




版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進(jìn)行舉報或認(rèn)領(lǐng)
文檔簡介
1、第四講 DSP外設(shè)應(yīng)用之系統(tǒng)時鐘 系統(tǒng)時鐘,即為各個模塊產(chǎn)生所需要的時鐘,如C55x core、慢速外設(shè)(Slow Peripherals),快速外設(shè)(Fast Peripherals)以及其它外設(shè)所需的基準(zhǔn)時鐘。系統(tǒng)時鐘的設(shè)置是任何一個可編程器件必須進(jìn)行的初始化操作。 在DSP5502中,系統(tǒng)的時鐘初始化語句為:PLL_setFreq(1, 0xC, 0, 1, 3, 3, 0);該語句為CSL(Chip Support Library)庫函數(shù)語句,在進(jìn)行時鐘設(shè)置時,系統(tǒng)調(diào)用該API初始化函數(shù),以完成系統(tǒng)設(shè)置,對于C55x 5502所涉及的時鐘寄存器如下表所示:系統(tǒng)涉及的函數(shù)原型為void
2、PLL_setFreq (Uint16 mode, Uint16 mul, Uint16 div0, Uint16 div1, Uint16 div2,Uint16 div3, Uint16 oscdiv);Uint16 mode / PLL mode /PLL_PLLCSR_PLLEN_BYP ASS_MODE /PLL_PLLCSR_PLLEN_PLL_MODEUint16 mul / Multiply factor, Valid values are (multiply by) 2 to 15.Uint16 div0 / Sysclk 0 Divide Down, Valid value
3、s are 0, (divide by 1) /to 31 (divide by 32)Uint16 div1 / Sysclk1 Divider, Valid values are 0, 1, and 3 corresponding/to divide by 1, 2, and 4 respectivelyUint16 div2 / Sysclk2 Divider, Valid values are 0, 1, and 3 /corresponding to divide by 1, 2, and 4 respectivelyUint16 div3 / Sysclk3 Divider, Va
4、lid values are 0, 1 and 3 /corresponding to divide by 1, 2 and 4 respectivelyUint16 oscdiv / CLKOUT3(DSP core clock) divider,Valid values are 0 /(divide by 1) to 31 (divide by 32)程序中,對于MODE,則5502有兩種模式:PLL旁路模式和PLL使能模式,前者是時鐘未經(jīng)PLL進(jìn)行倍頻,而后者使用PLL功能。由于目前無源晶振生產(chǎn)工藝限制,其所能產(chǎn)生的頻率超過30即會有較大的誤差,而5502最高可達(dá)到300M時鐘,一般需要
5、使能PLL功能。其它參數(shù)均為各除法器的值,查詢相應(yīng)的寄存器即可完成。表1 所涉及的PLL寄存器及其各相關(guān)位PLLCSR PLLEN, PLLPWRDN, OSCPWRDN, PLLRST, LOCK, STABLEPLLM PLLMPLLDIV0 PLLDIV0, D0ENPLLDIV1 PLLDIV1, D1ENPLLDIV2 PLLDIV2, D2ENPLLDIV3 PLLDIV3, D3ENOSCDIV1 OSCDIV1, OD1ENWAKEUP WKEN0, WKEN1, WKEN2, WKEN3CLKMD CLKMD0CLKOUTSRCLKOUTDIS, CLKOSEL圖1 系統(tǒng)時鐘
6、發(fā)生器圖2 晶振及其時鐘產(chǎn)生電路圖3 內(nèi)部時鐘頻率范圍值圖4 時鐘發(fā)生器寄存器附各個寄存器相關(guān)位說明(1) PLL Control / Status Register (PLLCSR) (0x1c80)nSTABLE6R1Oscillator output stable. This bit indicates if the OSCOUT output has stabilized. STABLE = 0: Oscillator output is not yet stable. Oscillator counter is not done counting 41,032 reference c
7、lock cycles.STABLE = 1: Oscillator output is stable. This is true if any one of the three cases is true:a) Oscillator counter has finished counting.b) Oscillator counter is disabled.c) Test mode.LOCK5R0Lock mode indicator. This bit indicates whether the clock generator is in its lock mode.LOCK = 0:
8、The PLL is in the process of getting a phase lock.LOCK = 1: The clock generator is in the lock mode. The PLL has a phase lock and the output clock of the PLL has the frequency determined by the PLLM register and PLLDIV0 register.PLLRST3R/W1Asserts RESET to PLLPLLRST = 0: PLL reset releasedPLLRST = 1
9、: PLL reset assertedOSCPWRDN2R/W0Sets internal oscillator to power-down modeOSCPWRDN = 0: Oscillator operationalOSCPWRDN = 1: Oscillator set to power-down mode based onstate of CLKMD0 bit of Clock Mode Control Register (CLKMD).When CLKMD0 = 0, the internal oscillator is set to power-down mode when t
