I would like to ask a question about the bootstrap capacitor of synchronous buck

小太阳yy

I would like to ask a question about the bootstrap capacitor of synchronous buck [Copy link]

 

 

As shown in the figure below, C is the bootstrap capacitor. When the low-side MOS tube is turned on, SW is 0, and the voltage on BOOT is provided by BOOT Charge. If it is 5V, the capacitor is charged. When the low-side MOS tube is turned off, the high-side MOS tube is turned on because the high-side Vgs>Vgs(th), so the high-side MOS tube can be turned on. As the high-side MOS tube is turned on, the voltage on SW will become VIN. If this C is not added, then when Vgs<Vgs(th), the high-side MOS tube will not be able to be turned on. After adding C, using the characteristic that the capacitor voltage cannot change suddenly, when SW becomes VIN, the voltage on BOOT will become VIN+5V, and at this time Vgs will be greater than Vgs(th), and the high-side MOS tube will be turned on.

Is my understanding correct?

maychang

[Is my understanding correct? ]

Generally correct.

However, you failed to explain clearly that "when the low-side MOS tube is turned on, SW is 0, and the voltage on BOOT is provided by BOOT Charge", while the high-side MOS tube is turned on, the voltage on SW becomes VIN, why BOOT (the upper end of the capacitor) does not discharge to BOOT Charge.

hjl2832

The principle is like this. It is commonly called a charge pump circuit, and the capacitor is called a pump capacitor. It uses the high and low level changes of PWM, combined with the reference voltage (that is, the 5V you mentioned), to make the charging voltage on the capacitor equal to VIN+5V.

小太阳yy

maychang posted on 2023-3-13 20:13 [Is my understanding correct? ] Generally correct. However, you failed to make it clear that "when the low-side MOS tube is turned on, SW is 0, BO ...

I don't understand why, teacher.

小太阳yy

maychang posted on 2023-3-13 20:13 [Is my understanding correct? ] Generally correct. However, you failed to make it clear that "when the low-side MOS tube is turned on, SW is 0, BO ...

Some people say that when the low-side MOS is turned off, the SW point is in a floating state. If no capacitor is added, no potential difference will be generated on the high-side MOS, and the high-side MOS cannot be turned on. Is this correct, teacher?

小太阳yy

maychang posted on 2023-3-13 20:13 [Is my understanding correct? ] Generally correct. However, you failed to make it clear that "when the low-side MOS tube is turned on, SW is 0, BO ...

I have another question here. When SW becomes VIN, how can the voltage on BOOT be VIN+5V? Doesn't the voltage of VIN pass through the BOOT charge? If the BOOT charge is 5V, shouldn't VGS be 5V+5V?

maychang

小太阳yy posted on 2023-3-13 22:25 And some people say that when the low-side MOS is turned off, the SW point is in a floating state. If no capacitor is added, no potential difference will be generated on the high-side MOS, and it will not be possible to make...

This capacitor is used in the circuit to power the driving circuit of the high-side MOS tube when the high-side MOS tube is turned on (SW is close to the power supply voltage). Without this capacitor, the driving circuit has no power supply and the high-side MOS tube cannot be turned on at all.

It is not accurate to say that "if no capacitor is added, no potential difference will be generated on the high-side MOS". The voltage between the drain and source of the high-side MOS is quite high when it is not conducting. This sentence does not specify which two electrodes they are. Without this capacitor, the gate of the high-side MOS cannot be higher than the source, so the high-side MOS cannot be turned on, which is correct.

maychang

Xiaoyangyy posted on 2023-3-13 22:12 I don’t quite understand why, teacher, please tell me why

You'd better post the complete schematic diagram, otherwise it's too much trouble to explain. This "bootstrap" capacitor is not only used in synchronous Buck circuits, but also in asynchronous Buck, and also in half-bridge and full-bridge. What you are discussing now is limited to a certain type of chip.

maychang

小太阳yy posted on 2023-3-13 22:56 I have another question here. When SW becomes VIN, how can the voltage on BOOT be VIN+5V? Isn’t the voltage of VIN charged by BOOT?

[When SW becomes VIN, how can the voltage on BOOT be VIN+5V?]

[When the low-side MOS is turned on, SW is 0, and the voltage on BOOT is provided by BOOT Charge. If it is 5V, the capacitor is charged], this is what you said in the first post. From this sentence, it is obvious that the voltage across the capacitor is 5V after charging, that is, the voltage between the BOOT and SW terminals is 5V. Then when the voltage of the SW terminal to the ground is not 0, but a voltage VX greater than 0, the BOOT terminal is obviously VX plus 5V. The high-side MOS is turned on, SW is close to VIN, then BOOT is of course VIN plus 5V (assuming the capacitor is not discharged).

maychang

小太阳yy posted on 2023-3-13 22:56 I have another question here. When SW becomes VIN, how can the voltage on BOOT be VIN+5V? Isn’t the voltage of VIN charged by BOOT?

[If the BOOT charge is 5V, shouldn’t VGS be 5V+5V?]

I don't know what the 5V in the front and the 5V in the back of your sentence [5V+5V] refer to.

