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CMOS characteristics; 2-input CMOS NOR

Short answer

CMOS gates feature near-zero static power, wide supply range (3-15V for 4000B), high noise margin (~30% VDD), and rail-to-rail output. A 2-input CMOS NOR uses series PMOS and parallel NMOS; output is high only when both inputs are low, otherwise low.

Explain the characteristics of CMOS gates and explain logic operation of CMOS 2-input NOR gate circuit with its truth table.

Characteristics of CMOS gates

CMOS (complementary MOS) gates use PMOS pull-up and NMOS pull-down transistors in pairs. Main characteristics (4000B and 74HC series):

CharacteristicTypical value / remark
Supply voltage3–15 V (4000B), 2–6 V (74HC)
Logic levels (5 V CMOS, typical)VIL≤1.5V_{IL} \le 1.5 V, VIH≥3.5V_{IH} \ge 3.5 V, VOL≈0.1V_{OL} \approx 0.1 V, VOH≈4.9V_{OH} \approx 4.9 V
Static poweralmost zero (nW per gate)
Dynamic powerPD=CLVDD2fP_D = C_L V_{DD}^2 f, rises with frequency
Noise marginhigh, about 30% of VDDV_{DD} (1.4 V at 5 V)
Propagation delay25–100 ns (4000B), about 8–10 ns (74HC)
Fan-outvery high (about 50) because MOS inputs draw almost no current
Handlinggate oxide is damaged by static charge; unused inputs must be tied to VDDV_{DD} or ground

2-input CMOS NOR gate

Two PMOS transistors P1P_1, P2P_2 are in series from VDDV_{DD} to the output, and two NMOS transistors N1N_1, N2N_2 are in parallel from the output to ground. Input A drives P1P_1 and N1N_1; input B drives P2P_2 and N2N_2.

            VDD
             |
     A --o| P1
             |
     B --o| P2
             |
             +---------- Y = (A+B)'
             |        |
     A ---| N1  B ---| N2
             |        |
            GND      GND

A PMOS is ON when its gate is 0; an NMOS is ON when its gate is 1.

Operation:

  • A = 0, B = 0: P1P_1, P2P_2 ON; N1N_1, N2N_2 OFF. Output connected to VDDV_{DD} → Y = 1.
  • A = 0, B = 1: N2N_2 ON pulls output to ground; P2P_2 OFF breaks the pull-up path → Y = 0.
  • A = 1, B = 0: N1N_1 ON, P1P_1 OFF → Y = 0.
  • A = 1, B = 1: both NMOS ON, both PMOS OFF → Y = 0.
ABP1P2N1N2Y
00ONONOFFOFF1
01ONOFFOFFON0
10OFFONONOFF0
11OFFOFFONON0

So Y=A+B‾Y = \overline{A + B}. Since the pull-up and pull-down networks are never ON together in steady state, no DC path exists from VDDV_{DD} to ground, which is why CMOS static power is so low.

✓ Answer checked by Claude Opus 5.5 (high effort)

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