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I have been trying to solve this question but I am getting messy eqyuations.

My Attempt:

For saturation mode of NMOS transistor , the following conditions should be satisfied;

$$v_{gs}\ge v_t $$ $$ v_d \ge v_g-v_t$$

Where v_d is drain voltage, v_gs is gate to source voltage, v_t is thershold voltage and v_g is the gate potential.

From the first equation,

$$5-I_d\times 1 \ge v_g-v_t$$ $$\implies 5-I_d \ge 0-v_t$$ $$\implies 5-I_d \ge 0-1$$ $$\implies I_d \le 6$$

Now I tried to express I_d as a function of R_s by using the current equations for drain and source current,but the equations kept getting messier and messier, eventually I gave up.

I think either the question is incorrect, this transistor always remains in saturation independent of R_s, or I have commited errors.

Help would be appreciatec either way.

  • \$\begingroup\$ you used 1 instead of 1K (1000 ohms) \$\endgroup\$
    – Marla
    Dec 5, 2015 at 18:19
  • \$\begingroup\$ I am taking current to be in milliamps. \$\endgroup\$ Dec 5, 2015 at 18:28

1 Answer 1


For the device to be in saturation, we know that the drain-to-source voltage \$ V_{ds}\$ must satisfy \$ V_{ds} \geq V_{gs}-V_t \$, where \$V_{gs}\$ and \$V_t\$ are the gate-to-source voltage and threshold voltage of the device, respectively. For this particular problem, we can simplify the above inequality to be dependent on only the drain voltage \$ V_d\$ and \$ V_t\$ such that \$ V_d \geq -V_t \$, which implies that the drain current for the device \$I_d\$ satisfies

\$ I_d \leq \frac{5+V_t}{1 k\Omega}= I_{d,max} = 6 mA \$.

Let's assume that, for \$ R_s\ = 0\Omega\$, the device is in saturation. Under this assumption, you get a drain current of \$I_d = 1.6 A\$, which violates the above constraint on \$I_d\$. Thus we know that we seek to find a lower-bound on \$R_s\$.

For a drain current of \$ I_d = I_{d,max} = 6 mA \$, you can make use of the \$I_d-V_{gs}\$ equation for a MOSFET in saturation to compute \$ V_{gs,max}\$ of the device to be \$1.245 V\$. By the use of KVL, we can obtain \$ V_{gs,max} + R_{s,min}I_{d,max} = 5\$. From there, you can compute the lower-bound on \$R_s\$ to be

\$R_{s,min} = (5 - V_{gs,max})/I_{d,max} = 625.84 \approx 626 \Omega\$.

A concluding remark:

So long as \$R_s\$ satisfies \$ R_{s,min} \leq R_s < \infty \$, the device will be in saturation; I encourage you to convince yourself that the device will also be on, i.e., \$ V_{gs} > V_t \$. (In fact, for this particular circuit, \$ V_{gs} > V_t \$ holds even if \$ 0 \leq R_s < R_{s,min} \$.)


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