.

maychang

Therefore, the circuit designed by the OP may not work properly at first. The reason why it cannot work properly is that there is a problem with the power supply of U1A, U1B and the complementary tube.

maychang

As for what the OP said, "the MOS tube burned out as soon as it was powered on", there may be other reasons, such as welding errors, or the MOS tube is damaged, etc.

led2015

As soon as the power is turned on, the MOS burns out. If the circuit is fine, it is a welding problem or a selection problem. If it is a circuit problem, it is so complicated that it still needs to be explained in detail.

1123445

maychang posted on 2023-8-7 15:39 The problem lies in capacitor C5. When capacitor C5 is charged, the voltage across the two ends will increase and is no longer zero voltage when J1 is powered. In a normal Buck circuit...

The actual circuit load is indeed connected to a large resistor

maychang

1123445 Published on 2023-8-7 21:40 The actual circuit load is indeed connected to a large resistor

I said on the 4th floor: "What load did you carry in your circuit during the experiment? You should always tell me clearly, right?"

The result still does not mention the load resistance value.

maychang

1123445 Published on 2023-8-7 21:40 The actual circuit load is indeed connected to a large resistor

Your circuit requires some major changes.

It is best to use 18V DC power directly for circuits such as U1A and U1B that generate rectangular waves. Your Buck circuit may not be able to drive the N-channel MOS tube into saturation conduction (called linear resistance region in textbooks) with an input 18V power supply. To improve the drive level, it may be safer to add a bootstrap circuit to drive the MOS tube after U1A and U1B generate rectangular waves.

It is safer to use a dedicated switching power supply control chip to generate rectangular waves. The dedicated switching power supply control chip can also achieve output voltage stability (DC negative feedback controls PWM duty cycle) and achieve overcurrent protection...

maychang

1123445 Published on 2023-8-7 21:40 The actual circuit load is indeed connected to a large resistor

Of course, if you use a dedicated switching power supply control chip, all the functions you need are integrated into the chip. This loses the fun of design. However, using discrete components to generate PWM and discrete components to form a negative feedback loop is really too troublesome.

A more convenient way is to use a dedicated Buck chip. A dedicated Buck chip has the power switch tube and freewheeling diode (possibly a MOS tube) built into the chip, which is very convenient. Of course, the fun of design is lost.

1123445

maychang posted on 2023-8-7 15:42 As for what the OP said, "the MOS tube burned out as soon as it was powered on", there may be other reasons, such as welding errors, or the MOS tube is damaged, etc.

The PCB board I made by myself may not be soldered properly. In addition, the MOS tube IRF740 has a Vgs withstand voltage of 20V and may break down as soon as it is powered on. In addition, when experimenting, a 12V output can be obtained using the breadboard J2

1123445

maychang posted on 2023-8-7 21:54 I said on the 4th floor: "What load did you bring to your circuit during the experiment? You should always tell me clearly, right?" In the end, I still didn't say what the load voltage was...

It should be a 10k resistor.