I have been trying to operate MSP430FR5969 launchpad at its lowest current with VLO (~10 kHz) and MCLK divider 32. So effectively it is running at 312.5 Hz. When the board is externally powered (1.9 V), it draws about 48 uA. Note that all J13 jumpers to debugger removed, Super cap is not used (J2 and J11). All port pins are are in the default high impedance state. Having achieved 0.4 uA current consumption for MSP430G2553 launchpad before at 1.5 kHZ MCLK and 1.9V, for MSP430FR5969 with 100 uA/MHz rated current consumption, I would expect a much lower current drawing(0.03 uA + minimum current leakages). Curious, whether has somebody tried or have any idea about the minimum power consumption at the lowest clock in active mode for MSP430FR5969?
Code is below.
Thanks!
#include <msp430.h>
void main(void)
{
WDTCTL = WDTPW + WDTHOLD; // Stop watchdog timer
CSCTL0_H = CSKEY >> 8; // Unlock clock registers
CSCTL5 &= ~LFXTOFFG;
CSCTL2 = SELM__VLOCLK | SELA__VLOCLK | SELS__VLOCLK;
do
{
CSCTL5 &= ~LFXTOFFG;
SFRIFG1 &= ~OFIFG;
} while (SFRIFG1&OFIFG);
__bis_SR_register(SCG1 + SCG0 + OSCOFF);
CSCTL3 = DIVM__32 | DIVA__32;
CSCTL4 = HFXTOFF | SMCLKOFF | LFXTOFF;
CSCTL0_H = 0;
/*M5CTL0 &= ~LOCKLPM5;
P4DIR = 0xFF;
P4OUT = 0x00;
P1DIR = 0xFF;
P1OUT = 0x00;
P2DIR = 0xFF;
P2OUT = 0x00;
P3DIR = 0xFF;
P3OUT = 0x00;
PJDIR = 0xFF;
PJOUT = 0x00;*/
for (;;)
{
_delay_cycles(100);
}
}