For the n-channel MOSFET shown in the figure given below, threshold voltage Vth = 2 V. The drain current ID of the MOSFET is 4 mA when the drain resistance RD is 1 kΩ. If the value of RD is increased to 4 kΩ, then the drain current ID will become :
1.4 mA
Read the connection first — it is what makes the problem solvable. The gate is tied to the drain, so
\(V_{GS}=V_{DS}\)
A diode-connected MOSFET is always in saturation, because \(V_{DS}=V_{GS} \gt V_{GS}-V_{th}\) is satisfied automatically. So the square law applies throughout:
\(I_D=K\left(V_{GS}-V_{th}\right)^{2}\)
Step 1 — use the first condition to find K. With RD = 1 kΩ and ID = 4 mA, the drain (and hence the gate) sits at
\(V_{GS}=V_{DS}=10-(4\ \text{mA})(1\ \text{k}\Omega)=10-4=6\ \text{V}\)
\(K=\dfrac{I_D}{(V_{GS}-V_{th})^{2}}=\dfrac{4}{(6-2)^{2}}=\dfrac{4}{16}=0.25\ \text{mA/V}^{2}\)
Step 2 — write the load line for the new resistor. With RD = 4 kΩ and ID in mA,
\(V_{GS}=10-4I_D\)
Step 3 — solve the two together.
\(I_D=0.25\left(10-4I_D-2\right)^{2}=0.25\left(8-4I_D\right)^{2}=4\left(2-I_D\right)^{2}\)
\(I_D=4\left(4-4I_D+I_D^{2}\right)\)
\(4I_D^{2}-17I_D+16=0\)
\(I_D=\dfrac{17\pm\sqrt{289-256}}{8}=\dfrac{17\pm5.745}{8}\)
giving \(I_D=2.84\ \text{mA}\) or \(I_D=1.41\ \text{mA}\).
Step 4 — reject the spurious root. A quadratic from a square law always returns two roots and only one is physical. Test each against \(V_{GS} \gt V_{th}\):
| Root | VGS = 10 − 4ID | Valid? |
|---|---|---|
| 2.84 mA | −1.37 V | Below Vth — the device would be off, contradiction ✗ |
| 1.41 mA | 4.37 V | Above the 2 V threshold ✓ |
So \(I_D\approx1.4\ \text{mA}\) — option 3.
The sanity check. Raising RD from 1 kΩ to 4 kΩ must reduce the current, since the load line becomes steeper and intersects the square-law curve lower down. The current falls from 4 mA to 1.4 mA ✓. Note that it does not fall by the full factor of four, because as the current drops the drain voltage rises, which raises VGS and partly compensates — the self-limiting behaviour of the diode connection.
Why this configuration matters. A diode-connected MOSFET behaves as a non-linear two-terminal resistor and is the standard active load and current-mirror input device in analogue CMOS, precisely because its operating point is set by the current rather than by a fragile bias voltage.
Hence, the drain current becomes 1.4 mA.
Similar Questions
In enhancement mode MOSFET the saturation (drain) current is given by
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(b) \(K\dfrac{W}{L}(V_{gs}-V_{th})(1+\lambda V_{ds})\)
(c) \(K\dfrac{W}{L}(V_{gs}-V_{th})^{2}(1+\lambda V_{ds})\)
(d) \(K\dfrac{W}{L}(V_{gs}-V_{th})^{2}(1-\lambda V_{ds})\)
Out of these
For an n-channel MOS transistor with $\mu_n$ = 600 cm2/Vs, Cox = 7 x 10-8 F/cm2, W = 40 $\mu_m$, L = 4$\mu_m$ and VTO=1.0 V, the value of K parameter is:
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A. The substrate fermi potential ϕF is negative in NMOS
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