The circuit in Fig. 7.3(a) is fabricated in a 0.18 -um CMOS...

The circuit in Fig. $7.3(a)$ is fabricated in a 0.18 -um CMOS technology for which $\mu_{n} C_{o x}=387 \mu \mathrm{AV}$ $\mu_{p} C_{o x}=86 \mu \mathrm{A} V^{2}, V_{t n}=-V_{t p}=0.5 \mathrm{V}, V_{A n}^{\prime}=5 \mathrm{V} / \mathrm{m}$ $\left|V_{A P}^{\prime}\right|=6 \mathrm{V} / \mu \mathrm{m},$ and $V_{D D}=1.8 \mathrm{V} .$ It is required to design the circuit to obtain a voltage gain $A_{v}=-40$ V/V. Use devices of equal length $L$ operating at $I=100 \mu \mathrm{A}$ and $\left|V_{O V}\right|=0.2 \mathrm{V}$ Determine the required values of $V_{G}, L,(W / L)_{1},$ and $(W / L)_{2}$

The circuit in Fig. $7.3(a)$ is fabricated in a 0.18 -um CMOS technology for which $\mu_{n} C_{o x}=387 \mu \mathrm{AV}$ $\mu_{p} C_{o x}=86 \mu \mathrm{A} V^{2}, V_{t n}=-V_{t p}=0.5 \mathrm{V}, V_{A n}^{\prime}=5 \mathrm{V} / \mathrm{m}$
$\left|V_{A P}^{\prime}\right|=6 \mathrm{V} / \mu \mathrm{m},$ and $V_{D D}=1.8 \mathrm{V} .$ It is required to design the
circuit to obtain a voltage gain $A_{v}=-40$ V/V. Use devices of equal length $L$ operating at $I=100 \mu \mathrm{A}$ and $\left|V_{O V}\right|=0.2 \mathrm{V}$ Determine the required values of $V_{G}, L,(W / L)_{1},$ and $(W / L)_{2}$

Key Concepts

CMOS Device Parameters

CMOS technology relies on parameters such as carrier mobility, oxide capacitance (??C?? and ??C??), threshold voltage (V?), and the channel-length modulation parameter. These process-specific values determine the operating characteristics of both NMOS and PMOS transistors and are essential for calculating bias currents, transconductance, and output impedance in analog circuit design.

Transistor Operating Regions and Overdrive Voltage

Ensuring a transistor operates in the saturation region is vital for amplifier performance. The overdrive voltage (V_OV), defined as the excess of the gate-to-source voltage over the threshold voltage, ensures that the device remains in saturation. This condition is crucial for linear amplification and directly impacts the device’s transconductance and, consequently, the voltage gain.

Voltage Gain in Amplifier Design

The voltage gain of a common-source or differential amplifier is primarily the product of the device's transconductance and its load resistance. A target gain necessitates careful balancing of design parameters such as bias current, overdrive voltage, and the aspect ratio of the transistors, ensuring that the calculated g? and load resistance yield the desired gain.

Device Sizing and Aspect Ratios

The aspect ratio (W/L) of a MOS transistor is a key design parameter that influences its current carrying capability and transconductance. By adjusting the width-to-length ratio, designers can set the operating current, control the overdrive voltage, and meet specific performance targets like desired voltage gain, making it an essential consideration in CMOS circuit design.

Channel-Length Modulation

Channel-length modulation describes how the effective channel length of a MOS transistor decreases as the drain-to-source voltage increases, thereby affecting the drain current. This phenomenon, quantified by the channel-length modulation parameter, impacts the output resistance and overall amplifier gain, and must be taken into account when designing for precise analog performance.

Transcript

0:00 Hi there.

00:01 So for this problem, we have the circuit that is shown in this figure and contains two resistors.

00:07 Resistor 1 is equal to 2 kilo -onms.

00:13 And resistor 2 is equal to, sorry, resistor 2 is equal to 3 kiloons.

00:24 So, and we have two capacitors.

00:28 The first one, it has a capacitance of, of 2 microphorets and the capacitor 2 has a capacitance of 3 micro ferrets and they are connected to a battery with an nf value of 120 bulls so if there are no charges on the capacitors before the switch s is closed we need to determine the charges q1 and q2 on the capacitor c1 and c2 respect as functions of time after the switch is closed.

01:08 So, the first thing that we are going to determine is the total resistance in the circuit.

01:15 As you can see, these resistors right here are in a parallel configuration.

01:22 So we're going to have that the resistance, the equivalent resistance for this, is 1 over the resistance 1 plus 1 over the resistance 2, and all of these elevated to the minus 1.

01:33 So in here we're going to have 1 over 2 kilooms plus 1 over 3 kiloons.

01:47 And all of this elevated to the minus 1.

01:51 So from this we obtain a value of 1 .2 kiloons.

01:55 And the total capacitance, as you can see, because the capacitors are in parallel, the total capacitance is simply the sum of the capacitance.

02:07 So we will have c1 plus c2.

02:10 So we substitute those values that is 2 microphorets plus 3 microphorates.

02:16 So we will obtain 5 microphoress for the total capacitance.

02:21 Then the maximum charge that we can have in here is the capacitance times the potential difference.

02:30 So this is going to be 5 microseau.

02:34 4x times 120 bolts.

02:37 So from this we obtain a maximum charge of 600 microcolops.

02:43 And we also can obtain the time constant because in this case we are dealing with an rc circuit.

02:54 This is going to be just the product between the resistance and the capacity...

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