Boost circuit problem

hau30729

I use USB power supply and want to increase the voltage to about 11V to power the OPA in the following circuit. I
encountered a very strange situation. After using FP6291 for a period of time (the time is not certain), the input will be equal to the output, that is, vin=5V vout=5v.
I would like to ask if there is any solution to this? Is it a welding problem, IC quality problem, or layout circuit problem?

fp6291v070.pdf (375.99 KB, downloads: 9)

2019-7-23 11:53 上传

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maychang

The Boost circuit "input equals output", which means the chip is not working at all.

It is likely that the chip is damaged or the solder joint is broken and open.

hau30729

maychang posted on 2019-7-23 12:19 The Boost circuit "input equals output" means the chip is not working at all. It is likely that the chip is damaged or the solder joint is broken...

There is a problem that is not mentioned here. After the power is turned on again, the circuit can work normally for a period of time and reach the expected output. But after 5-10 minutes, the problem will occur again.

A colleague said that it might be caused by the large reverse leakage current of SB240, and SR24 might be the solution? I don't quite understand why.

PowerAnts

It would be a hooligan if you don't mention the nature of the load and the current.

hau30729

PowerAnts posted on 2019-7-23 12:59 If you don't mention the nature of the load and the current, you are just a hooligan

The load is behind the OPA's VCC, because the OPA needs to output 10V
current. The above circuit is set to output 12V 1A.

PowerAnts

hau30729 posted on 2019-7-23 13:04 The load is behind the VCC of the OPA, because the OPA needs to output 10V current. The above circuit is set to output 12V 1A

Have you checked how much current the USB can provide? Is it enough?

hau30729

PowerAnts posted on 2019-7-23 13:13 Can you check how much current the USB can provide? Is it enough?

I have changed the USB. I used 5V 1A before and then I used a 5V2A 2.5A one. The same problem still occurs.

qwqwqw2088

Power on normal, and then it is equal to input?

Protected? Check what caused the protection

hau30729

qwqwqw2088 posted on 2019-7-23 14:56 It was normal when powered on, but then it was equal to input? Is it protected? I need to find out what caused the protection

I have done the following:
1. Reduced the output voltage from 12V to 11V (to prevent surges from causing OVP)
2. Removed the circuits that need to be loaded, but the problem still occurs.
3. Measured the temperature of FP6291 and found no increase
4. Replaced SB240 with SR24.
I don't know if there is any other way to detect...

topwon

If the protection is not caused by overload heating, then you should first check whether there is a big problem with your laundry, causing the switching circuit to work unstably.

qwqwqw2088

hau30729 posted on 2019-7-23 15:21 Currently, the following things have been done: 1. Reduce the output voltage from 12V->11V (to prevent surges from causing OVP) 2. Remove the circuits that need to be loaded at the back, and this situation will still occur...

The operating frequency of FP6291 is 1MHZ. The layout of surrounding components and feedback pin routing, as well as the position of inductor layout should be noted. You can refer to the official reference board

伟林电源

1. Check the input power load capacity;

2. Is the output normal when no-load? Is it normal at the beginning when no-load, and will the output voltage be the same as the input voltage after a while?

3. If no-load is normal, then connect an electronic load or adjustable resistor to the output, load it slowly, and monitor the waveform at the same time to see what is going on?

hau30729

Wei Lin Power published on 2019-7-23 20:47 1. Check the input power load capacity; 2. Is the output no-load normal? Is it normal at the beginning when no-load, but after a while, the output and input will be normal...

It is not normal when no load.
Yesterday I changed the circuit to the same as the Spec reference circuit, removed C10 and now the 7 PCBs are working normally for one night. I don't know why...

hau30729

qwqwqw2088 posted on 2019-7-23 16:31 FP6291 operating frequency 1MHZ, surrounding component layout and feedback pin routing, inductor layout position should be noted, you can refer to the official reference board

The component placement is copied from the official SPEC, so there shouldn't be any problems.
Yesterday I changed the circuit to be the same as the Spec reference circuit, removed C10, and now the 7 PCBs are working fine for one night. I don't know why...

hau30729

topwon posted on 2019-7-23 16:16 If it is not a protection caused by overload heating, then you should first check whether there is a big problem with your layout, causing the switch circuit to work unstable.

The component placement and layout are copied from the official SPEC, so there shouldn't be any problems.
Yesterday I changed the circuit to be the same as the Spec reference circuit, removed C10, and now the 7 PCBs are working fine for one night. I don't know why...

伟林电源

I have carefully read your data. The original reference circuit does not have this capacitor. It is preliminarily estimated that the leakage current of this capacitor caused the feedback misjudgment.

qwqwqw2088

For this type of switch chip, components and wiring near the feedback pin are very important.

C10 is not listed in the manual.

hau30729

qwqwqw2088 posted on 2019-7-24 09:58 For this type of switch chip, the components and wiring near the feedback pin are very important. C10 is not in the manual

I would like to ask what is the impact of removing C10?

hau30729

Wei Lin Power posted on 2019-7-24 09:39 I carefully read your data. The original reference circuit does not have this capacitor. It is preliminarily estimated that the leakage current of this capacitor caused the feedback misjudgment.

Capacitor leakage current?
The original intention of C10 should be to speed up the reaction speed? But it caused leakage current and caused the IC to misjudge OCP/OVP protection?

qwqwqw2088

hau30729 posted on 2019-7-24 11:13 I would like to ask what is the impact of removing C10?

The impact is undoubtedly a misjudgment. The manual does not explain why this capacitor should be added, especially on the feedback pin.

To analyze the chip, check the internal structure diagram in the manual.

The feedback pin generally feeds back the output voltage to the comparator inside the chip to obtain the parameters of the current output voltage, automatically adjust the output voltage in a closed loop, and stabilize the DC working voltage.