GATE EC Digital Circuits Formula Sheet

Organised by the official GATE 2027 EC syllabus, Section 5: Digital Circuits. Each worked example was checked in code. Practise these: GATE EC Digital Circuits previous year questions.

Number representations and combinatorial circuits

Syllabus: Binary, integer and floating-point numbers. Boolean algebra, minimization of functions using Boolean identities and Karnaugh map, logic gates and their static CMOS implementations, arithmetic circuits, code converters, multiplexers, decoders.

Two's complement

−2n−1≤x≤2n−1−1,−x=xˉ+1-2^{n-1} \le x \le 2^{n-1} - 1, \qquad -x = \bar{x} + 1
−2n−1≤x≤2n−1−1,−x=xˉ+1\begin{aligned} -2^{n-1} &\le x \\ &\le 2^{n-1} - 1, \\ -x &= \bar{x} + 1 \end{aligned}
Symbols
  • nn: number of bits
  • xˉ\bar{x}: x with every bit inverted
Applies when
Signed integers in n bits. The leading bit has weight −2^(n−1). Adding two numbers of the same sign overflows if the result's sign differs.
Common mistake
Thinking the range is symmetric: 8 bits go from −128 to +127, and −128 has no positive partner.
Worked example
The 8-bit pattern 11110110: -10

IEEE 754 single precision

x=(−1)s×1.f×2 e−127x = (-1)^s \times 1.f \times 2^{\,e - 127}
x=(−1)s×1.f×2 e−127\begin{aligned} x &= (-1)^s \\ &\quad \times 1.f \\ &\quad \times 2^{\,e - 127} \end{aligned}
Symbols
  • ss: sign bit
  • ee: 8-bit biased exponent, 1 to 254 for normal numbers
  • ff: 23-bit fraction
Applies when
Normal numbers. e = 0 gives zero and subnormals (0.f × 2^(−126)); e = 255 gives infinity and NaN.
Common mistake
Forgetting the hidden leading 1, or the bias of 127.
Worked example
s = 0, e = 1000 0001, f = 0100 0...0: 5 (1.25 × 2²)

Boolean identities

A+B‾=Aˉ Bˉ,AB‾=Aˉ+Bˉ,AB+AˉC+BC=AB+AˉC,A+AˉB=A+B\overline{A + B} = \bar A\,\bar B,\quad \overline{AB} = \bar A + \bar B,\quad AB + \bar A C + BC = AB + \bar A C,\quad A + \bar A B = A + B
A+B‾=Aˉ Bˉ,AB‾=Aˉ+Bˉ,AB+AˉC+BC=AB+AˉC,A+AˉB=A+B\begin{aligned} \overline{A + B} &= \bar A\,\bar B, \\ \overline{AB} &= \bar A + \bar B, \\ AB + \bar A C + BC &= AB + \bar A C, \\ A + \bar A B &= A + B \end{aligned}
Symbols
  • A,B,CA, B, C: Boolean variables
Applies when
Any Boolean expressions. The third is the consensus theorem: the BC term is redundant.
Common mistake
Applying De Morgan without changing the operator: the complement of a sum is a product of complements.
Worked example
Rows of the truth table where AB + A′C + BC equals AB + A′C: 8 (all 8)

Full adder and ripple-carry delay

S=A⊕B⊕Cin,Cout=AB+Cin(A⊕B),tripple≈n tcarryS = A \oplus B \oplus C_{in}, \qquad C_{out} = AB + C_{in}(A \oplus B), \qquad t_{ripple} \approx n\, t_{carry}
S=A⊕B⊕Cin,Cout=AB+Cin(A⊕B),tripple≈n tcarry\begin{aligned} S &= A \oplus B \oplus C_{in}, \\ C_{out} &= AB + C_{in}(A \oplus B), \\ t_{ripple} &\approx n\, t_{carry} \end{aligned}
Symbols
  • Cin,CoutC_{in}, C_{out}: carry in and out
  • tcarryt_{carry}: carry delay of one stage (s)
  • nn: number of bits
Applies when
A ripple-carry adder's worst case is a carry that travels through every stage; a carry-lookahead adder avoids it.
Common mistake
Counting the sum delay of every stage: only the carry ripples; the last sum adds one gate delay at the end.
Worked example
8-bit ripple-carry adder with 2 ns per carry stage: 16 ns