小太阳yy

maychang posted on 2023-3-14 10:35 You'd better post the complete schematic diagram, otherwise it's too much trouble to explain. This "bootstrap" capacitor is not only used in synchronous Buck circuits...

The teacher will post the schematic diagram when he returns home in the evening.

小太阳yy

maychang posted on 2023-3-14 10:43 [When SW becomes VIN, how can the voltage on BOOT be VIN+5V?] [When the low-side MOS tube is turned on, SW is 0, and the voltage on BOOT is from BOO...

Can you see the circuit diagram? It is a synchronous BUCK step-down

小太阳yy

maychang posted on 2023-3-14 10:44 [If the BOOT charge is 5V, then shouldn’t VGS be 5V+5V?] I don’t know what the 5V in front of your [5V+5V] refers to, ...

I have the above questions, can you help me answer them?

小太阳yy

maychang posted on 2023-3-14 10:43 [When SW becomes VIN, how can the voltage on BOOT be VIN+5V?] [When the low-side MOS tube is turned on, SW is 0, and the voltage on BOOT is from BOO...

Another question is, as you said, won't the bootstrap capacitor discharge? If it discharges, won't the voltage drop?

maychang

Xiao Taiyangyy posted on 2023-3-14 22:05 I have the above questions, can the teacher help me answer them?

It seems that you still have problems understanding the concept of "voltage".

Let's not talk about the physical definition of voltage for now (examining the ratio of the work done by the electric field on the charge when the charge moves from one point to another in the electrostatic field to the amount of charge), and just talk about some regulations of voltage in the circuit. Voltage must refer to the voltage between two points. For example, the Vgs of a MOS tube is the voltage between the gate of the MOS tube and the source. If only one point is mentioned, such as the voltage of the SW point, it refers to the voltage of this point to the ground in the circuit.

Therefore, when we say that the voltage across CBOOT refers to the voltage between the BST point and the SW point, it is equal to the BST voltage to ground minus the SW voltage to ground, that is, the difference between the BST voltage (to ground!) and the SW voltage (to ground!). Pay attention! Make sure you understand it!

maychang

Xiao Taiyangyy posted on 2023-3-14 22:05 I have the above questions, can the teacher help me answer them?

(Continued from 15th floor) If the voltage across CBOOT is V (charging is complete), note that it is the difference between the voltages at BST and SW. Then when the voltage at SW (to ground) changes, because the voltage across the capacitor cannot change suddenly and remains at V, the voltage at BST must change. Before the change, it was SW plus V, and after the change, it is still SW plus V. SW was originally close to zero, and BST was originally V (SW plus the voltage across the capacitor V). Now SW changes to be close to VIN, and BST must now change to VIN plus V (the voltage across the capacitor is still V, but SW becomes VIN). The voltage at BST (to ground) is higher than VIN, V higher.

maychang

Xiao Taiyangyy posted on 2023-3-14 22:05 I have the above questions, can the teacher help me answer them?

This solves the problem you marked in red: "A bit confused, I understand that after CBOOT is charged, the voltage is VCC". No! After CBOOT is charged, the voltage [at both ends] is V (I don't know how much, you said it is 5V), because SW rises to VCC, the voltage at the top of CBOOT will exceed VCC.

Another point to note: Do not change the numbers in the same circuit diagram or discussion at will. The VCC you mentioned in the red field actually refers to the VIN mentioned earlier. This cannot be changed casually, otherwise it will easily confuse you.

maychang

Xiao Taiyangyy posted on 2023-3-14 22:05 I have the above questions, can the teacher help me answer them?

Now let's talk about the first question in red: "What is the voltage at point G of the MOS tube?" Because only one point is mentioned, it refers to the voltage at that point to ground. This voltage (to ground) is obviously different when SW is zero and when SW is VIN (how much is unknown, and the HS DRIVER state must also be considered). If we talk about the voltage at point G to point S or the voltage between G and S, then it has nothing to do with the voltage at SW to ground and is completely determined by the HS DRIVER.

maychang

Xiao Taiyangyy posted on 2023-3-14 22:05 I have the above questions, can the teacher help me answer them?

Now let’s talk about the third question in red.

There are two dry cell batteries, each with a voltage of 1.5V across its two ends. When you connect the negative pole of the first cell to the positive pole of the second cell, the voltage between the positive pole of the first cell and the negative pole of the second cell is 3V, which is the sum of the voltages of the two cells. When you connect the positive pole of the first cell to the positive pole of the second cell, and the negative pole of the first cell to the negative pole of the second cell, the voltage between the positive pole of the first cell and the negative pole of the second cell is 1.5V. The same battery, connected in series and in parallel, this is something that is taught in junior high school. Obviously you did not understand, otherwise you would not have made the mistake of "adding two VCC voltages".

maychang

Xiaoyangyy posted on 2023-3-14 22:07 Another question is, as you said, won’t the bootstrap capacitor discharge? If it discharges, won’t the voltage drop?

This bootstrap capacitor will of course discharge. But as long as the charge is fast and the discharge is slow, the voltage across the capacitor (do not understand it as the voltage to the ground!) can be kept near full, and the voltage across the capacitor will not drop too much.