LED blinker draws high current during switching, how to prevent current spikes

Question

I have the circuit below:

Now focus your attention on TP4056 charging and protection module because it has a DW01A battery protection circuit in it which detects over current at 3A. Also focus your attention to SX1308 boost converter which boosts single cell voltage to 24V which means it draws high current. Also beware that J10 will be connected with an alarm which draws around 140mA and the flasher LED which draws 125mA current in steady state. The flasher flashes at 2.5Hz (200ms on 200ms off) rate. Also notice that I've added a 470uF capacitor between pin 3 and 4 of SX1308 to lower the voltage ripple, because originally the SX1308 had a 33uF capacitor at the output.

Problem:

When alarm starts to ring (when thr microcontroller activates Q2) in a really brief moment the alarm and also the whole system suddenly stops. When I remove the 470uF capacitor it rings longer but the sound is discrete and eventually the system completely turns off.

The boost converter draws:

(140+125) * 8 * 100/90 = 2355 mA

Note: "8" comes form the power equality V1 I1 = V2 I2 and we are boosting 3V to 24V. "100/90" comes because of efficiency.

That is less than 3A. When I remove the flasher LED, everything works fine. When the flasher LED is connected aline, it works also fine.

My diagnostic:

During the turn on of the flasher LED it discharges the capacitor instantly. When it is turned off the capacitor is quite empty and it draws high current during charging. If this were to happen once, the battery protector wouldn't make it a problem because it has a high current tolerance duration around 50ms, but this high current is constantly being drawn several times a second and it triggers the protection circuit thus shuts the system down.

Question1:

How can I lower the inrush current of the output capacitor of a boost converter which has a switching load which causes current spikes?

Question2:

Can I use pin 3 and 4 of U1 to power the system and leave 5-6 floating to disable current protection circuit but only keeping the charging circuit part?

Someone asked for a picture but I don't know what exactly they wanted, anyway the picture:

Boost converter is the one at front left.

Answer 1

• A boost converter is best controlled by its "enable" pin. You can leave its input powered always on and switch only the enable pin. The converter will take care about starting and stopping gracefully.

Of course, check its power drain in "disabled" state (it may or may not fit your use case).

Be also aware that a disabled boost converter usually has its input voltage (unboosted) at the output. This also may or may not work for you.

• Another approach would be to get rid of it and use different LED topology in order to power your LEDs with the available battery voltage.

• If you still insist converting your power twice (first in the boost converter and then in the LED driver) you can add a capacitor between Q2 source and gate in order to introduce some gradual switching on at Q3. 1uF will probably do.

Answer 2

First, IRLML6402 has quite high RdsON at low Vgs, so it introduces an unnecessary voltage drop at the input of the boost converter, which will cause it to draw more current. You should remove the FET and use the enable input of the boost converter, according to the datasheet it draws less than 1µA when disabled.

Since both the alarm and the flasher are powered from 24V, I'd use one MOSFET switch on each, which offers two features:

• Both loads can be completely cut off, this is useful when the boost is disabled, as it will still output the input voltage.

• Flasher and alarm will not be on at the same time, instead they should be powered one at a time. Flash, beep, flash, beep. This should reduce your peak current by half and solve your problem.

You can use two low side MOS switches, it's simpler.

Answer 3

Anyway, the problem wasn't the inrush current. When a high current is drawn from boost converter even nothing is being turned on and off, it needs to pull very high current from input. Either 2x2000 mAh parallel Lithium-ion batteries having hard time supplying that voltage or the boost converter doesn't support 2.5 ish Ampers currents. I will buy a bigger module with higher ratings or I will design a boost converter myself with an ASIC. I will also use 2 series lithium batteries at the input to make the boost process easier (draws less current). I will supply my MCU with a seperate single Li-ion battery. There will be 3 batteries (2 in series, 1 alone) I will control the alarm circuit (high voltage high current) with optocoupler from MCU (Low voltage low current). This project taught me many things and there are still many unknows. For example, the output current of Sx1308 boost converter module is rated with 2A current while I am drawing only 0.3 Amps from output. It is also rated to boost from 2V to 24 which I am doing is form 3.5 to 24. So I am not going over limits. Also I can pull 8A from batteries if we follow 2*C rule. I only draw 2.5A. Not over limits.

Consequently, there are things I couldn't figure out but I can take extra precautions to make it work. I hope. Still you can guide me, thanks.