TTL GATES. INEL4207 Digital Electronics

TTL GATES INEL4207 Digital Electronics

Simple pseudo-ttl Inverter V CC =+5V R R C v O v IN Q 1

V CC =+5V V CC =+5V R R C i B1 R i E1 i C2 R C Q 1 v O =V L + 0.7V - Q 1 i B2 + 0.7V - v O =V L v IN =V H =V CC Q 1 : reverse-active mode

V CC =+5V v IN =V L Q 1 : saturation mode v IN =V L 0.7V - + i B1 vo=vh=vcc R B R C i C2 =0 Q 1 i E1 i C1 =i B2 =0 : cutoff

V CC v i2 C L v i1 Q 3

TTL Inverter V CC =+5V R C R S R B 4K 1.6K 130 v IN Q 1 Q 3 D1 v OUT R E 1.0K

V CC =+5V R C R S R B 4K 1.6K 130 v IN 5V Q 1 Q 3 D1 v OUT Q1 works in reverse active mode R E 1.0K

ib1=(5v-2.1v)/4kω=0.73ma V CC =+5V R C R S R B 4K 1.6K 130 2.1V βrib1=.02 0.73mA=15μA v IN 5V + 0.7 - Q Q 3 1 + 0.7 - D1 v OUT Q1 works in reverse active mode! βr = 0.02 ic1=ib3 0.73mA R E 1.0K + 0.7 -

V CC =+5V 5-2.1 4 0.73mA IB I C = v IN 5V R =0.02 R C R S R B 4K 1.6K 130 5-0.9 1.6 0.9V 0 R I C βrib I ih =15uA Q 1 2.1V 1.4V 0.73mA 0.7mA 2.6mA (sat) 3.3mA 0.7V 2.6mA Q 3 (OFF) D1(OFF) (sat) v OUT 0.2V R E 1.0K

V CC =+5V 5-0.9 4 ~ 1mA R B R C R S 4K 1.6K 130 0.9V 0.2V v IN Q Q (OFF) 3 1 ~ 1mA 0 D1 i L v OUT I il R E 1.0K (OFF)

1.Assuming Q4 has β=50 and that VCE,Sat=0.3V, find value of il at which Q4 saturates. 2.Assuming vbe,4=vd=0.7v when i=1ma, find vo when il=1ma and 10mA. 3.Find the max il that can be sourced by a TTL gate while VOH > 2.4V (guaranteed value).

V CC =+5V 5-0.9 4 ~ 1mA R B R C R S 4K 1.6K 130 0.9V 0.2V v IN Q Q (OFF) 3 1 ~ 1mA 0 D1 i L v OUT I il R E 1.0K (OFF) Assuming Q4 has β=50 and that VCE,Sat=0.3V, find value of il at which Q4 saturates

0.2V v IN 5-0.9 4 ~ 1mA R B R C R S 4K 1.6K 130 Q Q (OFF) 3 1 ~ 1mA 0 I il 0.9V R E 1.0K D1 V CC =+5V (OFF) i L v OUT KVL por el colector de Q4:! ve4 = 5V - 130Ω ic4-0.3v (1)! KVL por base de Q4, con ib4 = ic4/β! ve4 = 5V - 1.6kΩ ic4/50-0.7v (2)! Igualando (1) y (2) 5V - 130Ω ic4-0.3v = 5V - 1.6kΩ ic4/50-0.7v! (130Ω - 1.6kΩ /50) ic4 = 0.7V - 0.3V = 0.4V ic4 = 0.4V (130Ω - 32Ω) = 4.08µA! il = 51 ic4/50 = 4.16µA

