Hello community,
I would really appreciate any insight with reducing the power while doing a cold start for the KL03.
Some background:
I am using a KL03, custom board. I want to minimize the power used during a cold start as Vdd rises to settle at 1.8V. In order to quantify this, I didn't populate any components other than a 10k resistor between Vdd and Vss and a 1uF between Vdd and Vss. Then I would start with Vdd=0V and raise Vdd on 0.2V increments. I would look then at the power draw using an ammeter (I dont know the voltage drop of the ammeter). I also looked at the voltage on the reset pin
Here are my results:
Vdd | I_vdd | V_reset
0 0 0
0.2 0 0.17
0.4 6u 0.4
0.6 33u 0.6
0.8 180u 0.1
1.1 300u 0
1.2 440u 0
1.4 370u 0
1.6 320u 0
1.7 20u 1.7 <<--- application is supposed to use 30uA, so we are good here
Then, I removed the resistor between Reset and Vdd ( there is an internal 20k-50k pup resistor there anyway).
Vdd | I_vdd |
0 0
0.2 0
0.4 0u
0.6 40u
0.8 130u
1.0 200u
1.2 210u
1.4 150u
1.6 370u
1.7 20u
So, removing the resistor between Reset and vdd seems to help a bit at the risk of having noise on Reset. I would like to reduce the power consumption even more _WITHOUT_ external active components (resistors, caps, etc are ok).
Any suggestions on what to do would be greatly appreciated..
I don't think it's relevant, but upon powering up, the chip sets LVDV on, sets it's clock at 60kHz and then toggles an LED at ~2Hz while entering and leaving VLPS (using LPTM to wake up). The results posted are WITHOUT the LED, but the functionality was verified with the LED
