MOSFET biasing: assume, solve, check the boundary
도입됨| English | 의미 |
|---|---|
| MOSFET | 금속 산화물 반도체 전계 효과 트랜지스터 |
| threshold voltage | threshold voltage |
| triode region | triode region |
| saturation region | 포화 영역 |
| square law | square law |
Billions of switches, no gate current
- A phone's processor holds billions of MOSFETs. Each one is controlled by a voltage on its gate, which draws almost no current.
- The same device makes an amplifier when it is biased correctly. The method in this lesson is: assume a region, solve, then check that the assumption holds.
The voltage-controlled transistor
- The MOSFET 金属氧化物半导体场效应晶体管 has an insulated gate, so the gate draws no DC current. The gate-source voltage $v_{GS}$ controls the drain current.
- Below the threshold voltage 阈值电压 $V_{t}$ the device is off (cut-off). Above it, a channel forms and current can flow from drain to source.
An n-channel MOSFET with V_t = 1.5 V and v_GS = 1.0 V conducts a small drain current set by the square law.
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$v_{GS} < V_{t}$, so no channel forms: the device is cut off. The square law applies only above threshold, in saturation.
Triode and saturation
- Triode region 三极管区 ($v_{DS} < v_{GS} - V_{t}$): the channel behaves like a resistor controlled by $v_{GS}$.
- Saturation region 饱和区 ($v_{DS} \ge v_{GS} - V_{t}$): the current stops depending on $v_{DS}$, and the square law 平方律 holds: $i_{D} = \tfrac{1}{2}k_{n}\dfrac{W}{L}(v_{GS} - V_{t})^{2}$. This is the amplifier region.
v_GS = 4 V, V_t = 1 V and v_DS = 2 V. Which region?
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The boundary is $v_{DS} = v_{GS} - V_{t} = 3$ V. At 2 V the device is below it, in triode: a voltage-controlled resistor, not an amplifier.
Worked example: assume saturation
- $V_{DD} = 10\,\text{V}$, $\tfrac{1}{2}k_{n}\dfrac{W}{L} = 0.5\,\dfrac{\text{mA}}{\text{V}^{2}}$, $V_{t} = 1.5\,\text{V}$, $R_{D} = 3\,\text{k}\Omega$, gate at 3 V. Saturation: $I_{D} = 0.5 \times 1.5^{2} = 1.125\,\text{mA}$.
- $V_{DS} = 10 - 1.125 \times 3 = 6.6\,\text{V}$. Check: $6.6 \ge 3 - 1.5 = 1.5$, so saturation holds and the Q-point is (1.125 mA, 6.6 V).
V_DD 10 V, 0.5 mA/V² device, V_t 1.5 V, R_D 3 kΩ, gate at 3 V. I_D and V_DS are
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$I_{D} = 0.5(3 - 1.5)^{2} = 1.125$ mA and $V_{DS} = 10 - 3.375 = 6.6$ V, which is at least 1.5 V: the region check passes.
In saturation with ½k_n(W/L) = 2 mA/V², V_t = 0.8 V and v_GS = 1.8 V, what is i_D, in mA?
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$2 \times (1.8 - 0.8)^{2} = 2$ mA. Confirm saturation with the circuit: $v_{DS}$ must be at least 1.0 V.
Worked example: the check fails
- Raise the gate to 5 V. Saturation gives $I_{D} = 0.5 \times 3.5^{2} = 6.1\,\text{mA}$ and $V_{DS} = 10 - 18.4 = -8.4\,\text{V}$. Impossible: the assumption is wrong.
- Re-solve in triode, $i_{D} = 0.5(2 \times 3.5\,v_{DS} - v_{DS}^{2})$ mA, with the load line $10 = 3i_{D} + v_{DS}$. The quadratic gives $v_{DS} = 1.0\,\text{V}$, $i_{D} = 3.0\,\text{mA}$; and $1.0 < 3.5$, so triode holds.
Same circuit with the gate at 5 V. The saturation assumption gives V_DS = −8.4 V. The right move is
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A negative drain voltage is impossible, so the assumption is wrong. The triode equation with the load line gives 1.0 V and 3.0 mA, and $1.0 < 3.5$ confirms triode.
Biasing with a divider
- Because the gate draws no current, a gate divider is unloaded: $V_{G} = V_{DD}\dfrac{R_{2}}{R_{1} + R_{2}}$ exactly, even with megohm resistors.
- That keeps the bias network from loading the signal source. A source resistor adds feedback, as $R_{E}$ does for the BJT.
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 a BJT base, whose current pulls the divider down.
Three common slips
- Skipping the region check. Checking only $V_{DS} > 0$ can pass a wrong answer; check $v_{DS} \ge v_{GS} - V_{t}$.
- Mixing up the names. MOSFET saturation is the amplifier region; BJT saturation is the switch-on region.
- Including gate current. At DC it is zero.
Match each region to what the device does there.
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The word "saturation" means opposite things for the two devices; check which device a question is about.
Recap
- Saturation: $i_{D} = \tfrac{1}{2}k_{n}\dfrac{W}{L}(v_{GS} - V_{t})^{2}$, valid when $v_{DS} \ge v_{GS} - V_{t}$.
- Assume a region, solve, check; if the check fails, re-solve in the other region.
- No DC gate current, so gate dividers are unloaded.