Interrupts and timers
Introduit| English | Sens |
|---|---|
| interrupt/ˈɪntərʌpt/ | Interrupteur |
| interrupt service routine/ˈɪntərʌpt ˈsɜːvɪs ruːˈtiːn/ | Routine de traitement d'interruption |
| timer | Minuteur |
| prescaler | prédiviseur |
A week or a year
- The sensor node must run for a year on one battery. If the CPU polls all day, the battery is empty in about a week.
- Interrupts let the CPU sleep until something needs it. This lesson shows how they work and what they cost.
The interrupt
- An interrupt 中断 stops the main program where it is. The hardware saves the return state and jumps to an interrupt service routine 中断服务程序 (ISR).
- The NVIC (nested vectored interrupt controller) picks the ISR and ranks priorities. The ISR runs, clears its flag and returns.
Put the handling of one interrupt in order.
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Save, jump, serve, clear, return. Forget the clear and the last step never comes.
The timer
- A timer 定时器 is a counter clocked through a divider, the prescaler 分频器. It raises an interrupt each time it reaches its reload value.
- Two divisions turn a fast clock into a slow, exact tick.
The timer equation
- The interrupt rate is $f_{\text{int}} = \dfrac{f_{\text{clock}}}{\text{prescaler} \times \text{reload}}$.
- Both factors live in 16-bit registers, so each must be at most 65 535. A design is valid only when both fit and the product equals the clock divided by the rate.
A timer runs from 84 MHz with prescaler 84 and reload 1000. What is the interrupt rate in Hz?
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$84 \times 10^{6} / (84 \times 1000) = 1000$ Hz: a 1 ms tick, with both factors inside 16 bits.
Worked example: a 1 Hz heartbeat
- From 84 MHz, prescaler 8400 gives 10 kHz. Reload 10 000 then gives exactly 1 Hz.
- Check: 8400 and 10 000 both fit 16 bits, and $8400 \times 10\,000 = 84 \times 10^{6}$. A prescaler of 84 000 would not fit.
From an 84 MHz clock, prescaler 4200 and reload 20 000 give an interrupt at
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$4200 \times 20\,000 = 84 \times 10^{6}$, the clock, so exactly 1 Hz; and 20 000 ≤ 65 535 fits a 16-bit reload.
Prescaler 1 and reload 84 000 is a valid 1 kHz design on 16-bit timer registers.
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The product is right, but 84 000 is more than 65 535, so the reload does not fit. Prescaler 84 with reload 1000 works.
Worked example: interrupt or poll?
- An ISR with 12 cycles of entry, 30 instructions and 10 cycles of return costs 52 cycles, or 0.62 µs at 84 MHz.
- That cost is paid on every event: 0.06% of the CPU at 1 kHz of events, but 62% at 1 MHz. At very high rates, polling wins.
A 52-cycle ISR at 84 MHz, serving 1 MHz of events, uses about
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$52 \times 10^{6}$ cycles a second out of $84 \times 10^{6}$ is 62%. The same ISR at 1 kHz uses only 0.06%: the event rate decides.
The same 52-cycle ISR serves 10 kHz of events at 84 MHz. What percentage of the CPU does it use? Give 1 decimal place.
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$52 \times 10\,000 = 520\,000$ cycles a second; $520\,000 / 84 \times 10^{6} = 0.62\%$.
Three common slips
- Forgetting to clear the flag. The CPU returns and enters the ISR again, for ever.
- Long ISRs. Do the minimum, set a flag, and let the main loop do the work.
- Reading a 16-bit timer in two halves. A carry between the reads corrupts the value; read the high byte twice or use a latch.
An ISR that forgets to clear its interrupt flag causes
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The flag still shows pending when the ISR returns, so the CPU jumps straight back in. Clear the flag and keep the ISR short.
Recap
- Interrupt: vector, short ISR, clear the flag.
- Timer: $f_{\text{clock}}$ divided by prescaler × reload, with each factor inside its register.
- Interrupts win at low event rates; polling wins at very high ones.