Question about whether the voltages across the capacitor are equal when Q2 is turned on and Q1 is turned off

tongshaoqiang

Question about whether the voltages across the capacitor are equal when Q2 is turned on and Q1 is turned off [Copy link]

 

As shown above, it is a lithium battery charge and discharge management circuit. U3 is the battery power display through the LED light. If you have any questions, please take a look.

My analysis of the role of Q1 and Q2 is that as long as USB5V is connected, U3 will not work:

(1). When USB is connected, Q2 is turned on, S_CTRL is turned on, Q1 is turned off, and pin 2 of U3 is left floating, U3 does not work;

(2) When USB is connected, Q2 is turned on, S_CTRL is not connected, Q1 is turned off, pin 2 of U3 is suspended, and U3 does not work;

question:

1. Since U3 doesn't work, why are the lights D2, D3, and D4 not on, but the light connected to D1 is on?

2. In the above situation, sometimes the voltage at point A is 3.49V (at this time, the LEDs connected to D2, D3, and D4 are all off), and sometimes it is 0.42V/0.38V (at this time, the LEDs connected to D2, D3, and D4 are all on). Why are there different situations?

My understanding is that as long as Q1 is turned off, the voltage across C5 is equal, that is, the voltage at point A is equal to BAT_IN.

Thanks~

maychang

[1. Since U3 does not work, why are the lights D2, D3, and D4 off, but the light connected to D1 is on? ]

The light connected to D1 is on, which may be due to the low level of pin 1 of U1. The current flows from V_LED through the red light-emitting tube, R12, diode D1, and pin 1 of U1 to the ground.

maychang

[Sometimes the voltage at point A is 3.49V (at this time, the LEDs connected to D2, D3, and D4 are all off), and sometimes it is 0.42V/0.38V (at this time, the LEDs connected to D2, D3, and D4 are all on). Why are there different situations? ]

The voltage at point A is a few tenths of a volt, which means Q1 is turned on, and of course Q2 is turned off at this time.

二月树影

This is more difficult

tongshaoqiang

maychang posted on 2024-8-19 15:26 [Sometimes the voltage at point A is 3.49V (at this time, the LEDs connected to D2 D3 D4 are not lit), sometimes it is 0.42V/0.38V (at this time, the LEDs connected to D2 D3 are not lit)...

Thanks to maychang for the guidance!

What I don't understand is:

1. Same operation: connect USB, turn on S_CTRL, at this time Q2 pin 1 is 5V, BAT_IN is 4.2V, why is the voltage at point A different? (Sometimes the voltage at point A is 3.49V, sometimes 0.38V)

2. My understanding is that as long as the USB is connected, Q2 is turned on and Q1 is in the off state. Is this correct?

maychang

tongshaoqiang posted on 2024-8-19 18:25 Thank you for the guidance of maychang senior! What I don’t understand is: 1. The same operation: connect USB, turn on S_CTRL, at this time Q2’s 1 pin is 5 ...

[2. My understanding is that as long as the USB is connected, Q2 is turned on and Q1 is turned off. Is this correct? ]

I think it is correct.

maychang

tongshaoqiang posted on 2024-8-19 18: 25Thanks to maychang for the guidance! What I don't understand is: 1. The same operation: connect USB, turn on S_CTRL, at this time Q2's 1 pin is 5...

[1. Same operation: connect USB, turn on S_CTRL, now pin 1 of Q2 is 5V...]

This is not right. When USB is connected, Q2 should be turned on and Q1 should be turned off, so point A cannot be at a low level. Moreover, according to you, two LEDs are on.

tongshaoqiang

maychang posted on 2024-8-19 18:34 [1. The same operation: connect USB, turn on S_CTRL, at this time Q2's 1 pin is 5V...] This is wrong. USB connected, Q2 ...

Thanks to maychang for his guidance.

Theoretical analysis shows that when Q1 is turned off, point A should not be at a high level. However, the current level measured at point A is 0.42V. Only in this way can U3 work and the LED light up.

According to theoretical analysis, the A point is high level (the same board was measured earlier and recorded as 3.49V), U3 does not work, and the LED light is not on. This situation has not been measured yet, so there is doubt.

maychang

tongshaoqiang posted on 2024-8-20 11:22 Thanks to Mr. Maychang for his guidance. Theoretical analysis shows that when Q1 is turned off, point A should not be at a high level, but the current level measured is 0.42V, so U ...

【Theoretical analysis: Q1 is turned off, point A should not be a high level】

wrong.

When Q1 is turned off, point A should be high level. Its drain is powered by pin 2 (GND) of U3.

tongshaoqiang

maychang posted on 2024-8-20 11:45 [Theoretical analysis: Q1 is turned off, point A should not be high level] No. Q1 is turned off, point A should be high level. Its drain is from pin 2 of U3 (G ...

Teacher Maychang:

1. When Q1 is turned off, the voltage at point A is high, which is roughly the voltage at the top of C5. Is this correct?

2. I just did a few experiments. The working state of U3 is also related to the power of the battery at the BAT end. But after measuring a few groups, I found that the point A is 0.3V~0.6V. I will explore it again.

Thanks again for your old reply.

maychang

tongshaoqiang posted on 2024-8-20 12:15 Teacher maychang: 1. When Q1 is turned off, the voltage at point A is high, which is roughly the voltage at the top of C5. Is this correct? 2. I just did a few experiments...

[Q1 is turned off, and the voltage at point A is high, which is roughly the voltage at the top of C5. Is this correct? ]

Yes.

gmchen

It is possible that the leakage current of Q1 causes the change in the potential at point A mentioned above.

According to the data sheet of 2N7002, when VGS=0, the maximum value of the drain current IDSS at 25° is 1 microampere, but this current varies greatly with temperature, and can reach a maximum of 500 microamperes at 125°. The normal operating current of HM1160 is only 10 microamperes, so when the temperature rises, it is likely that the leakage current of Q1 is enough for HM1160 to start working.

gmchen

In order to check or confirm the above hypothesis, you can heat Q1 slightly (bring it close to the soldering iron or blow hot air) and observe the potential at point A.

gmchen

If it is confirmed that the leakage current of Q1 is the cause, the modification method is to connect a suitable resistor in parallel between pins 1 and 2 of U3 (HM1160) so that the voltage drop of the leakage current of Q1 on this resistor is lower than the starting voltage of U3.

se7ens

You should not use a multimeter to test the voltage at point A.

This is equivalent to connecting a resistor to ground at point A.

The capacitor can be charged, so the potential at point A is of course low.