Fundamentals of Digital Systems and logic families
Q1) Explain the Characteristics of digital Logic Families?
A1)
Speed of a logic circuit is determined by the time between the application of input and change in the output of the circuit.
Every gate is connected to a power supply VCC (VDD in the case of CMOS).
A certain amount of current is drawn during its operation. Since each gate can be in a High, Transition or Low state, hence there are three types of currents drawn from power supply:
For TTL,
Average Power Dissipation = Vcc * (ICCH + ICCL)/2
For CMOS,
Average Power Dissipation = Vcc * ICCT.
For TTL logic family, power dissipation does not depend on frequency of operation.
For CMOS, the power dissipation depends on the frequency of operation.
Power Dissipation:
Power dissipation of a circuit defines its battery life: the greater the power dissipation, the shorter the battery life.
It is directly proportional to the heat generated by the chip or system hence excessive heat dissipation may increase operating temperature and cause gate circuitry to drift out of its normal operating range.
Total power dissipation = static power dissipation + dynamic power dissipation.
Figure of merit
Figure of merit is a product of propagation delay and power dissipation. It is measured in terms of Pico-Joules (ns ´mW = pJ). Current and voltage parameters define the minimum and maximum limit of current and voltage for input and output of a logic family.
Fan in
Fig.: Effects of Fan-in
Fan out
Fig.: Effects of Fan-out
Current
Voltage
They are the voltage levels for logic high and logic low.
Noise immunity
Gate circuits are made to sustain variations in input and output voltage levels.
Variations are usually the result of various factors.
Q2) What do you understand by TTL logic?
A2) They are built only with the help of transistors.
It has been improved to meet performance requirements.
TTL family comprises of:
Q3) Write the brief note on Operation of TTL NAND gate?
A3) NAND
In the single-input (inverter) circuit, grounding the input resulted in an output that assumed the “high” (1) state. In the case of the open-collector output configuration, this “high” state was simply “floating.”
Allowing the input to float (or be connected to Vcc) resulted in the output becoming grounded, which is the “low” or 0 state. Thus, a 1 in resulted in a 0 out, and vice versa.
Q4) Write the brief note on Operation of TTL AND gate?
A4) To create an AND function using TTL circuitry, we need to increase the complexity of this circuit by adding an inverter stage to the output, just like we had to add an additional transistor stage to the TTL inverter circuit to turn it into a buffer:
Of course, both NAND and AND gate circuits may be designed with totem-pole output stages rather than open-collector. I am opting to show the open-collector versions for the sake of simplicity.
Q5) Explain the work done by CMOS INVERTER?
A5)
Fig: CMOS inverter
A | Y = A’ |
0 | 1 |
1 | 0 |
Q6) Draw and explain CMOS Two input NOR Gate.
A6) The circuit consists of a parallel-connected n-net and a series-connected complementary p-net. The input voltages VX and VY are applied to the gates of one nMOS and one pMOS transistor.
When either one or both inputs are 1, then p-net is 0 (cut—off).
If both input voltages are 0, then the p-net conducts between the output node and the supply voltage.
Fig: Layout of CMOS NOR Gate
Q7) Draw and explain CMOS Two input NAND Gate.
A7) Here, the n – net consists of two series connected nMOS transistor conducts between the output node and the ground, if both input voltages are 1. Both of the parallelly connected pMOS transistor in p-net will be off.
For all other input combination, either one or both of the pMOS transistor will be turn ON, while p – net is cut off, thus, conducts between the output node and the power supply voltage.
Fig: Layout of CMOS NOR Gate
Q8) What do you understand by Interfacing of CMOS and TTL?
