Question

7. Use the circuit below to answer the three questions. LED1 is Yellow and LED2 is Amber and typically, V. = 1.8 V @ 10 mA. The NOR gates are 7402 with 0.35 V as typical VO Show complete work in the space provided [4, 6, 4] a. If A= 1 and B = 0, which LED is on? b. Calculate a typical value for Rg. (Round up calculated resistor value to the nearest standard 5% resistor; refer to attached data sheet) c. If A is now 0 while B stays at 0, compare and contrast the output here with your answer to (a). Tv LEDIY LED2

Answer 1

2 LEDs and S-R The giver Latch wring circuit Nor includes gate. Frost we will discuss about Nor gute Latch = Truth Table of wor hate x 4 12 o al ç xlooma Toro NOR B? Cayeli) Let A=0 a Bio and consider O=1+0=0 Input of "NORB" will be cool) se, co zo s Using abwe) Input of Nor" will be (@o) So 0=1 Table Now Consider ce=0 & ã=1 Input of Nord" (oo) 0= 1 Input of "NORA" (0,D 0=0 So, at AB anti o o an 2 Previous state

- CASE C) Now, A= L B=0 and ansider azotaal Input of Nors" Loo 0= 1 Inpid of "Nora" (Li) azo Case Lim Now, Azo: B= 1 and a = 1 a ceto Input of "Nora" (1, 1) ozo Input of (Nora's (go) s Q = 1 Case (6) Now Azis B:1 Imped of 'NOR B” (Isc) = Q = 0 L Input of Nora (La) cãão J. NOTE: If any of the input is logic ", output will be 360. NOT Valid A B 0 I anti ant IO Memory . و | | | اک بے lolot Reset Not valid Now hiver circuit ; & +5V q+5v LEDIG I SLED2 LED, Y Chow LEPz Angler Typical value of LED is VF-1.8V @loma VOL = 0.35 V for Nor gate

- a). If ASI 4 11:0 From the table, a= o ; e = 1 The circuit will look like, LED BY y LED2 M50 50.35V 65V Now at Aald Boo, the most of the supply voltage will be across LED and will be greater than the forward voltage. so, LED, will be on. In case of LEDa, Voltage difference is almost 3000 LED2 will not conduct & LED, will be off. so bl Now, Let the same condition, a -14 B0 LEDI will be un. So 学生 If = luma Apply k V2 at top o +5- VF - Ifx Ry – VOLEO 0.35 V SV Ry= 5-VE-UL If Rg = 5- 1-8-0.15 lox103

. The value of Rg = 285n] c) Now, B is some and in in chimsed to o iss Aco Bio - The Cunditium Azod Beo indicate the previous output and prehead and the that is flora curred ouped is - Anti - On The previous oupul was cho because of Azi theo. So, at A= B:0 Chri= th=0 . Then again LEDI will be conducting and LEDE will be conducting. NON- LED, SON LEDą = OFF, If we compare the output at A=B=0 with answer S of Al; Boo output will be tame. As at A=Bco, it provides the previous output is LED, 3 un 4 LED, 5 Off.