Digital Electronics for GATE EE
Number systems, Boolean algebra and K-maps, logic gates and families, combinational circuits (adders, multiplexers, decoders), flip-flops, counters, shift registers, and DAC and ADC.
GATE EE includes a compact digital electronics section. This lesson collects what it needs; the Digital Logic track on this site covers the same topics in more depth.
Number systems and Boolean algebra
- Binary, octal, hex conversions; 2's complement for signed numbers (range −2ⁿ⁻¹ to 2ⁿ⁻¹ − 1).
- De Morgan: (AB)′ = A′ + B′, (A + B)′ = A′B′.
- K-maps group 1s in powers of two to minimise sum-of-products expressions; don't-cares can be used as 0 or 1.
- NAND and NOR are universal gates.
Logic families
| Family | Notes |
|---|---|
| TTL | Bipolar; totem-pole output; fan-out about 10 |
| ECL | Fastest (non-saturating), high power |
| CMOS | Very low static power; power ∝ C V² f |
Noise margins: NM_H = V_OH − V_IH, NM_L = V_IL − V_OL.
Combinational circuits
- Half adder: S = A ⊕ B, C = AB. Full adder: S = A ⊕ B ⊕ C_in, C_out = AB + C_in(A ⊕ B).
- Multiplexer: 2ⁿ inputs, n select lines; any n-variable function can be built with a 2ⁿ:1 MUX (or a 2ⁿ⁻¹:1 MUX with one variable on the inputs).
- Decoder: n inputs → 2ⁿ outputs. Encoder, priority encoder, comparator.
Sequential circuits
| Flip-flop | Next state |
|---|---|
| SR | Q⁺ = S + R′Q (S = R = 1 not allowed) |
| JK | Q⁺ = JQ′ + K′Q (J = K = 1 toggles) |
| D | Q⁺ = D |
| T | Q⁺ = T ⊕ Q |
- Ripple (asynchronous) counter: n flip-flops count mod 2ⁿ; delays add up.
- Synchronous counter: all flip-flops on one clock.
- Ring counter with n flip-flops: n states. Johnson (twisted ring) counter: 2n states.
- Maximum clock of a ripple counter ≈ 1/(n × t_pd).
Data converters
- n-bit DAC: resolution = V_ref/(2ⁿ − 1) (or V_ref/2ⁿ depending on convention). R-2R ladder DACs need only two resistor values.
- ADCs: flash (fastest, 2ⁿ − 1 comparators), successive approximation (n clock cycles), dual slope (noise immune), counter type.
Worked example
An 8-bit DAC has a 5 V reference.
- Resolution = 5/255 = 19.6 mV; input 1000 0000 gives 128 × 19.6 mV = 2.51 V.
Practice
How many states does a Johnson counter with 4 flip-flops have?
Show answer and explanation
Answer: 8A Johnson counter with n flip-flops has 2n = 8 states.
A ripple counter uses 4 flip-flops, each with 25 ns delay. What is the maximum clock frequency in MHz?
Show answer and explanation
Answer: 10f_max = 1/(4 × 25 ns) = 10 MHz.
A JK flip-flop has J = K = 1 and Q = 0. What is Q after the clock edge?
Show answer and explanation
Answer: 1With J = K = 1 the flip-flop toggles: Q⁺ = Q′ = 1.
What is the 8-bit 2's complement representation of −5?
Show answer and explanation
Answer: A5 = 00000101; inverting gives 11111010; adding 1 gives 11111011.