Question

1. Explain why a full adder requires a carry input, opposed to just a carry output.

(I need short explain in a lab report)

2. Design (configure) an Astable Multivibrator using the 555 with the following characteristics: Astable Q T1 T2 Page 3 of 5 f = 1 kHz DUTY CYCLE = 60% As in industry, all values of resistors and capacitors are not available for implementing your design; only a discrete set of values are manufactured, and an even smaller subset is available in your lab kit. From the 555 Design Equations, it is clear that there are two independent variables and three unknowns (two resistors and one capacitor). Hence as the designer, you are allowed to arbitrarily select one of the unknowns. Since there are not as many available capacitor values as resistor values, it is common practice to select the capacitor and calculate the resistor values (the probability that a calculated resistor value is, in fact, available is higher than that for a calculated capacitor value). Also if a calculated resistor value is not available, combinations (series, parallel) of available resistor values may be used. And finally the selection of capacitor value is not purely arbitrary-the selected value should produce "reasonable" resistor values (for example if a selected capacitor value of 10pf produced resistor values in the 10 M? range, one would probably redo the design using a significantly larger capacitor value so that the resistors would be in the 10 k? range). Your design should be within 10% of the theoretical calculations (neglecting component tolerances). Verify your design using Multisim.

Answer 1

1. If addition is performed on 1-bit operands then there is no need of full adder as usually carry input is 0.

But if we perform multi-bit addition, then there may be carry propagated from lower bit addition and we need to consider it for next bit addition.

Assume A = 101 and B = 111, Here LSB bits added first are 1 + 1 = 0 and Carry out = 1. Now next higher bits added will be 0 and 1, but previous carry out generated is must to be taken care. Hence here we need to add 1 + 0 + 1 = 0 and Carry out = 1. This neccessiates a full adder usage.

2. Frequency of Oscillation = 1 KHz, Hence Time Period of Oscillation = 1/f = 1ms

Duty Cycle = 60% = T_ON / T

T_ON = 60% of T = 0.6 ms

T_OFF = T - T_ON = 0.4 ms

From the circuit diagram shown below:

T_ON = 0.693 (R₁ + R₂) C and T_OFF = 0.693 R₂C

Assume C = 1 uF

T_OFF = 0.4ms = 0.693 R₂C

R₂ = 0.4ms / (0.693 x 1 uF) = 577 ohm

As we know, T_ON = 0.693 (R₁ + R₂) C and substitute R₂ = 577 ohm

R₁ + R₂ = T_ON / (0.693 x 1 uF) = 865.8 ohm

or R₁ = 865.8 ohm - 577 ohm = 288.8 ohm