芯片是6733，用XT1的32768外设晶体，设主频25M，仿真状态下运行，发现XT1LFOFFG = 0，但DCOFFG和OFIFG都 = 1，感觉很奇怪，为什么XT1错误标志没有反应，但DCO错误置位了，而且在时钟中断里，看下来也是已经在25M下运行没有错。不用XT1，用REFOCLK就没有这个问题
还有个奇怪的现象是如果我在启动晶振的程序while (SFRIFG1&OFIFG) // Test oscillator fault flag 处设断点，程序就进入死循环，因为测到OFIFG = 1了，但如果一开始不设断点，直接运行，OFIFG 还是= 1，但会跑进主程序正常运行，这是仿真器的问题吗？
另外问一下各位高手，6733能不能接XT2？还是只能接XT1？（芯片引脚只有XIN，XOUT，没有XT2IN和XT2OUT），如果只能接XT1，能不能直接上25M的晶体？因为我看到XT1 有High-frequency 模式，XT1DRIVE设为11b时的描述是 Maximum drive capability and maximum current consumption for XT1 LF mode. XT1 oscillator operating range in HF mode is 24 MHz to 32 MHz.
下附晶振初始化的代码，请看看问题出在哪里？谢谢！
void setup_FLL(void)
{//启动晶振和时钟；
// Open PMM registers for write
PMMCTL0_H = PMMPW_H;
// Set VCore to new level
PMMCTL0_L |= PMMCOREV_3;
// Wait till new level reached
if ((PMMIFG & SVMLIFG))
while ((PMMIFG & SVMLVLRIFG) == 0);
// Lock PMM registers for write access
PMMCTL0_H = 0x00;
//用TX1做FLLREFCLK，主频选DCO
// Initialize DCO to 25MHz
UCSCTL6 &= ~(XT1OFF); // XT1 On
do
{
UCSCTL7 &= ~XT1LFOFFG; // Clear XT1 fault flags
} while (UCSCTL7 & XT1LFOFFG); // Test XT1 fault flag
__bis_SR_register(SCG0); // Disable the FLL control loop
UCSCTL3 = SELREF__XT1CLK; // Set DCO FLL reference = TX1
UCSCTL0 = 0x0000; // Set lowest possible DCOx, MODx
UCSCTL1 = DCORSEL_7; // Select DCO range 50MHz operation
UCSCTL2 = FLLD_0 | 762; // Set DCO Multiplier for 25MHz
// (N + 1) * FLLRef = Fdco
// (762 + 1) * 32768 = 25MHz
// Set FLL Div（DCOCLKDIV） = fDCOCLK
__bic_SR_register(SCG0); // Enable the FLL control loop
// Worst-case settling time for the DCO when the DCO range bits have been
// changed is n x 32 x 32 x f_MCLK / f_FLL_reference.. See UCS chapter in 5xx
// UG for optimization.
// 32 x 32 x 25 MHz / 32,768 Hz = 781250 = MCLK cycles for DCO to settle
__delay_cycles(781250);
while (SFRIFG1&OFIFG) // Test oscillator fault flag
{
UCSCTL7 &= ~(XT2OFFG + XT1LFOFFG + DCOFFG); // Clear XT2,XT1,DCO fault flags
SFRIFG1 &= ~OFIFG; // Clear fault flags
}
UCSCTL4 |= SELA__XT1CLK+SELS__DCOCLKDIV + SELM__DCOCLK;//ACLK = XT1CLK;SMCLK = DCOCLKDIV;MCLK = DCOCLK
}
