Nuts and Volts - February 2009

■ FIGURE 7

'*  Notes :The program generates a binary 3 digit sequence to *
'* :switch the multiplexer every 30 seconds.  A second *
'* :port outputs a 3 digit sequence every 60 seconds to *
'* :keep track of which cylinder is being monitored. *
'* :A ground on GPIO.5 will stop the count as long as *
'* :the button is pressd.
'***************************************************
 REM  DEVICE = 16F627A
 REM  DISABLE LOW VOLTAGE PROGRAMMING
 REM  CONFIGURATION:  INTOSC CLOCKOUT, WDT DISABLED, PWR UP ENABLED,
 REM  MCLR ENABLED, BROWNOUT DISABLED, NO PROTECTION

the needed circuit.

— Arsenio C. Problete, Jr.

REM  PORT B ARE OUTPUT, HOLD INPUT ON PORTB.3 (PIN 9)

REM  RA5 (PIN4) IS MASTER CLEAR, RA6 (PIN15) IS CLOCK OUT TEST POINT.
CMCON = 7
TRISA = %00100000 'PORTA.5 IS MASTER CLEAR
TRISB = %00001000 'PORTB.3 IS HOLD INPUT
CNT VAR BYTE
DIGIT VAR BYTE

START:

 FOR CNT =0 TO 7
 IF CNT<2 THEN DIGIT =1
 IF CNT = 2 OR CNT = 3 THEN DIGIT =2
 IF CNT = 4 OR CNT = 5 THEN DIGIT =3
 IF CNT = 6 OR CNT = 7 THEN DIGIT =4
 PORTA = DIGIT
 PORTB = CNT
 IF PORTB.3 = 0 THEN GOSUB HOLD
 PAUSE 30000 '30 SECONDS
 NEXT CNT
GOTO START

HOLD:

IF PORTB.3 = 0 THEN GOTO HOLD
RETURN
END

■ FIGURE 8

AA 40 watt fluorescent tube gives a lot more light than a 100 watt bulb because of its greater surface area; 10,000 mcd ( 10 candles, equivalent to 126 lumens) is an intensity measure and LEDs of that intensity usually have viewing angles of 20 to 30 degrees. You’ll need 12 to cover 360 degrees and many more to cover a hemisphere. Without going through the math, I am going to estimate 31 LEDs in the hemisphere. If these are connected in series, the total voltage is: 31 3. 5 = 108.5 volts. The 250 VAC source will give 336 volts DC. The total power input at 20 mA is 6. 7 watts which is not bad for 126 lumens of light. That is assuming you use an 11K, five watt resistor in series. Since you will only save a couple of watts by using PWM, I think the resistor is the best way to go.