Consider the sense-amplifier circuit of Fig. 16.20 during the reading of a 1. Assume that the storage cell provides a voltage increment on the B line of ΔV(1) = 0.1 V. If the NMOS devices in the amplifiers have (W/L)n = 0.54 μm/ 0.18 μm and the PMOS devices have (W/L)p = 2.16 μm/0.18 μm, and

Question

Consider the sense-amplifier circuit of Fig. 16.20 during the reading of a 1. Assume that the storage cell provides a voltage increment on the B line of ΔV(1) = 0.1 V. If the NMOS devices in the amplifiers have (W/L)_n = 0.54 μm/ 0.18 μm and the PMOS devices have (W/L)_p = 2.16 μm/0.18 μm, and assuming that V_DD = 1.8 V, V_tn = |V_tp| = 0.5 V, and μ_nC_ox = 4 μ_pC_ox = 300 μA/V², find the time required for v_B to reach 0.9 V_DD. Assume C_B = 1 pF.

Accepted Answer

First, we determine the transconductances g_mn and g_mp

[latex]g_{mn} = μ_{n}C_{ox} \left(\frac{W}{L}\right)_{n} (V_{GS} - V_{t})[/latex]

[latex]= 300 × \frac{0.54}{0.18}(0.9 - 0.5) [/latex]

= 0.36 mA/V

[latex]g_{mp} = μ_{p}C_{ox} \left(\frac{W}{L}\right)_{p} (V_{GS} - |V_{t}|)[/latex]

[latex]= 75 × \frac{2.16}{0.18}(0.9 - 0.5) = 0.36 mA/V[/latex]

Thus, the inverter G_m is

G_m = g_mn + g_mp = 0.72 mA/V

and the time constant τ for the exponential growth of v_B will be

[latex]τ ≡ \frac{C_{B}}{G_{m}} = \frac{1 × 10^{−12}}{0.72 × 10^{−3}} = 1.4 ns[/latex]

Now, the time, Δt, for v_B to reach 0.9 V_DD can be determined from

0.9 × 1.8 = 0.9 + 0.1 e^Δt/1.4

resulting in

Δt = 2.8 ns

Question 16.2: Consider the sense-amplifier circuit of Fig. 16.20 during th...

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