ECE Test #1: Name

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1 ECE Test #1: Name Closed Book, Closed Notes. Calculators Permitted. September 23, < ) Binary Inputs: Design a circuit to detect when the sun is out and when it's cloudy. Assume a light sensor has the following light vs. resistance relationship R = 100,000 Lux Ω Design a circuit which Outputs 5V when the light level is more than 100 Lux, Outputs 0V when the light level is less than 80 Lux, and Remains unchanged in-between 80 and 100 Lux. 5V 2.5V 10k 180k 1k 10k 180k 5V 0V 2.5V 10 Lux 2.778V 80 Lux Assume a 1k resistor for the volage divider: 100 Lux = 1000 Ohms = 2.5V 80 Lux = 1250 Ohms = V Gain = (5V - 0V ) / (2.778V - 2.5V ) = 18.0 When the input is large ( 5V or 0 Lux), the output is small. Connect to the minus input When the output is 0V, you switch at 2.5V. The offset is 2.5V

2 2) Binary Outputs. Design a circuit which allows a PIC to turn on and off a 10W LED at 1A. Assume the characteristics of the LED are Vf = 2A A Assume an NPN Zetex transistor with the following specifications: max(ic) = 4A Vce:sat = 200mV β=300 Assume the PIC can outputs 5V with a maximum current of 25mA Assume a +10V power supply (anything larger than 5.2V works) R c = 10V 5V 0.2V 1A = 4.8Ω To saturate the transistor βi b > I c Let Ib = 5mA I b > 1A 300 = 3.33mA R b = 5V 0.7V 5mA = 860Ω 10V V - PIC 860 1A LED + 5V V - Ib = 5mA 172 < Rb < mA < Ib < 3.33mA

3 3) Software: For a stoplight, a wait routine which takes 5 seconds to execute is desired (50 million clocks). 3a) Determine how many clocks the following wait routine takes. ans = 132,977,404 3b) Modify this routine so that it takes 50 million clocks (5 seconds) show on program Assembler Code 18 lines of assembler Wait5s: Number of Clocks 18 lines of assembler 2 b) Change to 96 movwf CNT2 C Code 89 lines of assembler void Wait(unsigned int SEC) unsigned int i; L2: movwf CNT1 2 * 255 unsigned long int j; for (i=0; i<sec; i++) for (j=0; j<130000; j++); L1: movwf CNT0 2 * 255 * 255 L0: decfsz CNT0,F goto L0 8 * 255 * 255 * 255 decfsz CNT1,F goto L1 3 * 255 * 255 decfsz CNT2,F goto L2 3 * 255 return 2

4 4) Software: Assume your PIC is connected to a stop light with RB0 = Red Light ( logic 1 = on ) RB1 = Yellow Light ( logic 1 = on ) RB2 = Green Light ( logis 1 = on ) Also assume you have a wait routine (i.e. reuse problem 3) Wait5s: which takes burns 5 seconds (50 million clocks) then returns. Write an assembler program to implement a stoplight: Assembler 37 Lines of Assembler (incl problem 3) #include <p18f4620.inc> Loop: org 0x800 clrf TRISB movlw 0x0F movwf ADCON1 movlw 0x04; movwf PORTB movlw 0x02; movwf PORTB C 169 Lines of Assembler (incl problem 3) // Global Variables unsigned char GREEN = 4; unsigned char YELLOW = 2; unsigned char RED = 1; // Subroutine Declarations #include <pic18.h> void Wait(unsigned int SEC) unsigned int i; unsigned long int j; for (i=0; i<sec; i++) for (j=0; j<130000; j++); void main(void) TRISB = 0; ADCON1 = 0x0F; while(1) PORTB = GREEN; Wait(20); Start Initialize Light = Green Wait 20 sec movlw 0x01; movwf PORTB PORTB = YELLOW; Wait(5); PORTB = RED; Wait(20); Light = Yellow Wait 5 sec goto Loop end Light = Red Wait 20 sec

5 5) Software: Write the assembler code which corresponds to the following flow chart: If RB1 (PortB pin 1) is pressed, N is increased by one, stopping at 255. If RB0 (PortB pin 0) is pressed, N is decreased by one, stopping at zero. Assume N is an 8-bit number. Assembler 11 lines of assembler C 20 Lines of Assembler // global variables unsigned char N; Start Start: void Prob5(void) btfss PORTB,1 goto B4 if (RB1) if (N < 255) N += 1; RB1 = 1? if (RB0) cpfseq N if (N > 0) N -= 1; N = 255? incf N,F N = N + 1 B4: btfss PORTB,0 goto B7 RB0 = 1? movlw 0 cpfseq N N = 0? decf N,F N = N - 1 B7: return Return

6 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 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 carry RRCF PORTA,F rotate right through carry RRNCF PORTA,F rotate right 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 t zero INCFSZ PORTA,F increment, skip if zero INFSNZ PORTA,F increment, skip if t 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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