Please help analyze how the following negative voltage is generated and the direction of current flow, thank you

cxq742536574

Please help analyze how the following negative voltage is generated and the direction of current flow, thank you [Copy link]

 

maychang

D2D4C20C33 constitutes a voltage doubler rectifier circuit, whose output (both ends of C33) is negative to ground.

But the position of D5 is not to output a stable -7V negative voltage, on the contrary, all the instability of C33 is applied to the -7V output terminal.

maychang

『How the following negative voltage is generated and the current flows』

As for the direction of current flow, the direction of current in D2D4 can be determined by the direction of the diode. The direction of current in C20C33 must be bidirectional, and the amount of charge passing through the capacitor in the two half-cycles of AC must be the same in magnitude but opposite in direction.

captz

Judging from the resistor R34, the output current of the -7V power supply is only tens of uA, so there is no need to make the circuit so complicated.

深度迷茫

I understand that negative voltage is caused by a sudden change in the voltage at the front end of C20. First, charge both ends of C20 to a certain voltage, such as 16v, and then reduce the voltage at the front end of C20 to 5v. The voltage at the back end of C20 will then decrease by 16-5=11v. Originally, the voltage at the back end of C20 is 0v, and after decreasing by 11v, it becomes -11v. This is just a guess.

maychang

Deep Confusion posted on 2022-5-12 14:52 I understand that negative pressure is caused by a sudden change in the voltage at the front end of C20. First, charge both ends of C20 to a certain voltage, such as 16v, and then reduce the voltage at the front end of C20 to 5v. C20 ...

The lower end of C20 is connected to the SW pin of the chip. This circuit is a Boost circuit. There is a power switch tube inside the chip pin 1. The voltage of pin 1 to ground always jumps between close to zero and the rated output voltage (16V in the figure).

深度迷茫

maychang posted on 2022-5-12 16:14 The lower end of C20 is connected to the SW pin of the chip. This circuit is a Boost circuit. There is a power switch tube inside the chip pin 1. The voltage of pin 1 to ground is always close to zero...

Oh, that is, the front end of C20 jumps between 0-16v, and the peak value of negative voltage is -16v

cxq742536574

maychang published on 2022-5-12 09:34 D2D4C20C33 constitutes a voltage doubler rectifier circuit, and its output (both ends of C33) is negative to ground. However, the position of D5 is not a stable output of -7V negative voltage, so...

I actually tested that AP1522 jumps between 0 and 16V, and the direction is opposite to the positive pole of D2, jumping between -16V and 0V, and becoming a stable 16V at the positive pole of D4, so the final output is still -7V. I just don't understand why the positive pole of D2 is in the opposite direction.

maychang

cxq742536574 posted on 2022-5-19 17:09 I actually tested that AP1522 jumps between 0 and 16V, and the direction is opposite to the positive pole of D2, jumping between -16V and 0V, and becomes stable at the positive pole of D4...

"But I still don't understand why the positive pole of D2 is in the opposite direction."

The direction of D2 is not reversed. If the directions of D2 and D4 are reversed, it will become a positive voltage output.

cxq742536574

maychang posted on 2022-5-19 18:42 "But I still don't understand why the positive direction of D2 is reversed." The direction of D2 is not reversed. If the directions of D2D4 are reversed, it will become a positive voltage output...

I don't understand why the voltage directions at both ends of C20 are different.

maychang

cxq742536574 posted on 2022-5-20 09:22 I don’t understand why the voltage directions at both ends of C20 are different

Then I will explain step by step from the change of the potential of the chip's SW pin. I will ask you a question for each step, and I can only talk about the next step if you answer correctly.

You already know that the pulse at SW is 0 to 16 V. We start with SW being 0 and assume that the voltage across C20 is zero at the beginning.

In the first pulse, SW rises from 0V to 16V. Question: Is D2 conducting at this time? Is current flowing through C20? What is the direction of the voltage across C20?

cxq742536574

maychang posted on 2022-5-20 09:58 Then I will talk about it step by step from the change of the potential of the SW pin of the chip. I will ask you a question for each step. Only when you answer correctly can I talk about the next step...

When the SW pin goes from 0V to 16V, D2 is turned on, C20 is charged, and the positive electrode of D2 goes from 0 to 0.7V. When the SW pin goes from 16V to 0V, C20 starts to discharge, D4 is turned on, and the positive electrode of D2 drops from 0.7V to -16V. C33 is charged and discharged to convert the voltage into DC. Is that right?

maychang

cxq742536574 posted on 2022-5-20 12:02 When the SW pin goes from 0V to 16V, D2 is turned on, C20 is charged, and the positive electrode of D2 goes from 0 to 0.7V; when the SW pin goes from 16V to 0V, C20 starts to discharge, D4 is turned on, and the positive electrode of D2...

"is that so"

almost.

The positive electrode of D2 cannot drop to 16V at the end of a cycle, because C20 is charged to 16V at the first pulse, and it is impossible for C20 to discharge and fully charge C33 after the pulse disappears. After the first pulse disappears, C33 can only be charged to about 0.16V with negative at the top and positive at the bottom. It takes many cycles (more than hundreds) to charge C33 to nearly 16V with negative at the top and positive at the bottom (it is actually impossible to achieve, but it is close).

cxq742536574

maychang posted on 2022-5-20 13:02 『Is that so?』 Almost. It is impossible for the positive electrode of D2 to drop to 16V at the end of a cycle, because C20 is charged to 16V in the first pulse...

Thank you for your advice, sir.