MOSFET biasing: assume, solve, check the boundary
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| MOSFET | MOSFET: transistor trường hiệu kim loại-oxit-bán dẫn |
| saturation region | saturation region: vùng bão hòa |
| square-law model | square-law model: mô hình bình phương |
| threshold/ˈθreʃəʊld/ | threshold: ngưỡng |
MOSFET biasing: assume, solve, check the boundary
- The gate draws no DC current (megohm dividers allowed), and in the saturation region — the amplifier region, the opposite naming to the BJT — the square-law model gives $i_D = \tfrac{1}{2}k_n(W/L)(v_{GS} - V_t)^2$, with threshold $V_t$.
- The region test is one comparison, $v_{DS} \ge v_{GS} - V_t$ — the boundary parabola on the characteristics plane. The model is an assumption; the check is what makes it legal.
$V_{DD} = 10$ V, $\tfrac{1}{2}k_n(W/L) = 0.5$ mA/V², $V_t = 1.5$ V, $R_D = 3$ kΩ. Gate at 3 V: saturation gives $I_D = 1.125$ mA, $V_{DS} = 6.6$ V $\ge 1.5$ ✓ — stands. Gate at 5 V: saturation would need $V_{DS} = -8.4$ V — impossible; re-solve in triode: $v_{DS} = 1.00$ V, $i_D = 3.00$ mA, and the triode equation returns $0.5(2 \times 3.5 \times 1 - 1) = 3.0$ mA ✓.

{exercise:uni-ee5x-mosfet}
VDD 10 V, 0.5 mA/V2 device, Vt 1.5 V, RD 3 k, gate 3 V. ID and VDS are
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ID = 0.5(3-1.5)^2 = 1.125 mA, VDS = 10 - 3.375 = 6.6 V >= 1.5 - the boundary check confirms saturation.
Same circuit with the gate at 5 V. The saturation assumption gives VDS = -8.4 V, so
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A negative drain voltage is impossible - the assumption died. The triode quadratic with the load line gives vDS = 1.0 V, iD = 3.0 mA, and the triode equation returns 3.0 mA - consistent.
At DC, the MOSFET gate divider can use megohm resistors because
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The insulated gate draws nothing at DC, so the divider is unloaded - unlike the BJT whose base current loads the divider (the 11% error of unit 3).