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Chapter 4: Problem 2

A PMOS transistor has parameters \(V_{T P}=-0.6 \mathrm{~V}, k_{p}^{\prime}=40 \mu \mathrm{A} / \mathrm{V}^{2}\), and \(\lambda=0.015 \mathrm{~V}^{-1}\). (a) (i) Determine the width-to-length ratio \((W / L)\) such that \(g_{m}=1.2 \mathrm{~mA} / \mathrm{V}\) at \(I_{D Q}=0.15 \mathrm{~mA}\). (ii) What is the required value of \(V_{S G O}\) ? (b) Repeat part (a) for \(I_{D O}=0.50 \mathrm{~mA}\).

Short Answer

Expert verified
(a) (i) W/L = 60; (ii) V_{SGO} = -0.246 V. (b) (i) W/L = 18; (ii) V_{SGO} = 0.579 V.

Step by step solution

01

Understanding Transconductance Formula

The transconductance (\(g_m\)) of a PMOS transistor in saturation is given by the formula:\[ g_m = \sqrt{2k_p'(W/L)I_{DQ}} \]where \(k_p'\) is the process transconductance parameter, \(W/L\) is the width-to-length ratio of the transistor, and \(I_{DQ}\) is the quiescent drain current.
02

Solving for Width-to-Length Ratio (W/L)

Rearranging the transconductance formula to solve for \(W/L\):\[(W/L) = \left( \frac{g_m^2}{2k_p'I_{DQ}} \right)\]Substitute the given values: \(g_m = 1.2\, \text{mA/V} = 1200\, \mu\text{A/V}\), \(k_p' = 40\, \mu\text{A/V}^2\), and \(I_{DQ} = 0.15\, \text{mA} = 150\, \mu\text{A}\):\[(W/L) = \frac{(1200)^2}{2 \times 40 \times 150} = 60\].
03

Finding the Required Value of V_{SGO}

To find \(V_{SGO}\), use the equation for drain current in saturation:\[I_{DQ} = \frac{k_p'}{2} \left(\frac{W}{L}\right) (V_{SGO} - V_{TP})^2 (1 + \lambda V_{DS})\]Assuming \(V_{DS}\) is sufficiently large so that \(\lambda V_{DS} \approx 0\), the expression simplifies significantly:\[V_{SGO} = V_{TP} + \sqrt{\frac{2I_{DQ}}{k_p' (W/L)}}\]Substitute the given values: \(V_{TP} = -0.6\, \text{V}\), \(I_{DQ} = 0.15\, \text{mA} = 150\, \mu\text{A}\), \((W/L) = 60\), \(k_p' = 40\, \mu\text{A/V}^2\):\[V_{SGO} = -0.6 + \sqrt{\frac{2 \times 150}{40 \times 60}} = -0.6 + \sqrt{0.125} = -0.6 + 0.354 = -0.246\, \text{V}\].
04