A8) CMOS driving TTL:
Following figure shows a CMOS gate driving N TTL gates. For such an arrangement to operate properly the following conditions are required to be satisfied,
VOH (CMOS) ≥ VIH(TTL)
VOL (CMOS) ≤ VIL(TTL)
– IOH (CMOS) ≥ NIIH(TTL)
IOL (CMOS) ≥ – NIIL(TTL)
The first possible type of CMOS-to-TTL interface is the one where both ICs are operated from a common supply. We have read in earlier sections that the TTL family has a recommended supply voltage of 5 V, whereas the CMOS family devices can operate over a wide supply voltage range of 3†“18 V. In the present case, both ICs would operate from 5 V. As far as the voltage levels in the two logic states are concerned, the two have become compatible. The CMOS output has a VOH (min.) of 4.95V (for VCC =5 V) and a VOL (max.) of 0.05 V, which is compatible with VIH (min.) and VIL (max.) requirements of approximately 2 and 0.8V respectively for TTL family devices. In fact, in a CMOS-to- TTL interface, with the two devices operating on the same VCC, voltage level compatibility is always there. It is the current level compatibility that needs attention. That is, in the LOW state, the output current-sinking capability of the CMOS IC in question must at least equal the input current-sinking requirement of the TTL IC being driven. Similarly, in the HIGH state, the HIGH output current drive capability of the CMOS IC must equal or exceed the High-level input current requirement of TTL IC. For a proper interface, both the above conditions must be met. As a rule of thumb, a CMOS IC belonging to the 4000B family (the most widely used CMOS family) can feed one LS TTL or two low-power TTL unit loads. When a CMOS IC needs to drive a standard TTL or a Schottky TTL device, a CMOS buffer (4049B or 4050B) is used. 4049B and 4050B are hex buffers of inverting and noninverting types respectively, with each buffer capable of driving two standard TTL loads.
Figure below shows a CMOS-to-TTL interface with both devices operating from 5V supply and the CMOS IC driving a low-power TTL or a low-power Schottky TTL device.
Figure below shows a CMOS-to-TTL interface where the TTL device in use is either a standard TTL or a Schottky TTL.
The CMOS-to-TTL interface when the two are operating on different power supply voltages can be achieved in several ways. One such scheme is shown below. In this case, there is both a voltage level as well as a current level compatibility problem.
TTL Driving CMOS:
For TTL gate driving N CMOS gates arrangement to operate properly, the following conditions are required to be satisfied:
VOH (TTL) ≥ VIH(CMOS)
VOL (TTL) ≤ VIL(CMOS)
– IOH (TTL) ≥ NIIH(CMOS)
IOL (TTL) ≥ – NIIL(CMOS)
In the TTL-to-CMOS interface, current compatibility is always there. The voltage level compatibility in the two states is a problem. VOH (min.) of TTL devices is too low as regards the VIH (min.) requirement of CMOS devices. When the two devices are operating on the same power supply voltage, that is, 5 V, a pull-up resistor of 10 k_ achieves compatibility as shown in below figure.
The pull-up resistor causes the TTL output to rise to about 5V when HIGH. When the two are operating on different power supplies, one of the simplest interface techniques is to use a transistor (as a switch) in-between the two, as shown below.
Another technique is to use an open collector type TTL buffer as shown below.
Q9) Compare Analog and digital signals.
A9)
Q10) Calculate the Decimal Equivalent of binary number 101112
A10)
Step | Binary Number | Decimal Number |
Step 1 | 101012 | ((1 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20))10 |
Step 2 | 101012 | (16 + 0 + 4 + 2 + 1)10 |
Step 3 | 101012 | 2310 |
Q11) Calculate the Decimal Equivalent of Hexadecimal Number: 19FDA16
A11)
Step | Hexadecimal Number | Decimal Number |
Step 1 | 19FDA16 | ((1 × 164) + (9 × 163) + (F × 162) + (D × 161) + (A × 160))10 |
Step 2 | 19FDA16 | ((1 × 164) + (9 × 163) + (15 × 162) + (13 × 161) + (10 × 160))10 |
Step 3 | 19FDA16 | (65536 + 36864 + 3840 + 208 + 10)10 |
Step 4 | 19FDA16 | 10645810 |
Q12) Explain XS-3 and Gray Code.
A12) Excess-3 code
Example
Fig.: BCD to XS 3 conversion (Ref. 1)
Gray Code
Fig.: Gray codes (Ref. 1)
Q13) Explain BCD code with advantages and disadvantages.
A13)
Fig.: BCD codes (Ref. 1)
Advantages of BCD Codes
Disadvantages of BCD Codes
Q14) Draw and explain basic gates.
A14) The basic gates are namely AND gate, OR gate & NOT gate.
AND gate
It is a digital circuit that consists of two or more inputs and a single output which is the logical AND of all those inputs. It is represented with the symbol ‘.’.
