Chapter 4: Problem 6
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)?
Short Answer
Step by step solution
Understanding MOSFET Output Resistance
Calculate \(r_o\) for \(I_D = 50 \mu \text{A}\)
Calculate \(r_o\) for \(I_D = 500 \mu \text{A}\)
Determine Change in \(I_D\) with Increase in \(V_{DS}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Channel-Length Modulation
Why does this matter? Well, it affects the MOSFET's output resistance. The parameter that characterizes channel-length modulation is \(\lambda\), which has units of V^{-1}. A nonzero \(\lambda\) indicates that I_{D} increases as V_{DS} increases, even if slightly. Recognizing channel-length modulation helps in understanding MOSFET behavior in real applications where ideal conditions don't always apply.
Drain Current
In the saturation region, the drain current achieves a stable and quasi-constant level given traditional parameters are held steady. However, due to channel-length modulation, I_{D} can continue to increase beyond this point if V_{DS} is increased. To compute I_{D}, engineers often consider the factor \(\lambda\) and its interplay with \(V_{DS}\).
The expression for I_{D} in the saturation region, considering channel-length modulation, is given by:
- \(I_D = I_{D0}(1 + \lambda V_{DS})\)
- Where I_{D0} is the ideal drain current without modulation effect
Saturation Region
- The MOSFET is fully "ON," allowing maximum current to flow from drain to source.
- It does not behave like a resistor; instead, it exhibits a constant current level.
- The condition for the saturation region is \(V_{DS} > V_{GS} - V_{th}\).
Despite its name, the current isn't truly flat due to channel-length modulation.
With a rise in V_{DS}, channel-length modulation may cause a slight increase in current, reflecting a non-ideal behavior that designers need to account for, especially in precise applications.
Percentage Increase in Current
- The change in current is calculated as \(\Delta I_{D} = I_{D0} \times \lambda \times \Delta V_{DS}\).
- The percentage increase is given by \(\frac{\Delta I_{D}}{I_{D0}} \times 100\%\).
For practical applications, designers calculate this percentage to ensure that their system performs within desired thresholds even when subject to voltage variations. In the given exercise, a 1 V increase in V_{DS} resulted in a 2% increase in I_{D}, based on the \(\lambda\) value of 0.02 V^{-1}. Understanding this relationship aids in the design and robustness of electronic circuits.