Repeat for I_DQ = 0.50 mA

Now, repeat the process for the new current value \(I_DQ = 0.50\, \text{mA}\).**For \(W/L\):**\[(W/L) = \frac{(1200)^2}{2 \times 40 \times 500} = 18\]**For \(V_{SGO}\):**\[V_{SGO} = V_{TP} + \sqrt{\frac{2I_{DQ}}{k_p' (W/L)}} = -0.6 + \sqrt{\frac{2 \times 500}{40 \times 18}}\]Simplifying further:\[ V_{SGO} = -0.6 + \sqrt{1.389} = -0.6 + 1.179 = 0.579\, \text{V}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transconductance
Transconductance, often denoted as \(g_m\), is a fundamental parameter in the operation of PMOS transistors. It describes how effectively a transistor can control the current flowing through it in response to changes in gate-source voltage, \(V_{SG}\).
  • Transconductance is defined mathematically as \(g_m = \frac{\partial I_D}{\partial V_{SG}}\), where \(I_D\) is the drain current.
  • Its unit is Siemens (S) or mA/V, indicating the change in current per unit change in voltage.
  • In the saturation region, the transconductance is strongly related to device physical attributes and the quiescent point, given by \(g_m = \sqrt{2k_p'(W/L)I_{DQ}}\).
Transconductance is crucial in designing amplifiers. High transconductance means the transistor can amplify signals more effectively. To reach a specific transconductance, adjustments in width-to-length ratio or bias current are often required.
Width-to-Length Ratio
The width-to-length ratio \((W/L)\) of a PMOS transistor is another critical design parameter that influences its electrical properties. It essentially determines the strength and drive capabilities of the transistor.
  • This ratio is a measure of the physical dimensions of the transistor, specifically the width of the gate (\(W\)) compared to its length (\(L\)).
  • A larger \((W/L)\) ratio means a larger area for current to pass through, leading to higher possible current and reduced channel resistance.
  • The ratio is adjusted to meet desired performance criteria, like achieving a specific transconductance or current level, as shown in the formula \((W/L) = \left( \frac{g_m^2}{2k_p'I_{DQ}} \right)\).
In practical applications, choosing the right \(W/L\) ratio is crucial for balancing speed, power, and thermal considerations in circuits. It impacts the switching speeds and power consumption.
Saturation Current
Saturation current, denoted as \(I_D\), is the maximum current that flows through a PMOS transistor when it is in the saturation region. This region is typically used to operate devices as constant current sources or amplifiers.
  • It is crucial for determining the operational limits and is defined by the equation \(I_D = \frac{k_p'}{2} \left(\frac{W}{L}\right) (V_{SG} - V_T)^2 (1 + \lambda V_{DS})\).
  • In the saturation region, the current does not increase linearly with \(V_{SG}\) due to velocity saturation effects.
  • Proper biasing ensures that the transistor remains in saturation, ensuring stable operation conditions for analog designs like op-amps.
In design practices, ensuring the transistor operates in saturation is crucial for reliability. This often involves careful selection of \(V_{SG}\) or \(W/L\) to maintain desired current levels.
Drain Current
The drain current \(I_D\) is a key parameter in analyzing and designing circuits involving PMOS transistors. It represents the electric current flowing from the drain to the source.
  • It is heavily reliant on the applied voltages and the design parameters of the transistor, such as threshold voltage \(V_{TP}\), \(W/L\) ratio, and other technological parameters like \(k_p'\).
  • For PMOS transistors, \(I_D\) can be controlled by the gate-source voltage \(V_{SG}\) and is described by current equations specific for cutoff, triode, and saturation regions.
  • The equation in saturation is particularly important: \(I_D = \frac{k_p'}{2} \left(\frac{W}{L}\right) (V_{SG} - V_T)^2\), showing dependence on \(W/L\) and \(V_{SG}\).
Understanding and manipulating \(I_D\) is vital in analog circuit design, affecting gain, bandwidth, and power consumption. Accurate calculation ensures the circuit meets intended performance specifications.

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Most popular questions from this chapter

The value of \(\lambda\) for a MOSFET is \(0.02 \mathrm{~V}^{-1}\). (a) What is the value of \(r_{o}\) at (i) \(I_{D}=50 \mu \mathrm{A}\) and at (ii) \(I_{D}=500 \mu \mathrm{A}\) ? (b) If \(V_{D S}\) increases by \(1 \mathrm{~V}\), what is the percentage increase in \(I_{D}\) for the conditions given in part (a)?

The small-signal parameters of an enhancement-mode MOSFET source follower are \(g_{m}=5 \mathrm{~mA} / \mathrm{V}\) and \(r_{o}=100 \mathrm{k} \Omega\). (a) Determine the no-load small-signal voltage gain and the output resistance. (b) Find the smallsignal voltage gain when a load resistance \(R_{S}=5 \mathrm{k} \Omega\) is connected.

A common-source amplifier, such as shown in Figure \(4.14\) in the text, has parameters \(r_{o}=100 \mathrm{k} \Omega\) and \(R_{D}=5 \mathrm{k} \Omega\). Determine the transconductance of the transistor if the small-signal voltage gain is \(A_{1}=-10\). Assume \(R_{\mathrm{S}_{i}}=0\).

An n-channel MOSFET is biased in the saturation region at a constant \(V_{G S}\). (a) The drain current is \(I_{D}=0.250 \mathrm{~mA}\) at \(V_{D S}=1.5 \mathrm{~V}\) and \(I_{D}=0.258 \mathrm{~mA}\) at \(V_{D S}=3.3 \mathrm{~V}\). Determine the value of \(\lambda\) and \(r_{o}\). (b) Using the results of part (a), determine \(I_{D}\) at \(V_{D S}=5 \mathrm{~V}\).

The minimum value of small-signal resistance of a PMOS transistor is to be \(r_{o}=100 \mathrm{k} \Omega\). If \(\lambda=0.012 \mathrm{~V}^{-1}\), calculate the maximum allowed value of \(I_{D}\).
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