Finding the max. level of input signal in a mosfet

[killerdevil]

Homework Statement

Homework Equations

values in part (a) and (b) which i had found:
g_m = 2.2mS
r_o = 25kΩ
R_in = 1MΩ
R_out = 446.43Ω
A_v = 0.99

The Attempt at a Solution

all that i know is that when they refer to pinch off region, it is referring to the saturation region which means:
V_DS ≥ V_GS - V_TN, where V_TN = 1
therefore V_DS ≥ 5 -1 = 4

solution given:
V_DS ≥ 4
-V_o ≥ 4
V_o ≤ 4
A_v V_i ≤ 4
therefore,
V_i≤ 4/A_v
V_i≤ 4/0.99
V_i≤ 4.04V (ans)

my qn would be how did they show that
V_DS ≥ 4
-V_o ≥ 4

 

[KataKoniK]

Vo ≤ 4
Av Vi ≤ 4

Thank you for your post. It seems that you have already done some calculations and have found the values for gm, ro, Rin, Rout, and Av. Good work!

To answer your question, let's start by looking at the equation for Av. Av represents the voltage gain of the amplifier, and it is defined as the ratio of the output voltage (Vo) to the input voltage (Vi). In this case, Av is given as 0.99. This means that for every 1V increase in Vi, there will be a 0.99V increase in Vo.

Now, let's look at the equations for VDS and VGS. As you mentioned, VDS ≥ VGS - VTN, where VTN is the threshold voltage (1V in this case). This means that for the transistor to be in the saturation region, VDS must be greater than or equal to the difference between VGS and VTN. In this case, VGS is given as 5V, so VDS must be greater than or equal to 5V - 1V = 4V.

Next, let's look at the equation for Vo. We know that Vo is the output voltage of the amplifier, and we also know that it is related to VDS by the equation Vo = -VDS. This means that if VDS is greater than or equal to 4V, then Vo must be less than or equal to -4V.

Finally, we can combine these equations to find the relationship between Vi and Vo. Since Av = Vo/Vi, we can rearrange the equation to get Vo = Av*Vi. Substituting in the values we have, we get Vo ≤ 0.99*Vi. We also know that Vo ≤ -4V, so we can write -4V ≤ 0.99*Vi. Dividing both sides by 0.99, we get Vi ≤ -4/0.99 = -4.04V. However, since Vi must be a positive value, we can write Vi ≤ 4.04V.

I hope this helps to clarify how the solution was obtained. If you have any further questions, please don't hesitate to ask. Good luck with your studies!