Microcontroller architecture and programming model

Why it matters

Almost every modern instrument, from a smart pressure transmitter to a pulse oximeter, has a microcontroller at its heart. A microcontroller puts the CPU, program memory, data memory, timers, serial ports, ADC and I/O pins on one chip, so the programming model (which registers exist, where memory lives and how long instructions take) decides how you write the firmware and how fast it can respond.

Key ideas

Microcontroller versus microprocessor. A microprocessor is a CPU that needs external memory and peripherals; it is optimised for computing power. A microcontroller integrates CPU, flash/ROM, RAM, EEPROM, timers, UART/SPI/I²C, ADC, PWM and GPIO on one chip; it is optimised for low cost, low power and direct control of hardware. Most use a Harvard architecture, with separate program and data memories.

The 8051 programming model (classic reference).

• 8-bit CPU, 16-bit PC and DPTR (data pointer); 64 KB program address space and 64 KB external data space.
• On-chip: 4 KB ROM, 128 bytes RAM, four 8-bit ports P0–P3, two 16-bit timers/counters, one full-duplex UART, five interrupt sources (INT0, Timer 0, INT1, Timer 1, serial).
• Internal RAM map: 00H–1FH four register banks of R0–R7 (bank chosen by RS1 RS0 in the PSW); 20H–2FH bit-addressable area (128 bits); 30H–7FH general scratch-pad. SFRs (A, B, PSW, SP, DPTR, port latches, timer and serial registers) sit at 80H–FFH.
• SP resets to 07H, so the stack starts at 08H (inside bank 1); programs that use bank 1 move SP higher.
• PSW bits: CY (carry), AC (auxiliary carry), F0 (user flag), RS1, RS0 (bank select), OV (signed overflow), P (parity of A: 1 when A has an odd number of 1s).
• Timing: one machine cycle = 12 oscillator periods. At 12 MHz a machine cycle is 1 µs; at 11.0592 MHz (chosen for standard baud rates) it is 1.085 µs. Most instructions take 1 or 2 machine cycles; MUL and DIV take 4.
• Port 0 is open-drain and needs pull-ups when used as general I/O; it also carries the multiplexed low address/data bus for external memory, with ALE latching the address.

Addressing modes. Immediate (MOV A, #25H), register (MOV A, R3), direct (MOV A, 40H), register-indirect (MOV A, @R0), indexed (MOVC A, @A+DPTR for look-up tables in code memory), and bit addressing (SETB P1.0).

Modern families. AVR and PIC are 8-bit RISC Harvard machines executing most instructions in one or a few clocks; ARM Cortex-M parts are 32-bit with nested vectored interrupt controllers, DMA and much richer peripherals. The concepts (register file, memory map, peripheral registers, interrupts, timers) carry over directly.

Programming. Firmware is written in C or assembly, compiled to the program memory (flash), and talks to peripherals by reading and writing their special-function registers. Typical structure: initialise clocks, ports and peripherals; then a main loop plus interrupt service routines. A watchdog timer resets the chip if the firmware stops servicing it.

Formulas

T_MC = 12 / f_osc (classic 8051)

• T_MC: machine-cycle time (s); f_osc: crystal frequency (Hz).

t_instr = (machine cycles) × T_MC

t_instr = (clock cycles) / f_clk (single-clock RISC parts)

Register bank address of R0 = 8 × (2·RS1 + RS0) (in hex: 00H, 08H, 10H, 18H)

ADC step = V_ref / 2ⁿ (on-chip n-bit ADC)

Worked examples

Example 1 (standard). An 8051 runs from a 12 MHz crystal. Find the machine-cycle time and the time taken by DJNZ (2 machine cycles). Repeat for 11.0592 MHz.

1. T_MC = 12 / 12 MHz = 1 µs; DJNZ = 2 × 1 µs = 2 µs.
2. At 11.0592 MHz: T_MC = 12 / 11.0592 MHz = 1.085 µs; DJNZ = 2.170 µs.

Answer: 1 µs and 2 µs; 1.085 µs and 2.17 µs

Example 2 (GATE level: nested delay). At 12 MHz, find the delay of: MOV R2, #200 (1 MC); L1: MOV R3, #250 (1 MC); L2: DJNZ R3, L2 (2 MC); DJNZ R2, L1 (2 MC).

1. Inner loop: 250 × 2 = 500 MC.
2. One outer pass: 1 (MOV R3) + 500 + 2 (DJNZ R2) = 503 MC.
3. Total: 1 + 200 × 503 = 100 601 MC.
4. Delay = 100 601 × 1 µs = 100.6 ms.

Answer: about 100.6 ms

Example 3 (flags). Execute MOV A, #9AH then ADD A, #7BH. Find A, CY, AC, OV and P.

1. 9AH + 7BH = 115H, so A = 15H and CY = 1.
2. Low nibbles: AH + BH = 15H > FH, so AC = 1.
3. Signed: 9AH = −102, 7BH = +123; the sum +21 fits in −128 to +127, so OV = 0 (operands of opposite sign can never overflow).
4. A = 15H = 0001 0101 has three 1s (odd), so P = 1.

Answer: A = 15H, CY = 1, AC = 1, OV = 0, P = 1

Common mistakes

• Using the crystal period as the 8051 instruction time; one machine cycle is 12 oscillator periods.
• Forgetting that the stack starts at 08H and overwrites register bank 1.
• Confusing CY (unsigned carry) with OV (signed overflow).
• Leaving Port 0 pins without pull-up resistors when used as outputs.
• Writing delays in software without counting the loop-setup and the final non-taken branch.

For GATE IN

Questions give a short assembly program and ask for register, memory or flag contents afterwards, the execution time or delay of a loop for a given crystal, the memory map and register-bank addresses, or the purpose of a peripheral. Practise tracing programs instruction by instruction and counting machine cycles for nested loops.

Quick check

1. Machine-cycle time of an 8051 with a 24 MHz crystal?
2. With RS1 RS0 = 10, at which internal RAM address is R0?
3. What is SP after reset?
4. An instruction needs 4 clock cycles on a 16 MHz single-clock MCU. How long does it take?

Answers: 1. 0.5 µs 2. 10H 3. 07H 4. 0.25 µs