V CC =+5V 5-0.9 4 ~ 1mA R B R C R S 4K 1.6K 130 0.9V 0.2V v IN Q Q (OFF) 3 1 ~ 1mA 0 D1 i L v OUT I il R E 1.0K (OFF) Assuming vbe,4 = vd= 0.7V when i = 1mA, find vo when il = 1mA and 10mA

a) when il = 1mA V CC =+5V vo = 5V - il 1.6kΩ 51-0.7V - 0.7V = 3.6V - 1mA 1.6kΩ 51 3.6V b) when il = 10mA ic il = IS exp( vbe / VT ) (1) 1mA = IS exp( 0.7V / VT ) (2) (1) (2) ic / 1mA = exp( (vbe - 0.7V)/ VT) vbe = vd = 0.7 + (0.025V) ln (10) = 0.76V Assuming active region operation, 0.2V v IN 5-0.9 4 ~ 1mA ve4 = 5V - il 1.6kΩ 51-0.76V = 4.24V - 10mA 1.6kΩ 51 3.9V vc4 5V - 130Ω 10mA = 3.7V Q4 is saturated, and vo = vc4-0.3v - 0.76V = 2.64V R B R C R S 4K 1.6K 130 0.9V Q Q (OFF) 3 1 ~ 1mA 0 I il R E 1.0K D1 (OFF) i L v OUT

Find the max il that can be sourced by a TTL gate while VOH > 2.4V (guaranteed value). 0.2V v IN 5-0.9 4 ~ 1mA R B R C R S 4K 1.6K 130 0.9V Q Q (OFF) 3 1 ~ 1mA 0 I il R E 1.0K D1 V CC =+5V (OFF) i L v OUT From the previous problem, we expect Q4 to be saturated, and vo = 5V - 130Ω il - 0.3V - vd = 2.4V il = (2.3V - vd)/130ω Another equation is needed. Use il / 1mA = exp( (vd - 0.7V)/ VT) vd = 0.7V + VT ln (il / 1mA) and solve the two equations by recursive approximation. vd il 0.7V 12.3 ma 0.76V 11.8 ma 0.76V 11.8 ma

A better approx. not assuming ic = il: 0.2V v IN 5-0.9 4 ~ 1mA R B R C R S 4K 1.6K 130 0.9V Q Q (OFF) 3 1 ~ 1mA 0 I il vd R E 1.0K il 0.7V 13 ma 0.76V 12.5 ma 0.76V 12.5 ma D1 V CC =+5V (OFF) i L v OUT vo = 5V - 130Ω ic - 0.3V - vd = 2.4V ic = (2.3V - vd)/130ω = il - ib ib = (5V - 2.4V - vbe - vd)/1.6kω Let vbe = vd il = ic + ib = (2.3V - vd)/130ω + (2.6V - 2vD)/1.6kΩ il = 19.3mA - vd/111.8ω Another equation is needed. Use il / 1mA = exp( (vd - 0.7V)/ VT) vd = 0.7V + VT ln (il / 1mA) and solve the two equations by recursive approximation.

El siguiente diagrama muestra un invertidor de lógica TTL simplificada como los que discute el libro de texto. V CC R 1 R 2 v o v i Q 1 Use los siguientes datos: V CESAT = 0,15V, v BE = 0,7V para una junta PN polarizada hacia adelante (forward bias), β F = 20, β R = 0,1, R 1 = R 2 = 20kΩ y V CC = 2,5V. Determine: a) el voltaje v H en la salida cuando el nivel es alto (5 puntos); b) el voltaje v L en la salida cuando el nivel es bajo (5 puntos); c) la corriente en la base del transistor cuando v i = v H (5 puntos); d) la corriente en la base del transistor cuando v i = v L (5 puntos); e) si el transistor esta saturado (explique su respuesta en detalle) (5 puntos); f ) la potencia disipada cuando v i = v H (5 puntos); g) la potencia disipada cuando v i = v L (5 puntos);

