How to achieve normally closed switch

cxq742536574

How to achieve normally closed switch [Copy link]

 

 

When boosting, M1 needs to be normally on and M2 needs to be normally open. When bucking, M4 needs to be normally on and M3 needs to be normally open. How to achieve normally closed with a gate driver chip? Doesn't the input control signal of the gate driver chip need to be a PWM signal?

lansebuluo

This is the circuit diagram of the motor control. It is normally open and normally closed. For this kind of MOS, voltage drive is required.

tagetage

When boosting, M1 needs to be normally on and M2 needs to be normally open. When lowering, M4 needs to be normally on and M3 needs to be normally open. What is this? I don't understand. How can I understand it this way?

maychang

The original poster used three words with the character "通": 常通, 常开, 常闭. What is the difference between the three?

cxq742536574

maychang posted on 2024-9-20 11:29 The OP used three words with the character "通": 常通, 常开, 常闭. What is the difference between the three?

The statement is not clear. Both normally open and normally closed mean that the tube is always conducting, and normally open means that the tube is not conducting. This circuit is a buck-boost topology.

cxq742536574

tagetage posted on 2024-9-20 11:20 When boosting, M1 needs to be normally connected and M2 needs to be normally open. When reducing the voltage, M4 needs to be normally connected and M3 needs to be normally open. What is this? I don't understand. How can I understand it this way?

It is not a motor drive, it is a buck-boost topology

maychang

cxq742536574 posted on 2024-9-20 13:12 It is not a motor drive, it is a buck-boost topology

[It is buck-boost topology]

This is not a Buck-Boost topology. In the previous post, you said that the [Boost-Boost topology] is correct.

maychang

cxq742536574 posted on 2024-9-20 13:11 The statement is not clear. Normally open and normally closed both mean that the tube is always conducting, and normally open means that the tube is never conducting. This circuit is a buck-boost topology

Your circuit does not need a certain tube to be "normally closed", that is, it does not need a certain tube to be always on. Therefore, you can use a "bootstrap" driver chip.

cxq742536574

maychang posted on 2024-9-20 13:18 [It is a buck-boost topology] This is not a Buck-Boost topology. In the previous 5th floor, you said that [boost-boost topology] is correct.

For example, if you need to step down the voltage, does M4 need to be turned on all the time? Doesn't the gate drive require a PWM control signal and a bootstrap capacitor to achieve a boost? Then how can we achieve a continuous on state?

maychang

cxq742536574 Published on 2024-9-20 13:26 For example, if you need to reduce the voltage, does it require M4 to be turned on all the time? Doesn't the gate drive require a PWM control signal, and the bootstrap capacitor can achieve a boost? Then how...

[For example, if the voltage needs to be reduced, does M4 need to be turned on all the time?]

I have already said on the 8th floor that it is not necessary.

maychang

cxq742536574 Published on 2024-9-20 13:26 For example, if you need to reduce the voltage, does it require M4 to be turned on all the time? Doesn't the gate drive require a PWM control signal, and the bootstrap capacitor can achieve a boost? Then how...

This circuit is actually a Buck circuit plus a Boost circuit. The characteristics of this circuit are that it can boost or buck, and Vin and Vout have the same polarity (the four power switch tubes in the first post are N-channel MOS tubes, and Vin and Vout are both positive to ground). Buck-Boost circuit is different from this circuit. Buck-Boost circuit can boost or buck, but the output voltage direction is opposite to the input voltage direction, that is, reverse polarity.

maychang

cxq742536574 Published on 2024-9-20 13:26 For example, if you need to reduce the voltage, does it require M4 to be turned on all the time? Doesn't the gate drive require a PWM control signal, and the bootstrap capacitor can achieve a boost? Then how...

This circuit has another feature: energy can flow in both directions. In other words, the Vout terminal can be used as input and the Vin terminal can be used as output (the load is of course connected to the Vin terminal at this time). Therefore, this circuit is a type of bidirectional DC converter. Of course, the control of the four power switches is more complicated.

maychang

cxq742536574 Published on 2024-9-20 13:26 For example, if you need to reduce the voltage, does it require M4 to be turned on all the time? Doesn't the gate drive require a PWM control signal, and the bootstrap capacitor can achieve a boost? Then how...

An example of a situation where bidirectional energy transfer is needed is when an electric car accelerates, of course energy is transferred from the battery to the motor, and the converter can control the speed of the motor. When decelerating or going downhill, if a bidirectional converter is used, energy can be returned from the motor to the battery to charge the battery. Obviously, this can save energy.

maychang

cxq742536574 Published on 2024-9-20 13:26 For example, if you need to reduce the voltage, does it require M4 to be turned on all the time? Doesn't the gate drive require a PWM control signal, and the bootstrap capacitor can achieve a boost? Then how...

If bidirectional energy transmission is not required, this circuit has no advantages. This circuit requires the use of four power switches and is very complex to control. It is better to use a converter with a transformer.

beyond_笑谈

Do you want to use the full bridge to make Buck and Boost? The gate drive chip can be controlled by the microcontroller.

led2015

To keep devices such as M1 and M2 always on, it is necessary to ensure that the gate driver chip continuously provides sufficient positive voltage to the gates of these devices.

beyond_笑谈

If it is just normally on or normally closed and the switching frequency is low, PWM is not needed. The optocoupler can be controlled by high and low levels to drive the MOSFET, but the upper tube is replaced with P-MOSFET.

振动试验仪器

When boosting, M1 needs to be normally connected and M2 needs to be normally open. What is the status of M3 and M4 on the right? The method that cannot be explained is that all are normally open and all are disconnected.

When reducing the voltage, M4 needs to be normally connected and M3 needs to be normally open. What is the status of M1 and M2 on the left?

If M1\M3 and M2, M4 are not switched on and off alternately, how will this work?

振动试验仪器

From the circuit structure, it is an H-type bridge amplifier. SW1 and SW2 are the two output terminals, and the L inductor in the middle is the load. The OUT in the upper right corner should be the positive pole of the Vin power supply.

The motor is controlled by an H-type bridge amplifier. There is a lot of information on the Internet. The motor is controlled by a bridge amplifier, which can control forward and reverse rotation and speed regulation. Its working principle is the same as the power amplifier principle of the vibration table.

振动试验仪器

The switch tube amplifier of the vibration table, such as a 10KW amplifier cabinet, can be directly connected to control the DC motor. The maximum output voltage is positive and negative 90V, and the current rating can reach 100A. MOS tubes are generally used.

The new type of power amplifier can reach 40KW in a single box, and the output voltage can reach 300V. This one uses high-voltage IGBT.