ECE Test #1: Name

Transcription

1 ECE Test #1: Name Closed Book, Closed Notes. Calculators Permitted. February 12, 2016 Design an embedded system which is placed inside a refrigerator and, each time the door is opened The PIC increments a counter on PORTD, Beeps at a frequency of Hz (E4) for 100ms on RC0 The beep is made by an 8 Ohm speaker, driven with a square wave at 0 / 250mA Each problem corresponds to a different part of this system. 1) Hardware - Binary Inputs. To detect that the refrigerator door has been opened, a light sensor is used. Assume the light sensor has the following light vs. resistance relationship R 100,000 Lux Design a circuit which Outputs 5V when the light level is more than 35 Lux, Outputs 0V when the light level is less than 30 Lux, and Remains unchanged in-between 30 and 35 Lux. Assume a 3k resistor gain 35 Lux = 2587 Ohms = 2.439V 30 Lus = 3333 Ohms = Volts 5V0V 2.439V2.6316V V 2.439V 100k 2.59M 3k 100k 2.59M

2 2) Hardware - Binary Outputs. To produce the Hz signal, an 8 Ohm speaker is to be used. The requirements are When the PIC outputs 0V, the current to the speaker should be 0mA When the PIC outputs 5V, the current to the speaker should be 250mA. Design a circuit so that a PIC can turn on and off the 8 Ohm speaker at 0mA and 250mA. Assume A 5V DC power supply and A transistor with a current gain of 100 (3904) Assume a 5V power supply. To set the current to 250mA R total 5V0.2V 250mA R c To saturate a transistor with a gain of 100 Let Ib = 5mA I b I c 250mA I b 250mA mA R b 5V0.7V 5mA V Ic = 250mA 11.2 Ohms 8 Ohms 860 PIC NPN Ib = 5mA

3 3) Software: To play a note of Hz, you need a wait routine that burns ms (15,169 clocks). 3a) Determine how many clocks the following wait routine takes. Wait: movlw 53 movwf CNT1 L1: movlw 174 movwf CNT0 L0: decfsz CNT0,F goto L0 2 clocks executed 1 time = 2 clocks 2 clocks executed 53 times = 106 clocks 6 clocks 5 instructions plus a jump = 6 clocks executed 174 * 53 times = 6 * 174 * 53 = 55,332 clocks decfsz CNT1,F goto L1 3 clocks executed 53 times = 159 clocks return 2 clocks executed 1 time = 2 clocks Adding up the number of clocks gives 55,602 clocks 3b) Modify this code so that it takes ms (15,169 clocks) to execute. ( 6 * ) * = 15,169 Wait: L1: L0: movlw 10 movwf CNT1 movlw 251 movwf CNT0 decfsz CNT0,F goto L0 decfsz CNT1,F goto L1 return 2 clocks executed 1 time = 2 clocks 2 clocks executed 10 times = 20 clocks 6 clocks 5 instructions plus a jump = 6 clocks executed 251 * 10 times = 6 * 251 * 10 = 15,060 clocks 8 clocks executed 10 times = 80 clocks 7 clocks executed 1 time = 7 clocks

4 4) Software: Given a wait routine which takes ms to execute, write an assembler routine, BEEP, which outputs a Hz square wave on RC0 for 100ms when called. 4a) If RC0 is toggling at a frequency of Hz, how many times does it toggle in 100ms? N 100ms ms b) Write the routine, BEEP, which outputs a Hz square wave on RC0 for 100ms. Assume you have a wait routine, Wait:, which pauses ms (i.e. problem 3). Beep: Loop: movlw 66 movwf X btg PORTC,0 call decfsz goto Wait X,F Loop return

