Regarding the low power consumption problem of the domestic MCU HC32L110, please give me some advice.

pcf2000

Regarding the low power consumption problem of the domestic MCU HC32L110, please give me some advice. [Copy link]

 

 

I would like to ask: After HC32L110 enters deep sleep, the current is about 4uA, and it will reach 3.2uA after a long time; I plan to use the RST pin for power-on wake-up (no external reset circuit was used before), that is, the whole system restarts after power-on, and uses a 4.7K+0.1uF reset circuit (the circuit on the demo), which achieves the purpose of power-on wake-up, but the current after power-off and entering deep sleep reaches about 150uA. After analyzing the circuit, I don’t understand where this extra 100uA current comes from? ! The reset circuit is newly added

damiaa

Can't R31 be increased? Also look at the settings of other IOs. It is best to set them to fixed high or low during sleep. It is probably caused by external IO. You can check them one by one. You will find it quickly.

pcf2000

damiaa posted on 2020-7-17 17:08 Can't R31 be enlarged? Also look at the settings of other IOs. It is best to set them to a fixed high or low during sleep. . It is probably caused by external IO. You can check them one by one...

The external 5V power supply is gone, and only a button battery is used to power the MCU. All IOs are set to output mode and output 0, entering deep sleep mode.

When no external reset circuit is used, the sleep current is less than 4uA.

damiaa

pcf2000 published on 2020-7-17 17:15 The external 5V power supply is gone, only a button battery is used to power the MCU, all IOs are set to output mode and output 0, and enter deep sleep mode...

Increase the resistor of the reset circuit. 100K? Try it first. Are all other circuits disconnected? Add a diode 1N4148 in reverse parallel to C1. This way, it can be reset reliably after powering off and then on again.

littleshrimp

Is the 5V connected directly to the outside world through the terminal? Are there any other circuits?

When you say power-on wake-up, do you mean inserting 5V to wake up the MCU after it loses power?

Measure the voltage of the reset pin in power-down mode of the two solutions to see what the difference is

pcf2000

damiaa posted on 2020-7-17 17:17 pcf2000 posted on 2020-7-17 17:15 The external 5V power supply is gone, only a button battery is used to power the MCU, and all IOs are set to input...

Increasing the resistance of the reset circuit can reduce the sleep current. A 200K resistor was used, and the current was around 15uA, but the system could not be reset after the main power was restored.

After C1 is connected in parallel with a diode 1N4148, the system can be reset, but the sleep current is about 160uA, and 1N4148 is equivalent to a small resistance resistor.

There is no external 5V, and RST is in input mode, so how can there be current consumption? RST has an internal pull-up resistor, is it caused by the internal pull-up resistor? !

chunyang

The built-in pull-up resistor will discharge leakage current to the external 5V input side through R31. You should carefully check the manual to see whether the RST pin can be configured as a high-impedance input state.

chunyang

The OP should measure the voltage across C1 after the device has been in sleep mode for a long enough time. This is the key to determining whether RST can be used to wake it up.

chunyang

pcf2000 posted on 2020-7-17 17:45 Increasing the resistance of the reset circuit can reduce the sleep current. A 200K resistor is used, and the current is about 15uA, but the system cannot be restored after the main power is restored...

The circuit should be reset at a low level, but C1 is wrong in reverse connection with a diode. Are you sure?

damiaa

damiaa posted on 2020-7-17 17:17 pcf2000 posted on 2020-7-17 17:15 The external 5V power supply is gone, only a button battery is used to power the MCU, and all IOs are set to input...

The typical reverse current leakage current of 1N4148 is 25nA at reverse voltage. Calculate its reverse resistance. A small resistor, connected in the forward direction, right?

pcf2000

As shown in the figure, the connection method of the diode should be like this, which can quickly discharge the charge on the capacitor for the next reliable reset. However, since the diode can quickly discharge the charge on the capacitor, the external diode plus the pull-up circuit inside the MCU forms a current loop. It is not surprising that current is generated. The hateful thing is that the pull-up resistor inside RST (P00) cannot be modified (I haven't seen where it can be set in the current information). It seems that other types of external reset circuits should be considered.

pcf2000

chunyang posted on 2020-7-17 19:51 The built-in pull-up resistor will discharge leakage current to the external 5V input side through R31. You should carefully check the manual to see if the RST pin can be configured as a high-impedance input state.

"When the external reset pin detects a low level, a system reset will be generated. The reset pin has a built-in pull-up resistor and
an integrated glitch filter circuit. The glitch filter circuit will filter out glitch signals less than 20uS (typical value). Therefore
, the low-level signal added to the reset pin must be greater than 20uS to ensure reliable reset of the chip."

"－ This port has a built-in pull-up resistor, so even if this pin is used as an input port, it can only be used as an "input
port with a pull-up function."

This is the description of the RST pin in the official document. It should be fixed and cannot be changed.

chunyang

pcf2000 posted on 2020-7-18 12:59 "When the external reset pin detects a low level, a system reset is generated. The reset pin has a built-in pull-up resistor and integrates a rough...

Then you cannot use the RST reset method to exit the sleep mode. Check the device manual to see whether it supports external interrupt or IO level change wake-up function.

pcf2000

chunyang posted on 2020-7-18 15:35 Then you can't use the RST reset method to exit the sleep mode. Check the device manual to see if it supports external interrupts or IO level change wake-up functions.

At present, RST is still used to reset the system. A reset circuit built by ourselves can achieve the purpose, mainly considering that hardware reset is simple and reliable.

But what I can't understand is that when the main power is turned off and the device enters sleep mode, the current starts to fluctuate: it starts at 3.6uA, then jumps to 6 or 7uA, and after a night of power-on testing (always sleeping), it drops to 2.4uA. It feels like the current fluctuates when the device just enters sleep mode.

chunyang

pcf2000 posted on 2020-7-19 11:16 At present, RST is still used to reset the system. A reset circuit built by myself can achieve the purpose, mainly considering the hardware reset...

This requires analysis of the load components and their respective impacts, as well as reviewing your testing methodology.