Use transistor + MOS tube to control battery output

QWE4562009

Use transistor + MOS tube to control battery output [Copy link]

 

Use transistor + MOS tube to control battery output

1. When MCU_CTL high level 3.3V and battery 4.2V exist at the same time, can Q4 and Q2 be turned on reliably to output the battery voltage?

2. When MCU_CTL has a high level of 3.3V and the battery 4.2V is not present, can Q4 and Q2 be turned on reliably to output the battery voltage?

3. When MCU_CTL is at a low level of 0V and the battery is at 4.2V at the same time, can Q4 and Q2 be turned off reliably? Thus turning off the output battery voltage

仙猫

1. When MCU_CTL high level 3.3V and battery 4.2V exist at the same time, can Q4 and Q2 be turned on reliably to output the battery voltage?

　——It can be conducted.

2. When MCU_CTL has a high level of 3.3V and the battery 4.2V is not present, can Q4 and Q2 be turned on reliably to output the battery voltage?

　—— It will usually be turned on, but the base resistance is too large, so be careful whether the hFE of the transistor is large enough.

3. When MCU_CTL is at a low level of 0V and the battery is at 4.2V at the same time, can Q4 and Q2 be turned off reliably to turn off the output battery voltage?

　—— Cannot be turned off, because R12<<R19, the battery is much more powerful than MCT_CTL.

[Suggestion] If you want to consume less power and have reliable control, you can consider replacing Q4 with MOSFET.

elec32156

Whether it can be turned on or off reliably, just calculate the voltage at point A; when the voltage Ua at point A is greater than 0.7V ===> Q4 is turned on === > Q2 is turned on;

1. When 3.3V and 4.2V exist at the same time: (4.2 - Ua)/R12 = (Ua-3.3)/R19 + Ua/R20;

2. When 4.2V is not present, it depends on the internal pull-up resistor of the MCU, and Ua is calculated accordingly;

3. When MCU_CTL is pulled low: (4.2 - Ua)/R12 = Ua/R19 + Ua/R20;

QWE4562009

Xianmao published on 2021-8-16 16:34 1. When MCU_CTL high level 3.3V and battery 4.2V exist at the same time, can Q4 and Q2 be reliably turned on? So as to output the battery voltage. &mda ...

2------Because that line comes from the MCU, in order to reduce power consumption, a 200K resistor is used

QWE4562009

Xianmao published on 2021-8-16 16:34 1. When MCU_CTL high level 3.3V and battery 4.2V exist at the same time, can Q4 and Q2 be reliably turned on? So as to output the battery voltage. &mda ...

2----How to adjust the parameters to ensure conduction?

3------How to adjust parameters for reliable shutdown

The above 2 and 3 must be implemented at the same time

QWE4562009

elec32156 posted on 2021-8-16 16:40 To determine whether it can be turned on or off reliably, just calculate the voltage at point A; when the voltage Ua at point A is greater than 0.7V ===> Q4 is turned on === > Q2 is turned on; ...

Make the most of Kirchhoff's laws

I want to achieve the above three functions, how should I adjust the parameters?

QWE4562009

QWE4562009 posted on 2021-8-16 18:12 2----How to adjust the parameters to ensure conduction? 3------How to adjust the parameters to ensure reliable shutdown&n ...

Using MOS is also risky. I have thought about this because the battery of our product needs to work between 2-4.2V. Can 2V conduct MOS? I am afraid it is difficult to find such MOS and even if you find it, it is expensive.

仙猫

QWE4562009 posted on 2021-8-16 18:12 2----How to adjust the parameters to ensure conduction? 3------How to adjust the parameters to ensure reliable shutdown&n ...

This circuit is probably more than just adjusting parameters. In addition to the three points asked, should we also consider whether the battery voltage is turned off or output by default when the MCU is not working? This difference will affect the structure of the control circuit.

Another point that may need to be confirmed is whether the pin MCU_CTL is floating or pulled up by default when the MCU is just released from RESET. In order to prevent malfunction of the MCU at the moment of power-on, some circuit measures may be needed.

As for MOSFETs with small threshold voltages, they are now everywhere, 1V or even 0.x volts are very common, and there is no problem of difficulty in obtaining them or high prices.

QWE4562009

Xianmao published on 2021-8-17 09:30 This circuit is probably more than just adjusting parameters. In addition to the three points asked, should we also consider whether to let the battery voltage be turned off by default or input when the MCU is not working?

If the MCU doesn't work, the battery must be turned off...

QWE4562009

Xianmao published on 2021-8-17 09:30 This circuit is probably more than just adjusting parameters. In addition to the three points asked, should we also consider whether to let the battery voltage be turned off by default or input when the MCU is not working?

You will know the domestic low-voltage conduction MOS after using it

仙猫

QWE4562009 posted on 2021-8-18 14:12 If the MCU doesn’t work, the battery must be turned off. . .

