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问 利用dco能否实现串行9600的设计？？如果能，需要怎样做？？ 答 1: dco直接用于串行通讯的时钟不合适，还要有外部校正。这样还不如直接用外部晶振做时钟源。 答 2: 我用的是MSP430f1121，请问，要实现96000的串行通讯，外部的晶振最小能够用多大的？如何计算？谢谢！ 答 3: 用timera做只能整数分频，就是时钟除以波特率的结果是整数，小数部分不能产生。分频数越大误差越小。您可以参考下面ti给出的例子。
//******************************************************************************
// MSP-FET430x110 Demo - Timer_A UART 9600 Echo, HF XTAL ACLK
//
// Description: Use timer_A CCR0 hardware output modes and SCCI data latch to
// to implement UART function 9600 baud. Software does not directly read and
// write to RX and TX pins, instead proper use of output modes and SCCI data
// latch are demonstrated. Use of these hardware features eliminates ISR
// latency effects as hardware insures that output and input bit latching and
// timing are perfectly synchronised with timer_A regardless of other
// software activity. In the Mainloop the UART function readies the UART to
// receive one character and waits in LPM0 with all activity interrupt driven.
// After a character has been received, the UART receive function forces exit
// from LPM0 in the Mainloop which echo's back the received character.
// ACLK = MCLK = TACLK = HF XTAL = 3.579545MHz
// //*An external 3.579545Hz XTAL on XIN XOUT is required for ACLK*//
//
// MSP430F1121
// -----------------
// /|\| XIN|-
// | | | 3.58Mhz
// --|RST XOUT|-
// | |
// | CCI0A/TXD/P1.1|-------->
// | | 9600 8N1
// | CCI0B/RXD/P2.2|<--------
//
#define RXD 0x04 // RXD on P2.2
#define TXD 0x02 // TXD on P1.1

// Conditions for 9600 Baud HW/SW UART, ACLK = 3.579545MHz

#define Bitime_5 0x0BA // ~ 0.5 bit length
#define Bitime 0x175 // 104 us ~ 9596 baud

unsigned int RXTXData;
unsigned char BitCnt;

void TX_Byte (void);
void RX_Ready (void);

// M.Buccini
// Texas Instruments, Inc
// March 2002
//******************************************************************************

#include <MSP430x11x1.h>

void main (void)
{
unsigned int i;
WDTCTL = WDTPW + WDTHOLD; // Stop watchdog timer

BCSCTL1 |= XTS; // ACLK = LFXT1 = HF XTAL

do
{
IFG1 &= ~OFIFG; // Clear OSCFault flag
for (i = 0xFF; i > 0; i--); // Time for flag to set
}
while ( IFG1 & OFIFG ); // OSCFault flag still set?

BCSCTL2 |= SELM1+SELM0; // MCLK = LFXT1 (safe)

CCTL0 = OUT; // TXD Idle as Mark
TACTL = TASSEL0+MC1; // ACLK, continous mode
P1SEL = TXD; // P1.1/TA0 for TXD function
P1DIR = TXD; // TXD output on P1
P2SEL = RXD; // P2.2/TA0 as RXD input

// Mainloop
for (;;)
{
RX_Ready(); // UART ready to RX one Byte
_BIS_SR(CPUOFF+GIE); // Enter LPM0 Until character RXed
TX_Byte(); // TX Back RXed Byte Received
}
}


// Function Transmits Character from RXTXData Buffer
void TX_Byte (void)
{
BitCnt = 0xA; // Load Bit counter, 8data + ST/SP
CCR0 = TAR; // Current state of TA counter
CCR0 += Bitime; // Some time till first bit
RXTXData |= 0x100; // Add mark stop bit to RXTXData
RXTXData = RXTXData << 1; // Add space start bit
CCTL0 = OUTMOD0+CCIE; // TXD = mark = idle
while ( CCTL0 & CCIE ); // Wait for TX completion
}


// Function Readies UART to Receive Character into RXTXData Buffer
void RX_Ready (void)
{
BitCnt = 0x8; // Load Bit counter
CCTL0 = SCS+CCIS0+OUTMOD0+CM1+CAP+CCIE; // Sync, Neg Edge, Capture
}


// Timer A0 interrupt service routine
interrupt[TIMERA0_VECTOR] void Timer_A (void)
{
CCR0 += Bitime; // Add Offset to CCR0

// RX
if (CCTL0 & CCIS0) // RX on CCI0B?
{
if( CCTL0 & CAP ) // Capture mode = start bit edge
{
CCTL0 &= ~ CAP; // Switch from capture to compare mode
CCR0 += Bitime_5;
}
else
{
RXTXData = RXTXData >> 1;
if (CCTL0 & SCCI) // Get bit waiting in receive latch
RXTXData |= 0x80;
BitCnt --; // All bits RXed?
if ( BitCnt == 0)
//>>>>>>>>>> Decode of Received Byte Here <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
{
CCTL0 &= ~ CCIE; // All bits RXed, disable interrupt
_BIC_SR_IRQ(CPUOFF); // Clear LPM0 bits from 0(SR)
}
//>>>>>>>>>> Decode of Received Byte Here <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
}
}
// TX
else
{
if ( BitCnt == 0)
CCTL0 &= ~ CCIE; // All bits TXed, disable interrupt
else
{
CCTL0 |= OUTMOD2; // TX Space
if (RXTXData & 0x01)
CCTL0 &= ~ OUTMOD2; // TX Mark
RXTXData = RXTXData >> 1;
BitCnt --;
}
}
} 答 4: 谢谢刘工，我明白了！