Does shift registers consume less power when enabling 8 output pins altogether than enabling 7 output pins? I was measuring the current at Vcc pin of 74HC595 shift register with an LED connected to each output pins, and noticed that enabling all output pins drives IC more efficiently than enabling individual pins say seven out of eight. (Enabling seven LEDs consumes 44.31mA vs 33.87mA for eight)
I presume enabling all pins bypasses certain logics inside the chip but could not find related information in the datasheet. I am new to electronics and would like to hear from experts in this forum to see if this is the typical behavior of shift-registers (not just for NXP one, or for my testing specific).
# of enabled pins - Voltage on LED on PIN0 - Current at Vcc
1 - 2.96v - 7.56mA
2 - 2.91v - 14.75mA
3 - 2.89v - 22.12mA
4 - 2.88v - 28.25mA
5 - 2.86v - 34.23mA
6 - 2.84v - 39.54mA
7 - 2.82v - 44.31mA
8 - 2.86v - 33.87mA
* I am supplying 3.28v to the chip.
I presume enabling all pins bypasses certain logics inside the chip but could not find related information in the datasheet. I am new to electronics and would like to hear from experts in this forum to see if this is the typical behavior of shift-registers (not just for NXP one, or for my testing specific).
# of enabled pins - Voltage on LED on PIN0 - Current at Vcc
1 - 2.96v - 7.56mA
2 - 2.91v - 14.75mA
3 - 2.89v - 22.12mA
4 - 2.88v - 28.25mA
5 - 2.86v - 34.23mA
6 - 2.84v - 39.54mA
7 - 2.82v - 44.31mA
8 - 2.86v - 33.87mA
* I am supplying 3.28v to the chip.