MSP430F6726的SD24_B电压值转换问题

void Init_SD24_B(void)

{
WDTCTL = WDTPW | WDTHOLD; // Stop WDT
SD24BCTL0 = SD24REFS | SD24SSEL_1; // Select internal REF
// Select SMCLK as SD24_B clock source ----8M

SD24BCCTL0 = SD24SNGL | SD24SCS_4; // Single conversion, group 0
SD24BCCTL1 = SD24SNGL | SD24SCS_4; // Left-aligned, group 0

SD24BIE = SD24IE1; // Enable channel 1 interrupt
__delay_cycles(0x3600); 

SD24BCTL1 |= SD24GRP0SC; // Set bit to start conversion
}

#if defined(__TI_COMPILER_VERSION__) || defined(__IAR_SYSTEMS_ICC__)
#pragma vector=SD24B_VECTOR
__interrupt void SD24BISR(void)
#elif defined(__GNUC__)
void __attribute__ ((interrupt(SD24B_VECTOR))) SD24BISR (void)
#else
#error Compiler not supported!
#endif
{
switch (SD24BIV)
{
case SD24BIV_SD24OVIFG: // SD24MEM Overflow
break;
case SD24BIV_SD24TRGIFG: // SD24 Trigger IFG
break;
case SD24BIV_SD24IFG0: // SD24MEM0 IFG
break;
case SD24BIV_SD24IFG1: // SD24MEM1 IFG

results1 = SD24BMEMH0; // Save CH0 results (clears IFG)
results1 = results1<<16; // Save CH0 results (clears IFG)
results1 |= SD24BMEML0; // Save CH0 results (clears IFG)
valt1 = ((results1-8388608)*75)>>19;


printf("a %d ",valt1);

results2 = SD24BMEMH1; // Save CH1 results (clears IFG)
results2 = results2<<16; // Save CH1 results (clears IFG)
results2 |= SD24BMEML1; // Save CH1 results (clears IFG)
valt2 = ((results2-8388608)*75)>>19;
// printf("b %d\n ",valt2);

SD24BCTL1 &= ~SD24GRP0SC; // Clear bit for next conversion
break;
default:break;
}
}

请教一下，MSP430F6726的SD24_B的电压值转换公式是不是 (（SD24BMEMH1<<16）|SD24BMEML1)*1.2/2^24？但

我算出的值和测量值差很多，而用上面别的公式就很准，但是我没看懂他的公式是怎么来的