If the MCU is not working, how can we conclude that the battery is off? At least it cannot be seen from the top floor picture.

Without a complete description of the problem, it will be difficult for others to offer advice.

仙猫

QWE4562009 posted on 2021-8-18 14:12 You will know the domestic low-voltage conduction MOS after you use it

So what if it's made in China? The products and data sheets of regular manufacturers cannot contain nonsense about basic parameters.

se7ens

Why don't you use a dedicated battery protection solution?

DW01K+SC8205A/S

Or single solution DW06D/DW03D

Of course, you can use your own solution, you can change the circuit as follows

se7ens

se7ens posted on 2021-8-19 10:52 Why don't you use a dedicated battery protection solution DW01K+SC8205A/S or a single solution DW06D/DW03D? Of course, you have to use your own solution...

You can also use a simpler method to short-circuit D1D2 directly

QWE4562009

se7ens posted on 2021-8-19 10:57 You can also use a simpler method to short-circuit D1D2 directly

Hi, thank you very much!

The logic to be realized

1. To save power when entering the warehouse --- after receiving APP OFF (internal beta version), MCU_CTL outputs low level! Cut off the battery power supply



2. Customers use when leaving the warehouse --- regardless of the App. Direct charging, the MCU gets power, MCU_CTL outputs high level, so that the battery outputs power to the subsequent circuit. Even if the external power supply 4.2V (charger 5V) is unplugged, the battery can maintain the voltage output


Three points to be confirmed:

MCU_CTL output is already low level after entering the warehouse. When the charger is taken out from the warehouse and connected to the chip for power supply, there is no reset signal, no wake-up action, the chip gets the working voltage, and MCU_CTL can output high level?


Entering the warehouse requires the detachment from the base (which means that the 4.2V from the charging base is gone, or the 4.2V to 3.3V conversion is gone) Can MCU_CTL maintain a low level? At this time, the battery power supply has been cut off, so the MCU will not work? Can Q4 and Q2 be locked? That is, MCU_CRL is in a floating state, and there is no external 4.2V, so the battery has no output at this time!


When MCU_CTL does not receive APP OFF, it is always high level output, and when APP OFF is clicked, it is always low level. Can it be realized?


Notes------The 3.3V of the chip has two sources. One is the battery. After it is separated from the base, the battery is used to step down to 3.3V through LDO for power supply. When connected to the base, the 4.2V from the base charging chip is used to charge the battery (if Q4 and Q2 are turned on at this time); at the same time, the 4.2V is stepped down to 3.3V to power the chip. The external 4.2V is directly stepped down to 3.3V by LDO to power the chip.

se7ens

QWE4562009 posted on 2021-8-19 14:10 Hello, thank you very much! The logic I want to implement is 1. In order to save power, after receiving the APP OFF...

Your needs have changed a bit.

1. The storage power saving function can definitely be realized, but 4v2 cannot. MCU_CTL is forced to be pulled low by APP, and Q4 Q2 are both turned off at this time;

2. Before leaving the warehouse for use, you must first connect the charger, activate 4V2, and power the MCU. At this time, it can ensure that Q4 Q2 is turned on;

The circuit diagram is modified as follows

Three confirmation points:

1. MCU_CTL is low after storage, but how come MCU_CTL becomes high after storage? You need to confirm what operation or command can achieve this.

(ii) After storage, 4V2 is gone, and after MCU_CTL is forced low by APP, Q4 Q2 will definitely be shut down; the battery has no output;

(III) The high and low of MCU_CTL is controlled by APP OFF, which can be realized, provided that MCU can communicate with APP, that is, MCU must be working when controlled by APP. At this time, MCU may be powered by charger 4V2 or battery.

QWE4562009

se7ens posted on 2021-8-20 10:51 QWE4562009 posted on 2021-8-19 14:10 Hello, thank you very much! The logic I want to implement is 1. In order to enter the warehouse...

Is it possible to remove R12 and D2 (direct IO control)? By default, MCU_CTL is high level. It becomes low level as soon as it receives the command from the APP, and it is high level at all other times. After leaving the warehouse, the customer needs to activate it for use, put it in the base, do not operate the APP, the bracelet is charging, the MCU is powered, MCU_CTL returns to the default high level, Q4 and Q2 are turned on, and the bracelet works. Even if it is separated from the base, it can be powered by the battery at this time.

Is it possible?

se7ens

QWE4562009 posted on 2021-8-20 14:07 Is it possible to do this, remove R12 and D2 (direct IO control). By default, MCU_CTL is high level. As long as it receives the command sent by APP, it will become low level...

It is recommended to keep R12

QWE4562009

se7ens posted on 2021-8-23 14:34 It is recommended to keep R12

Why? But I have kept it.

QWE4562009

se7ens posted on 2021-8-23 14:34 It is recommended to keep R12

Would it be better to replace R20 with a capacitor?