The transistor used in a portion of the TTL gate shown in the figure has a β = 100. The base-emitter voltage of is 0.7 V for a transistor in active region and 0.75 V for a transistor in saturation. If the sink current I = 1mA and the output is at logic 0, find the current IA?
A TTL inverter prototype is shown in Figure 8 with the following values: VCC = 3.5 V, RB = 6kΩ, and RC = 3kΩ Figure 8. (a) When V1 = 0.15 V, Q2 is off and Q1 is in saturation with VBEsat = 0.75 V and VCEsat = 0.05 V. Calculate the available currents and the power dissipated in the circuit. (b) When vi = Vcc, Q1 is at reversed active mode with βR = 0.25 and Q2 is in saturation mode with VBEsat = 0.75 V and VCEsat = 0.15 V. Calculate all the available currents and the total power of dissipation.
For the low power TTL inverter of Figure P7.11, obtain the following: (a) Sketch the VTC. (b) Calculate the maximum fan-out = N = IOL/IIL. (c) Calculate the average power dissipation. Use βF = 90, βR = 0.05, VBE(FA) = VBC(RA) = 0.7 V, VBE(SAT) = 0.8 V, and VCT(SAT) = 0.2 V. Use σo = 0.85 for the output low state.
Answer the following questions about the TTL gate shown: a. Make a table that states whether each of the transistors Q1 . . . Q6 is on or for all combinations of HI and LOW inputs A and B. b. What logic function does this gate represent? c. Approximately, what current flows in an input line when the input is HIGH? What flows when the input is LOW? d. Without any load, the voltage when the output is HIHG is less than 5 volts. Explain why, and what voltage is it and why it is that voltage. e. If the output is HI, then the output current is 1.5 mA, what would the output voltage be? Explain and/or show a calculation that supports this.
Consider the following TTL circuit. Assume that the transistor parameters are that βF = 50, βR = 0.1 (for each emitter input), VBE( on ) = 0.7 V, VBE( sat ) = 0.8 V and VCE( sat ) = 0.1 V a) (15) Calculate the maximum fanout when vA = vB = 5 V. b) (15) Calculate the power dissipation when vA = vB = 5 V.
For the TTL inverter given below a) Find RC /RD ratio using the given voltage transfer characteristic. b) Find RC/RB ratio NOH,max is 10 with 0.6 V high noise margin. c) Find NOL,max. d) Average power dissipation of the given TTL is 5.05 mW when ten similar gates are connected, find RB, RC , and RD. VCC = 5 V βF = 45 αF ≈ 1 βR = 0.5 σ = 0.9 VBE(FA) = 0.7 V VBE(SAT) = 0.8 V VCE(SAT) = 0.2 V VBC(RA) = 0.7 V VD = 0.7 V
VCE(SAT) = 0.2V VBE(ON) ≈ VBE(SAT) = 0.7V VBE(CUT-IN) = 0.5V βF = 25, βL = 0.1, βR = 0.5. 1. Find worst case input sinking requirement. 2. Find Io(av) (available current) for output low condition ( VO=”L”). 3. Find fanout “0” for worst case input sinking requirement.
Consider the basic TTL inverter given in Figure 2. Assume that VBE = 0.7 V, VBE(SAT) = 0.8 V, VCE(SAT) = 0.2 V, VBC = 0.7 V, βf = 50, βR = 0.1. Use σ0 = 0.8 (saturation parameter of Q0) for the output low state. i) Find the operation mode of each BJT and IIN, IRB, IRC, IB2 when Vin = 0 V and Vin = 5 V. ii) Sketch the voltage transfer characteristics by finding VOL, VOH, VIL, VIH. iii) Find the noise margins. iv) Find the maximum fanout. What is the limiting case for determining the fan-out limit of this circuit?
In the TTL circuit in Figure below, the transistor parameters IOL = 40 mA, IOH = 10 mA and IIH = 0.1 mA, VBE sat = 0.8 V and VCEsat = 0.1 V : 1- Write the truth table of the circuit showing the values of logic “0” and Logic “1”. 2- Calculate the fan out NL and NH, which one you select as a circuit fan-out?
For the TTL gate shown, VBE (FA) = 0.7 V, VBC(RA) = 0.7 V, VBE(sat) = 0.8 V, VCE(sat) = 0.2 V, and VD(ON) = 0.7 V. Answer the following questions: The circuit above is a * (2 Points) QS conducts at VIL = * (2 Points) OR Gate 0.5 V NOR Gate 0.8 V None None AND Gate 1.4 V NAND Gate 1.2 V 20 VOH for the given circuit is * (1 Point) 22 QO conducts at Input break point VIB = * (2 Points) 0.2 V 1.2 V None 3.6 V 5 V None 1.4 V 0.8 V 0.7 V 0.7 V Output break point VOB is * (2 Points) 5 V None 3.6 V 1.6 V 2.48 V 24 VIH * (2 Points) 1.2 V 1.4 V 1.6 V None 5 V