Press button to trigger battery charge display

朝朝暮暮朝朝

Press button to trigger battery charge display [Copy link]

 

Hey guys, I am going to use CN1185 to make a battery level display. I want to add a display circuit like a power bank where you press a button and the LED lights up. What else should I add?

maychang

Add a button switch, press the chip and the LEDs to work, release the button and they will stop working. Is it possible?

朝朝暮暮朝朝

maychang posted on 2020-7-17 12:56 Add a button switch, press the chip and the LEDs to work, release it and they will stop working. Is it okay?

I want to make it like a power bank, where you press a button and it displays a few seconds after you release it.

dcexpert

朝朝暮暮朝朝 posted on 2020-7-17 13:37 I want to make it like a power bank, where you press a button and it will display for a few seconds after you release it.

Add a monostable, press it to turn on the MOS and power the chip; after releasing it, it will turn off the MOS after a few seconds. The time can be adjusted by parameters.

朝朝暮暮朝朝

dcexpert posted on 2020-7-17 14:02 Add a monostable, press it to turn on the MOS and power the chip; after releasing it, it will delay for a few seconds to turn off the MOS. The time can be adjusted through parameters

Can it be achieved by charging and discharging capacitors? MOS is a bit large, and the size of the board I want to make is required to be very small

maychang

朝朝暮暮朝朝 posted on 2020-7-17 13:37 I want to make it like a power bank, where you press a button and it will display for a few seconds after you release it.

Then add a delay circuit.

The delay circuit can be made of a monostable circuit as dcexpert on the 4th floor said.

maychang

朝朝暮暮朝朝 posted on 2020-7-17 14:06 Can it be achieved by charging and discharging capacitors? MOS is a bit big, and the size of the board I want to make is very small

The capacitor that can keep four LEDs glowing for one or two seconds has a relatively large capacitance and is not very small in size.

maychang

朝朝暮暮朝朝 posted on 2020-7-17 14:06 Can it be achieved by charging and discharging capacitors? MOS is a bit big, and the size of the board I want to make is very small

The four LEDs emit light with a current of only a few mA, and the power control tube does not need to pass a larger current, so the volume will not be large. If the monostable circuit uses a digital chip, the output of the monostable chip itself is sufficient to power CN1185 and the four LEDs.

朝朝暮暮朝朝

maychang posted on 2020-7-17 15:07 The capacitor that can keep four LEDs lit for one or two seconds has a relatively large capacitance and is not very small in size.

Can I just add an RC in front like this?

maychang

朝朝暮暮朝朝 posted on 2020-7-17 15:13 Can I just add RC in front like this?

The minimum power supply voltage (the voltage being monitored) is 2.7 V and the maximum is 6 V. If you use a resistor as large as 250 kilo-ohms, will the LED still light up?

maychang

朝朝暮暮朝朝 posted on 2020-7-17 15:13 Can I just add RC in front like this?

R10 is not needed and should be removed (short-circuited).

Four LEDs, assuming each needs 1mA, a total of 4mA. After the button is released, it is required to be able to maintain 1 second. Within this 1 second, the voltage across the capacitor is reduced to 75% of the original, that is, it is allowed to drop to 4.5V, so the capacitor needs to be 2700uF. This capacitor is not very small.

maychang

朝朝暮暮朝朝 posted on 2020-7-17 15:13 Can I just add RC in front like this?

If you want to test it in practice, you can replace CN1185 and four LEDs with a 1500 ohm resistor (6V divided by 4mA), and connect a 100uF capacitor in parallel to see how long the voltage across the capacitor can be maintained after the button is released. Can it reach half a second?

朝朝暮暮朝朝

maychang posted on 2020-7-17 15:24 R10 is not needed and should be removed (short circuit). Four LEDs, assuming each requires 1mA, a total of 4mA. After the button is released, it is required to be able to maintain 1 second. ...

. . . I looked at it and it was really too big if I used a capacitor. But if I used a chip, it would also take up a lot of space. I want to make it as small as possible.

maychang

朝朝暮暮朝朝 posted on 2020-7-17 15:32 . . . I looked at it and it was indeed too big if I used a capacitor. But if I used a chip, it would also take up a lot of space. I want to make it as small as possible

If we give up the requirement of "displaying for a few seconds after releasing the button", no circuit board area will be occupied.

朝朝暮暮朝朝

maychang posted on 2020-7-17 16:51 Give up the requirement of "displaying for a few seconds after releasing", then no circuit board area will be occupied.

Do you think this works? There is no need to increase the capacitance (the component parameters are wrong, don't worry about it)

maychang

朝朝暮暮朝朝 posted on 2020-7-17 19:14 Do you think this works? There is no need to increase the capacitor (the component parameters are wrong, ignore it for now)

Of course you can.

However, three resistors plus one capacitor plus two transistors take up quite a lot of area on the circuit board.

朝朝暮暮朝朝

maychang posted on 2020-7-17 19:47 Of course it is possible. However, three resistors plus one capacitor plus two transistors do not occupy a very small area on the circuit board.

It feels better than capacitors of more than 1,000 uf . I started to think about filling the board with capacitors...

maychang

朝朝暮暮朝朝 posted on 2020-7-17 20:09 It feels better than capacitors of more than 1,000 uf. I started to think about covering the board with capacitors. . .

"I started thinking about covering the board with capacitors..."

How much does it cost? Cost is still something to consider.

maychang

朝朝暮暮朝朝 posted on 2020-7-17 20:09 It feels better than capacitors of more than 1,000 uf. I started to think about covering the board with capacitors. . .

I don't know if you noticed: if there are two transistors, the current amplification factor of the first one is β1 and the current amplification factor of the second one is β2, then the total amplification factor after connecting them in Darlington is β1·β2. The effect of the control circuit in Figure 15 is equivalent to the capacitance C2 being increased to β1·β2 times.