Multiplexers and decoders

2n ⁣: ⁣1 MUX: n select lines;n-to-2n decoder: 2n outputs, one per minterm2^n\!:\!1 \text{ MUX: } n \text{ select lines}; \qquad n\text{-to-}2^n \text{ decoder: } 2^n \text{ outputs, one per minterm}
2n ⁣: ⁣1 MUX: n select linesn-to-2n decoder: 2n outputs,one per minterm\begin{gathered} 2^n\!:\!1 \text{ MUX: } n \text{ select lines} \\ n\text{-to-}2^n \text{ decoder: } 2^n \text{ outputs,} \\ \text{one per minterm} \end{gathered}
Symbols
  • nn: number of select or input lines
Applies when
Any function of n variables fits a 2^n : 1 MUX with constants on the inputs, or a 2^(n−1) : 1 MUX with the last variable (or its complement) on the inputs.
Common mistake
Thinking an n-variable function needs a 2^n : 1 MUX: one variable can go on the data inputs.
Worked example
A 16 : 1 multiplexer: 4 select lines

Dynamic power of CMOS logic

Pdyn=α CLVDD2fP_{dyn} = \alpha\, C_L V_{DD}^2 f
Symbols
  • α\alpha: activity factor: fraction of clock cycles with a 0 to 1 output transition
  • CLC_L: switched load capacitance (F)
  • ff: clock frequency (Hz)
Applies when
Switching power of static CMOS; static (leakage) and short-circuit power are extra.
Common mistake
Dropping the square on VDDV_{DD}: halving the supply cuts dynamic power to a quarter.
Worked example
α = 0.1, CLC_{L} = 10 pF, VDDV_{DD} = 1.8 V, f = 100 MHz: 0.324 mW

Sequential circuits

Syllabus: Latches and flip-flops, counters, shift-registers, finite state machines, propagation delay, setup and hold time, critical path delay.

Counters

mod-N: ⌈log⁡2N⌉ flip-flops;ring: N states with N;Johnson: 2N states with N\text{mod-}N:\ \lceil \log_2 N \rceil \text{ flip-flops}; \qquad \text{ring: } N \text{ states with } N; \qquad \text{Johnson: } 2N \text{ states with } N
mod-N: ⌈log⁡2N⌉ flip-flops;ring: N states with N;Johnson: 2N states with N\begin{aligned} \text{mod-}N:\ \lceil \log_2 N \rceil \text{ flip-flops}; \\ \text{ring: } N \text{ states with } N; \\ \text{Johnson: } 2N \text{ states with } N \end{aligned}
Symbols
  • NN: number of states, or of flip-flops for ring and Johnson counters
Applies when
A mod-N counter divides the clock frequency by N at its last output.
Common mistake
Using a ring counter's count for a Johnson counter: N flip-flops give 2N states in a Johnson counter.
Worked example
A decade (mod-10) counter: 4 flip-flops

Setup and hold timing

Tclk≥tclk→q+tcomb,max⁡+tsetup,tclk→q+tcomb,min⁡≥tholdT_{clk} \ge t_{clk\to q} + t_{comb,\max} + t_{setup}, \qquad t_{clk\to q} + t_{comb,\min} \ge t_{hold}
Tclk≥tclk→q+tcomb,max⁡+tsetup,tclk→q+tcomb,min⁡≥thold\begin{aligned} T_{clk} &\ge t_{clk\to q} \\ &\quad + t_{comb,\max} \\ &\quad + t_{setup}, \\ &t_{clk\to q} + t_{comb,\min} \\ &\quad \ge t_{hold} \end{aligned}
Symbols
  • tclk→qt_{clk\to q}: flip-flop clock-to-output delay (s)
  • tcombt_{comb}: logic delay between the flip-flops (s)
  • tsetup,tholdt_{setup}, t_{hold}: the receiving flip-flop's setup and hold times (s)
Applies when
Flip-flops on the same ideal clock. Clock skew adds to one side or the other.
Common mistake
Trying to fix a hold violation by slowing the clock: hold doesn't depend on the clock period.
Worked example
tclkt_{clk}→q = 1 ns, tcombt_{comb},max = 5 ns, tsetupt_{setup} = 0.5 ns: the highest clock frequency: 153.8 MHz

Ripple counter speed

fmax⁡=1n tpdf_{\max} = \frac{1}{n\, t_{pd}}
Symbols
  • nn: number of flip-flops
  • tpdt_{pd}: propagation delay of each (s)
Applies when
Asynchronous (ripple) counters, where each flip-flop clocks the next, and the count is read after it settles.
Common mistake
Using a single flip-flop's delay: the last output settles n delays after the clock edge.
Worked example
4-bit ripple counter, 10 ns per flip-flop: 25 MHz

Data converters

Syllabus: Sample and hold circuits, ADCs and DACs.

