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Some time ago, a question came up on the PICList asking if the timer input pin could be used as an interrupt source pin. The answer to this is yes if the timer (and prescaler) is set up so that the next transition will increment the timer and cause an interrupt. Here s some code to do it in a mid-range PIC microcontroller:

movlw B 11000000 ; First Setup with Instruction Clock option ; as TMR0 Source movlw B 11100000 ; Option Setup for TOCK1 TMR0 Source clrf TMR0 ; Set TMR0 to 0x0FF decf TMR0, f option ; Enable Timer on Outside Interrupt ; Edge ; NOTE - Executing this Instruction ; after decf will Load the ; Synchronizer with a 1

This code also can be used on a low-end PIC microcontroller to monitor when an input changes instead of continuously polling the input pin.

As much as you try to avoid it, sometimes you have to divide. Here s an algorithm from Andy Warren for dividing a positive value by 3 by knowing that divide by 3 can be represented by the series x/3 = x/2 x/4 + x/8 x/16 + x/32 x/64 The algorithm

Deciding whether to take questions during or after a presentation is often a challenge for speakers. The following suggestions will help you decide the best approach to take when addressing questions.

int DivideBy3( int Value ) Return it { int Quotient = 0; while (Value != 0) { Value = Int(Value / 2); // Quotient + 1/2 if (Value != 0) { Quotient = Quotient + Value; Value = int(Value / 2); // Quotient 1/4 if (Value != 0) Quotient = Quotient - Value; } } return Quotient; } // End DivideBy3 // Divide Value by 3 and

Eliminating e,

Div3: movwf clrf Div3_Loop: Dividend Quotient ; Loop Until the Dividend == 0 ; Divide Contents of w by 3

STATUS, C Dividend, f Dividend, w STATUS, Z Div3_Done Quotient Dividend, f Dividend, w STATUS, Z Div3_Done Quotient, f Div3_Loop

; Dividend /2 (ie x/2 in Series) ; Is it Equal to Zero ; If it is, then Stop ; Add the Value to the Quotient ; Dividend /2 (ie x/4 in Series)

; Quotient = Quotient-(Dividend / 4)

When I rst started working with the PIC microcontroller, I thought I was exceedingly clever when I came up with

movlw HiDlay movwf HiCount movlw LoDlay movwf LoCount Dlay: decfsz LoCount, f goto Dlay decfsz HiCount, f goto Dlay ; Load the Delay Values

movlw HiDlay ; Load the Delay Values movwf HiCount movlw (LoDlay ^ 0x0FF) + 1 Dlay: addlw 1 ; Increment the Counter by 1 btfsc STATUS, Z decfsz HiCount, f ; Decrement the High Counter goto Dlay

The advantage to taking questions throughout your presentation is that any confusion your listeners have can be clari ed immediately, and you can speak directly to their concerns. The major disadvantage is that the sequence, emphasis, and your major message can get lost when you give oor time to individual audience members. Any listener with a personal agenda can grandstand during your allotted time. Taking questions during your performance tends to give the presentation a more informal tone and may be the best technique to use with small groups or as you become a more experienced presenter. Taking questions during a talk requires exibility and often demands more time for the total presentation.

This loop takes ve cycles to execute regardless of whether or not HiCount is to be decremented (and uses one less le register than my method above). The actual time delay is calculated using the 16-bit number from

This formula is a lot easier to use than having to (correctly) gure out the delay for the Dlay loop code given above this. In the rst edition of this book, it took three printings to get the rst Dlay loop s delay printed correctly. The rst method is useful if you require approximately a 200,000-instruction-cycle delay. To implement this, rather than loading HiDlay and LoDlay with constants, simply clear them before entering the two decfsz instruction loops.

Next substitute for dc/dt:

This is an improvement on the 16-bit delay code that I presented in the rst edition of this book. PulseWidth is a 16-bit value that contains the number of times through the loop. The code that measures the pulse width for a high pulse is