A designer uses a BJT common-emitter amplifier at a quiescent collector current of 1 mA, giving g_m = 40 mA/V. She wants to achieve the same g_m using a MOSFET common-source amplifier. Which statement best describes the required bias current?
AThe same 1 mA — g_m scales with bias current the same way in both devices
BThe MOSFET will require less than 1 mA because its gate draws no current, reducing power
CThe MOSFET will require more than 1 mA because its g_m scales with the square root of drain current, not linearly, making it less efficient at achieving high g_m from low bias currents
DThe MOSFET cannot achieve the same g_m because its insulated gate limits transconductance
BJT transconductance is g_m = I_C / V_T, scaling linearly with collector current. MOSFET transconductance is g_m = √(2k_n'(W/L)I_D), scaling with the square root of drain current. To match a BJT's g_m, a MOSFET must typically run at a higher drain current. This square-root dependence means MOSFETs are less efficient at generating high transconductance from a small bias current — a fundamental design tradeoff when choosing between the two device types.
Question 2 Multiple Choice
In a circuit design, a high-impedance sensor output must drive a low-impedance load without significant voltage loss. Which MOSFET configuration is best suited, and why?
ACommon-source, because it provides the highest voltage gain
BCommon-gate, because it has the best high-frequency isolation
CCommon-drain (source follower), because it has near-unity voltage gain, high input impedance set by the bias network, and low output impedance (~1/g_m)
DCommon-source with a large R_D, because larger drain resistance increases input impedance
The source follower (common-drain) is a buffer: it presents a high impedance to the source (set by gate bias resistors, typically MΩ) and a low output impedance (~1/g_m, typically tens to hundreds of Ω). Its gain is approximately unity, so the signal is faithfully reproduced at the output without voltage loss despite the impedance mismatch. Common-source has high gain but also moderate output impedance and phase inversion — not suitable for buffering.
Question 3 True / False
A MOSFET's gate draws no DC current, which means the input impedance of a MOSFET amplifier circuit is essentially infinite.
TTrue
FFalse
Answer: False
The gate itself draws no DC current — the gate oxide is an insulator — but the actual circuit input impedance is set by the bias resistor network connected to the gate. Practical MOSFET amplifiers use resistor voltage dividers (e.g., two resistors from supply to ground) to set the DC gate voltage, and the parallel combination of these resistors determines the circuit's input impedance, typically hundreds of kΩ to a few MΩ. Confusing device physics (infinite gate impedance) with circuit impedance (finite, set by bias network) is a common design error.
Question 4 True / False
Doubling the quiescent drain current in a MOSFET common-source amplifier doubles the transconductance g_m.
TTrue
FFalse
Answer: False
MOSFET transconductance is g_m = √(2k_n'(W/L)I_D), so it scales with the square root of drain current. Doubling I_D increases g_m by a factor of √2 ≈ 1.41, not 2. This contrasts with BJTs, where g_m = I_C/V_T scales linearly with collector current. Designers who apply BJT intuition to MOSFET gain calculations will systematically overestimate the gain achieved by increasing bias current.
Question 5 Short Answer
Why does MOSFET transconductance scale with the square root of drain current rather than linearly, and what does this mean for amplifier design compared to BJTs?
Think about your answer, then reveal below.
Model answer: MOSFET drain current in saturation follows I_D = (k_n'/2)(W/L)(V_GS − V_th)², giving g_m = dI_D/dV_GS = k_n'(W/L)(V_GS − V_th) = √(2k_n'(W/L)I_D). This square-law relationship means g_m ∝ √I_D. In contrast, the BJT's exponential I_C–V_BE relationship gives g_m = I_C/V_T, scaling linearly. The consequence is that achieving high g_m in a MOSFET requires disproportionately large drain current — the MOSFET is less 'efficient' at converting bias current into transconductance, which matters for low-power design.
This difference in device physics shapes design strategy: BJTs are preferred when high transconductance from small bias current is critical (e.g., low-noise, low-power analog). MOSFETs dominate digital circuits (superior switching) and high-input-impedance applications. Understanding why the scaling differs — exponential vs. square-law transfer characteristic — is the foundation of transistor-level analog design intuition.