10、he clock generator is set to its idle mode CLKIS bit of the IDLE Status Register (ISTR) becomes 1.When CLKMD0 = 1, the internal oscillator is set to power-down mode immediately after the OSCPWRDN bit is set to 1.PLLPWRDN1R/W0Selects PLL power downPLLPWRDN = 0: PLL operational PLLPWRDN = 1: PLL place
11、d in power-down statePLLEN0R/W0PLL mode enable. This bit controls the multiplexer before dividers D1, D2, and D3.PLLEN = 0: Bypass mode. Divider D1 and PLL are bypassed. SYSCLK1 to 3 divided downdirectly from input reference clock. PLLEN = 1: PLL mode. Divider D1 and PLL are not bypassed. SYSCLK1 to
12、 3 divided down from PLL output.(2) PLL Multiplier Control Register (PLLM)15-54-0ReservedPLLMPLLM4:0R/W00000PLL multiplier-selectPLLM = 0000000001: ReservedPLLM = 00010: Times 2PLLM = 00011: Times 3PLLM = 00100: Times 4PLLM = 00101: Times 5PLLM = 00110: Times 6PLLM = 00111: Times 7PLLM = 01000: Time
13、s 8PLLM = 01001: Times 9PLLM = 01010: Times 10PLLM = 01011: Times 11PLLM = 01100: Times 12PLLM = 01101: Times 13PLLM = 01110: Times 14PLLM = 01111: Times 15PLLM = 1000011111: Reserved(3) PLL Divider 0 Register (PLLDIV0) (Prescaler)1514-54-0D0ENReservedPLLDIV0D0EN15R/W1Divider D0 enableD0EN = 0: Divi
14、der 0 disabledD0EN = 1: Divider 0 enabledPLLDIV04:0R/W00000Divider D0 ratioPLLDIV0 = 00000: Divide by 1PLLDIV0 = 00001: Divide by 2PLLDIV0 = 00010: Divide by 3.PLLDIV0 = 11111: Divide by 32(4)PLL Divider1 Register (PLLDIV1) for SYSCLK11514-54-0D1ENReservedPLLDIV1D1EN15R/W1Divider D1 enableD1EN = 0:
15、Divider 1 disabledD1EN = 1: Divider 1 enabledPLLDIV14:0R/W00011Divider D1 ratio (SYSCLK1 divider)PLLDIV1 = 00000: Divide by 1PLLDIV1 = 00001: Divide by 2PLLDIV1 = 00010: ReservedPLLDIV1 = 00011: Divide by 4PLLDIV1 = 0010011111: ReservedPLLDIV2和PLLDIV3的位定義與PLLDIV1完全一樣,在此不再重復(fù)寫了。(5) Oscillator Divider1
16、 Register (OSCDIV1) for CLKOUT31514-54-0OD1ENReservedOSCDIV1OD1EN15R/W0Divider D0 enableD0EN = 0: Divider 0 disabledD0EN = 1: Divider 0 enabledOSCDIV14:0R/W00000Divider D0 ratioPLLDIV0 = 00000: Divide by 1PLLDIV0 = 00001: Divide by 2PLLDIV0 = 00010: Divide by 3.PLLDIV0 = 11111: Divide by 32(6) CLKOUT3 Select Register (CK3SEL)15-43-0ReservedCK3SELCK3SEL3:0R/W1011Output on C
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時也不承擔(dān)用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。
最新文檔
- 裝箱運(yùn)輸合同樣本
- 分包單位的合同范本
- 代理銷售中介合同范本
- 農(nóng)民集資建房合同范本
- 弘揚(yáng)中華優(yōu)良傳統(tǒng)文化過中國人自己的傳統(tǒng)節(jié)日單元整體教學(xué)設(shè)計
- 做好班主任 做一名有智慧的班主任 校園廉潔 14
- 2025家庭居室設(shè)計施工一體化合同
- 2025機(jī)電安裝工程合同乙種本范本
- 2025YY年房屋租賃合同協(xié)議
- 語文核心素養(yǎng)的培育知到課后答案智慧樹章節(jié)測試答案2025年春湖南師范大學(xué)
- 第四課 人民民主專政的社會主義國家 課件-高考政治一輪復(fù)習(xí)統(tǒng)編版必修三政治與法治
- 2025年鄭州黃河護(hù)理職業(yè)學(xué)院單招職業(yè)適應(yīng)性考試題庫帶答案
- (完整版)特殊教育與隨班就讀
- 旋流風(fēng)口RA-N3選型計算表格
- 《VB程序結(jié)構(gòu)基礎(chǔ)》課件教程
- 個人房屋租賃合同標(biāo)準(zhǔn)版范本
- DBJ50-T-157-2022房屋建筑和市政基礎(chǔ)設(shè)施工程施工現(xiàn)場從業(yè)人員配備標(biāo)準(zhǔn)
- 2024年中考模擬試卷地理(湖北卷)
- 沙塘灣二級漁港防波堤工程施工組織設(shè)計
- 大學(xué)生心理健康教育知到智慧樹章節(jié)測試課后答案2024年秋長春醫(yī)學(xué)高等??茖W(xué)校
- 慢腎風(fēng)中醫(yī)辨證施護(hù)
評論
0/150
提交評論