a) el voltaje V H en la salida cuando el nivel es alto (5 puntos); RESPUESTA: V H = 2,5V b) el voltaje V L en la salida cuando el nivel es bajo (5 puntos); RESPUESTA: V L = V CESAT = 0,15V c) la corriente en la base del transistor cuando v i = V H (5 puntos); RESPUESTA: El voltaje en la base de Q 1 será más bajo que el del emisor, pero será suficientemente alto para hacer conducir la junta BC de Q 1 y la de BE de. Por lo tanto, i B1 V CC v BC1 v BE2 R 1 = 2,5V 1,4V 20kΩ = 55µA La corriente del emisor de Q 1 será β R i B1 = 0,1 55µA = 5,5µA, donde el signo negativo indica que la corriente entra al emisor. i B2 = i C1 = i E1 + i B1 = 60,5µA d) la corriente en la base del transistor cuando v i = V L (5 puntos); RESPUESTA: Si v i = V L = 0,15V, Q 1 estará saturado y el voltaje en la base de será de aproximadamente 0,3V. Por lo tanto, estar en la región de corte y la corriente será 0. e) si el transistor esta saturado cuando = (explique su respuesta en detalle) (5 puntos);

e) si el transistor esta saturado cuando v i = V H (explique su respuesta en detalle) (5 puntos); RESPUESTA: Si estuviera operando en la región activa, El voltaje en el colector de seria i C2 = β F i B2 = 20 60,5µA = 1,21mA v C2 = V CC i C2 R 2 = 2,5V 20kΩ 1,21mA = 21,7V lo cual no es posible. Esto demuestra que estará saturado. f ) la potencia disipada cuando v i = V H (5 puntos); RESPUESTA: P = 2,5V 2,5V 0,15V 20kΩ + 2,5V 55µA + 2,5V 5,5µA = 445µW g) la potencia disipada cuando v i = V L (5 puntos); P = (2,5V 0,15V ) 2,5V 0,7V 0,15V 20kΩ = 194µW

For the following circuit shown, β R =0.2V, β F =10,V CE,SAT =0.2V and V BE,ON = 0.7V.Findtheoutputvoltageifv i =0.2V for an inverter driving a load of 20 inverters like the one shown, connected to the output node. (20 points) +5V 4kΩ 1.6kΩ 130Ω v IN Q 1 Q 3 D 1 v OUT 1kΩ

ASWER: For the driver inverter, the input is 0.2V, Q 1 will be saturated, Q 3 and will be OFF and will be ON. For the load inverters, the input will be V H and transistor Q 1 will be operating in reverse-saturation mode, and Q 3 will be ON. The input current on each load inverter will thus be 5V 2.1V β R =0.2.725mA =145µA 4kΩ The total current in the emitter of is then i e4 =20.145mA =2.9mA The voltage at s collector will be 5V 2.9mA 130Ω =4.623V and the transistor will operate in active mode with a base current i b4 = 2.9mA =264µA. Thusthevoltage 11 at s base is 5V.264mA 1.6kΩ 4.6V and the output voltage is v O =4.6V 1.4V =3.2V

Find the power dissipated by the following circuit if v i values given in problem 1. (25 points) previous problem =2V. Use the parameters +5V 28kΩ 28kΩ v O v i Q 1

ANSWER: Q 1 must be operating in reverse-active mode, with it s base pinned-down at 1.4V by the two forward-biased PN junctions (Q 1 s base-collector and s base-emitter). Q 1 s base current is thus 5V 1.4V i b1 = =129µA 28kΩ Acurrentof0.2 129µA =25.8µA flows into the emitter of Q 1.Thecurrentflowing into the base of is i b2 =1.2 129µA =154.3µA This current is large enough to saturate transistor -otherwise,thedropacrossthe collector resistor would be larger than the supply voltage. Thus the output voltage equals V CE,SAT =0.2V and the current in s collector is i c2 = 5V 0.2V 28kΩ =171.4µA Finally, the dissipated power can be found to be P =5V 129µA +5V 171.4µA +2V 25.8µA =1.55mW

V CC =+5V 4K 1.6K 130 A B Q 1A,1B Q 3L AB 1.0K TTL NAND Gate

V CC =+5V 4K 1.6K 4K 130 AB+CD A Q 3L Q 3R C B D Q 1A,1B Q 1C,1D AND-OR-INVERT Gate 1.0K