The following is the truth table of 2-input AND gate.
A | B | Y = A. B |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Here A, B are the inputs and Y is the output of two input AND gate.
If both inputs are ‘1’, then only the output, Y is ‘1’. For remaining combinations of inputs, the output, Y is ‘0’.
The figure below shows the symbol of an AND gate, which is having two inputs A, B and one output, Y.
Fig.: AND gate (ref. 1)
Timing Diagram:
OR gate
It is a digital circuit which has two or more inputs and a single output which is the logical OR of all those inputs. It is represented with the symbol ‘+’.
The truth table of 2-input OR gate is:
A | B | Y = A + B |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
Here A, B are the inputs and Y is the output of two input OR gate.
When both inputs are ‘0’, then only the output, Y is ‘0’. For remaining combinations of inputs, the output, Y is ‘1’.
The figure below shows the symbol of an OR gate, which is having two inputs A, B and one output, Y.
Fig.: OR gate (ref. 1)
Timing Diagram:
NOT gate
It is a digital circuit that has one input and one output. Here the output is the logical inversion of input. Hence, it is also called as an inverter.
The truth table of NOT gate is:
A | Y = A’ |
0 | 1 |
1 | 0 |
Here A and Y are the corresponding input and output of NOT gate. When A is ‘0’, then, Y is ‘1’. Similarly, when, A is ‘1’, then, Y is ‘0’.
The figure below shows the symbol of NOT gate, which has one input, A and one output, Y.
Fig.: NOT gate (ref. 1)
Timing Diagram:
Q15) Draw and explain universal gates.
A15)
NAND gate
It is a digital circuit which has two or more inputs and single output and it is the inversion of logical AND gate.
The truth table of 2-input NAND gate is:
A | B | Y = (A.B)’ |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
Here A, B are the inputs and Y is the output of two input NAND gate. When both inputs are ‘1’, then the output, Y is ‘0’. If at least one of the inputs is zero, then the output, Y is ‘1’. This is just the inverse of AND operation.
The image shows the symbol of NAND gate:
Fig.: NAND gate (ref. 1)
NAND gate works same as AND gate followed by an inverter.
Timing Diagram:
NOR gate
It is a digital circuit that has two or more inputs and a single output which is the inversion of logical OR of all inputs.
The truth table of 2-input NOR gate is:
A | B | Y = (A+B)’ |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 0 |
Here A and B are the two inputs and Y is the output. If both inputs are ‘0’, then the output is ‘1’. If any one of the inputs is ‘1’, then the output is ‘0’. This is exactly opposite to two input OR gate operation.
The symbol of NOR gate is:
Fig.: NOR gate (ref. 1)
NOR gate works exactly same as that of OR gate followed by an inverter.
Timing Diagram:
Q16) Explain Ex-OR gate
A16) It stands for Exclusive-OR gate. Its function varies when the inputs have even number of ones.
The truth table of 2-input Ex-OR gate is:
A | B | Y = A⊕B |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
Here A, B are the inputs and Y is the output of two input Ex-OR gate. The output (Y) is zero instead of one when both the inputs are one.
Therefore, the output of Ex-OR gate is ‘1’, when only one of the two inputs is ‘1’. And it is zero, when both inputs are same.
The symbol of Ex-OR gate is as follows:
Fig.: XOR gate (ref. 1)
It is similar to that of OR gate with an exception for few combination(s) of inputs. Hence, the output is also known as an odd function.
Timing Diagram:
Q17) Explain Ex-NOR gate
A17) It stands for Exclusive-NOR gate. Its function is same as that of NOR gate except when the inputs having even number of ones.
The truth table of 2-input Ex-NOR gate is:
A | B | Y = A⊙B |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
Here A, B are the inputs and Y is the output. It is same as Ex-NOR gate with the only modification in the fourth row. The output is 1 instead of 0, when both the inputs are one.
Hence the output of Ex-NOR gate is ‘1’, when both inputs are same and 0, when both the inputs are different.
The symbol of Ex-NOR gate is:
Fig.: XNOR gate (ref. 1)
It is similar to NOR gate except for few combination(s) of inputs. Here the output is ‘1’, when even number of 1 is present at the inputs. Hence is also called as an even function.
Timing Diagram: