# Find the Range of Rs for which the transistor is in saturation 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.

• you used 1 instead of 1K (1000 ohms) – Marla Dec 5 '15 at 18:19
• I am taking current to be in milliamps. – Hashir Omer Dec 5 '15 at 18:28

## 1 Answer

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 $0 \leq R_s < R_{s,min}$.)