微控制器架构
引入| English | 释义 | 拼音 |
|---|---|---|
| microcontroller | 微控制器 | wēi kòng zhì qì |
| bus/bʌs/ | 总线 | zǒng xiàn |
| fetch-decode-execute cycle | 取指译码执行周期 | qǔ zhǐ yì mǎ zhí xíng zhōu qī |
| program counter/ˈprəʊɡræm ˈkaʊntə/ | 程序计数器 | chéng xù jì shù qì |
| clock/klɒk/ | 时钟 | shí zhōng |
| memory map | 存储器映射 | cún chǔ qì yìng shè |
A computer the size of a grain of rice
- A thermostat, an anti-lock brake controller and a wireless earbud each contain a complete computer on one tiny chip, costing less than the postage to ship it.
- This lesson opens that chip: what is inside, how it runs a program, and how fast.
Three parts on one chip
- A microcontroller 微控制器 puts flash memory (the program), SRAM (the data) and a CPU on one chip, together with peripherals for the outside world.
- Every transfer between them travels on a bus 总线: address lines say where, data lines carry what, control lines say read or write.
Match each part of a microcontroller to what it holds or does.
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All four sit on one chip, which is what makes a microcontroller cheap and small.
Fetch, decode, execute
- The CPU repeats the fetch-decode-execute cycle 取指译码执行周期. The program counter 程序计数器 (PC) holds the address of the next instruction; the CPU fetches it, decodes it and executes it.
- Then the PC moves on, or jumps for a branch. The clock 时钟 paces every step.
Put one instruction's cycle in order.
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The cycle then repeats at the new PC value, once per instruction.
The memory map
- An $N$-bit address reaches $2^{N}$ locations, so a memory of a given size needs $N = \log_{2}(\text{size})$ address lines.
- Each region of the memory map 存储器映射 sits at its own base address. Regions whose top address bits differ can never overlap.
Flash at 0x0000_0000 and SRAM at 0x2000_0000 can overlap if the flash is large enough.
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The regions differ in their top address bits, so no address can belong to both; the decoder separates them by those bits.
Worked example: sizing the map
- 64 KB of flash is $2^{16}$ bytes: 16 address lines, from
0x0000_0000to0x0000_FFFF. - 8 KB of SRAM is $2^{13}$ bytes: 13 lines, from
0x2000_0000to0x2000_1FFF. The top bits differ, so the two can never collide.
8 KB of SRAM needs how many address lines?
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$8\,\text{KB} = 8192 = 2^{13}$ bytes, so 13 lines; it spans 0x2000_0000 到 0x2000_1FFF.
32 KB of flash starts at address 0. What is its last address, in decimal?
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$32\,\text{KB} = 32\,768$ bytes, numbered 0 to 32 767, which is 0x7FFF.
Worked example: the time per instruction
- At 84 MHz and about one instruction per cycle, one instruction takes $\dfrac{1}{84 \times 10^{6}} = 11.9\,\text{ns}$.
- A 100-instruction control pass takes 1.19 µs, so a 1 ms control period has room for about 840 such passes. This budget arithmetic returns in every later lesson.
At 84 MHz, one instruction per cycle, a 100-instruction routine takes about
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$100 \times 11.9\,\text{ns} = 1.19\,\mu$s.
Three common slips
- Address lines for data width. Address lines count locations; data lines set how wide each one is.
- Bits for bytes. Flash is sized in bytes; mixing them is a factor of eight.
- Assuming a boot address. The vendor fixes the map; the linker script must match it.
A 48 MHz CPU runs a 240-instruction control routine (one instruction per cycle). How many times can it run per millisecond?
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$48\,000$ cycles per millisecond divided by 240 cycles per run gives 200 runs.
Recap
- Flash, SRAM and CPU on one chip, joined by the bus and paced by the clock.
- Fetch, decode, execute, advance the PC; repeat.
- $N$ address lines reach $2^{N}$ bytes; the clock turns instruction counts into time.