DAC output and resolution

Vout=Vref D2n,Δ=Vref2nV_{out} = V_{ref}\,\frac{D}{2^n}, \qquad \Delta = \frac{V_{ref}}{2^n}
Vout=Vref D2n,Δ=Vref2n\begin{aligned} V_{out} &= V_{ref}\,\frac{D}{2^n}, \\ \Delta &= \frac{V_{ref}}{2^n} \end{aligned}
Symbols
  • DD: input code, 0 to 2^n − 1
  • nn: number of bits
  • Δ\Delta: step size, one LSB (V)
Applies when
Binary-weighted or R-2R DAC with full scale VrefV_{ref}(1 − 2^(−n)). Some texts divide by 2^n − 1 instead; check which the question uses.
Common mistake
Using 2^n − 1 or 2^n inconsistently with the question's definition of full scale.
Worked example
8 bits, VrefV_{ref} = 5.12 V, D = 100: 2 V

ADC quantisation and SQNR

Δ=VFS2n,σq2=Δ212,SQNR=6.02n+1.76 dB\Delta = \frac{V_{FS}}{2^n}, \qquad \sigma_q^2 = \frac{\Delta^2}{12}, \qquad \text{SQNR} = 6.02n + 1.76\ \text{dB}
Δ=VFS2n,σq2=Δ212,SQNR=6.02n+1.76 dB\begin{aligned} \Delta &= \frac{V_{FS}}{2^n}, \\ \sigma_q^2 &= \frac{\Delta^2}{12}, \\ \text{SQNR} &= 6.02n + 1.76\ \text{dB} \end{aligned}
Symbols
  • VFSV_{FS}: full-scale range (V)
  • σq2\sigma_q^2: quantisation noise power
Applies when
The SQNR formula is for a full-scale sinusoid and a uniform quantiser.
Common mistake
Using 6n dB with no 1.76: it matters when solving for n from an SNR.
Worked example
A 10-bit ADC: 61.96 dB

ADC architectures

flash: 2n−1 comparators;successive approximation: n clock cycles;dual slope: up to 2n+1 cycles\text{flash: } 2^n - 1 \text{ comparators}; \qquad \text{successive approximation: } n \text{ clock cycles}; \qquad \text{dual slope: up to } 2^{n+1} \text{ cycles}
Flash: 2n−1 comparatorsSuccessive approximation:n clock cyclesDual slope: up to 2n+1 cycles\begin{gathered} \text{Flash: } 2^n - 1 \text{ comparators} \\ \text{Successive approximation:} \\ n \text{ clock cycles} \\ \text{Dual slope: up to } 2^{n+1} \text{ cycles} \end{gathered}
Symbols
  • nn: resolution in bits
Applies when
Flash is the fastest and largest; successive approximation does one bit per cycle; dual slope is slow but rejects noise.
Common mistake
Saying 2^n comparators for a flash ADC: the 2^n levels need 2^n − 1 thresholds.
Worked example
An 8-bit flash ADC: 255 comparators

Semiconductor memories and computer organisation

Syllabus: ROM, SRAM, DRAM. Machine instructions and addressing modes, ALU, data-path and control unit, instruction pipelining.

Memory size and address lines

capacity=2a×w bits,a=log⁡2(number of words)\text{capacity} = 2^a \times w \text{ bits}, \qquad a = \log_2(\text{number of words})
capacity=2a×w bitsa=log⁡2(number of words)\begin{gathered} \text{capacity} = 2^a \times w \text{ bits} \\ a = \log_2(\text{number of words}) \end{gathered}
Symbols
  • aa: number of address lines
  • ww: word length (bits)
Applies when
Each address selects one word. A 6-transistor SRAM cell holds 1 bit; a DRAM cell is one transistor and one capacitor, and needs refreshing.
Common mistake
Counting address lines from the total bits instead of the number of words.
Worked example
A 64K × 8 memory: 16 address lines

Pipeline speed-up

Tpipe=(k+n−1) τ,S=nkk+n−1→k as n→∞T_{pipe} = (k + n - 1)\,\tau, \qquad S = \frac{n k}{k + n - 1} \to k \text{ as } n \to \infty
Tpipe=(k+n−1) τ,S=nkk+n−1→k as n→∞\begin{aligned} T_{pipe} &= (k + n - 1)\,\tau, \\ S &= \frac{n k}{k + n - 1} \\ &\to k \text{ as } n \\ &\to \infty \end{aligned}
Symbols
  • kk: number of pipeline stages
  • nn: number of instructions
  • τ\tau: stage time (s)
Applies when
Equal stage times and no stalls. Hazards and unequal stages lower the speed-up.
Common mistake
Claiming a speed-up of k for a short run: the pipeline takes k − 1 cycles to fill.
Worked example
k = 5, n = 100: 4.81

Practise these: GATE EC Digital Circuits previous year questions

Source for the section headings: the official GATE 2027 EC syllabus, IIT Madras. The formulas, examples and notes are Studyloaf's own.

Other GATE EC formula sheets