5 5) Main Routine: Write the main routine which Monitors RB0 to detect when the refrigerator door has been opened When the door is opened, it - Beeps at Hz for 100ms (problem 4), - Displays on PORTD how many times the door has been opened, and - Clears the count (PORTD) if button RB1 is pressed Write the corresponding assembler code: org 0x800 ; Initialize Ports clrf TRISC clrf TRISD movlw 0xFF movwf TRISB movlw 0x0F movwf ADCON1 clrf PORTD Start Initialize ports PORTD=0 A1: btfss goto PORTB,0 A1 RB0 = 0? yes no call Beep Beep incf PORTD,F Inc PORTD A2: btfss goto PORTB,0 A1 no RB0 = 1? yes btfsc PORTB,1 no RB1 = 1? yes clrf PORTD Clear PORTD goto A2

6 Bonus! One clock on a PIC processor is 100ns. How fast do you have to travel for time to slow down 100ns every hour? Time slows down as 1 1 v2 c 2 1hr + 100ns 1hour 1 1 v2 2 c 2 using a Taylor's series 1 a 1 a 1 100ns 1hour 1 v2 2c 2 v 2 100ns 1 hour 2c2 v 2243 m s v 8044 km/h v 4998 mph note: 100ns in one year is equal to 23.8 m/s = 53 mph

7 Memory Read & Write MOVWF PORTA memory write PORTA = W MOVFF PORTA PORTB copy PORTB = PORTA MOVF PORTA,W memory read W = PORTA MOVLW 234 Move Literal to WREG W = 123 Memory Clear, Negation CLRF PORTA clear memory PORTA = 0x00 COMF PORTA toggle bits PORTA =!PORTA NEGF PORTA negate PORTA = -PORTA Addition & Subtraction INCF PORTA,F increment PORTA = PORTA + 1 ADDWF PORTA, F add PORTA = PORTA + W ADDWFC PORTA, W add with carry W = PORTA + W + carry ADDLW Add Literal and WREG DECF PORTA,F decrement PORTA = PORTA - 1 SUBFWB PORTA,F subtract with borrow PORTA = W - PORTA - c SUBWF PORTA,F subtract no borrow PORTA = PORTA - W SUBWFB PORTA,F subtract with borrow PORTA = PORTA - W - c SUBLW 223 Subtract WREG from # W = W Shift left (*2), shift right (/2) RLCF PORTA,F rotate left through carry (9-bit rotate) RLNCF PORTA,F rotate left no carry RRCF PORTA,F rotate right through carry RRNCF PORTA,F rotate right no carry Bit Operations BCF PORTA, 3 Bit Clear f clear bit 3 of PORTA BSF PORTA, 4 Bit Set f set bit 4 of PORTA BTG PORTA, 2 Bit Toggle f toggle bit 2 of PORTA Logical Operations ANDWF PORTA, F logical and PORTA = PORTA and W ANDLW 0x23 AND Literal with WREG W = W and 0x23 IORWF PORTA,F logical or PORTA = PORTA or W IORLW 0x23 Inclusive OR Literal W = W or 0x23 XORWF PORTA,F logical exclusive or PORTA = PORTA xor W XORLW 0x23 Exclusive OR Literal W = W xor 0x23 Tests (skip the next instruction if...) CPFSEQ PORTA Compare PORTA to W, skip if PORTA = W CPFSGT PORTA Compare PORTA to W, Skip if PORTA > W CPFSLT PORTA Compare PORTA to W, Skip if PORTA < W DECFSZ PORTA,F decrement, skip if zero DCFSNZ PORTA,F decrement, skip if not zero INCFSZ PORTA,F increment, skip if zero INFSNZ PORTA,F increment, skip if not zero BTFSC PORTA, 5 Bit Test f, Skip if Clear BTFSS PORTA, 1 Bit Test f, Skip if Set Flow Control GOTO Label Go to Address 1st word CALL Label Call Subroutine 1st word RETURN Return from Subroutine RETLW 0x23 Return with 0x23 in WREG RETFIE Return from Interrupt Other Stuff... NOP No Operation MULLW Multiply Literal with WREG MULWF PORTA multiply TSTFSZ PORTA test, skip if zero

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