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

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Prince Charming

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I have the circuit below: [moderator action: removed link to external file server] Now focus your attention on TP4056 charging and protection module because it has 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. Flasher flashes in 2.5Hz (200ms on 200ms off) rate. Also notice that I've added 470uF capacitor between pin 3 and 4 of SX1308 to lower the voltage ripple. Because originally SX1308 has 33uF capacitor at the output.

Problem: When alarm starts to ring (as 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. But 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.
which is les than 3A. When I remove the Flasher LED, everything works 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 for 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 in a second and it triggers the protection circuit thus shuts the system down.

Question1: How to lower the inrush current of 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?
 
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Are you sure the buzzer isn't overloading the MCU?
How does the function of Q2 and Q3 relate to the function of Q4?
I'm not familiar with that boost module, are pins 2 and 4 internally linked?

Brian.
 

Are you sure the buzzer isn't overloading the MCU?
Yes, buzzer draws 10 mA and I made billion test with it, no faulty yet.
How does the function of Q2 and Q3 relate to the function of Q4?
Q3 is a P-MOS and Q2 is N-MOS which pulls the gate of P-MOS to ground in order to turn on the Q3. When Q3 is on the boost converter is powered.
I'm not familiar with that boost module, are pins 2 and 4 internally linked?
Yes.
 

I fear the thread is useless without a schematic.
Thx to Klaus who deleted my schematic.
Anyway, the problem wasn't the inrush current. When a high current is drawn from boost converter, 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 Li-ion battery. 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 covnerter 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.
 
When I remove the Flasher LED, everything works fine.

No doubt your own diagnosis is correct as to shutdown caused by repeated inrush current. There's another factor caused by sudden removal of a load from the boost converter. Unless regulation is super-quick to reduce duty cycle, only a few cycles need occur to drive capacitor charge soaring to alarmingly high volt levels. It's easy to observe with an oscilloscope.

In any case consider making two stages of capacitive filtering, with resistor in between. The second capacitor varies widely as the load turns on and off. The first capacitor stays at more or less stable voltage, serving as a reservoir. Although the resistor wastes power (as heat), try adjusting values until your shutdown problem is cured, yet with acceptable power loss.

two stages capacitive smoothing for pulsating load.png
 

No doubt your own diagnosis is correct as to shutdown caused by repeated inrush current. There's another factor caused by sudden removal of a load from the boost converter. Unless regulation is super-quick to reduce duty cycle, only a few cycles need occur to drive capacitor charge soaring to alarmingly high volt levels. It's easy to observe with an oscilloscope.

In any case consider making two stages of capacitive filtering, with resistor in between. The second capacitor varies widely as the load turns on and off. The first capacitor stays at more or less stable voltage, serving as a reservoir. Although the resistor wastes power (as heat), try adjusting values until your shutdown problem is cured, yet with acceptable power loss.

View attachment 179251
So that when the load is disconnected, there will still be a load for a brief time, this load will go lighter and lighter over time and give enough time to boost converter to adjust its duty cycle. The load I am talking about here is the second capacitor. Well this is enlightening thank you.

But two questions in my mind.
1. I've seen filters which are similar to yours but there is an inductor instead of resistor in between capacitors. How about it? It will help simulate the load when real load is disconnected because it will continue to consume same current for some time and gradually decrease the load current. This current will charge the capacitor.

2. I think other reason for the shutdown could be the current drawn from the lithium ion batteries. It is actually 2.5A and it seems to not disturb 2 parallel 2000mAh. Mine is just a feeling. Maybe if I used 2 series batteries to boost from, there would be less current drawn batteries and the batteries would be able to supply enough current to the boost converter. Do you think that could be the case?
 

I have the circuit below: [moderator action: removed link to external file server] Now focus your attention on TP4056 charging and protection module because it has 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. Flasher flashes in 2.5Hz (200ms on 200ms off) rate. Also notice that I've added 470uF capacitor between pin 3 and 4 of SX1308 to lower the voltage ripple. Because originally SX1308 has 33uF capacitor at the output.

Problem: When alarm starts to ring (as 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. But the boost converter draws:



which is les than 3A. When I remove the Flasher LED, everything works 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 for 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 in a second and it triggers the protection circuit thus shuts the system down.

Question1: How to lower the inrush current of 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?
Compare Po/Pin = efficiency and see if more bulk Ec energy storage will help. Boost V regulators also boost source current and this transformer law is unavoidable.
 

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