H5P1 ZERO-OFFSET AMPLIFIER

In many amplifiers, we use dual power supplies so we can obtain a 0V offset at the output, that is, the DC operating point at the output is 0V. An example amplifier circuit that can achieve this is shown below.
In this circuit, VS+=1.0V, VS−=−1.0V, and the MOSFET parameters are K=1 mAV2 and VT=0.5V.
what formulas do i need to find Vin (while operating in saturation), RL, and max Vin (while still in saturation)

Question

H5P1 ZERO-OFFSET AMPLIFIER

In many amplifiers, we use dual power supplies so we can obtain a 0V offset at the output, that is, the DC operating point at the output is 0V. An example amplifier circuit that can achieve this is shown below.
In this circuit, VS+=1.0V, VS−=−1.0V, and the MOSFET parameters are K=1 mAV2 and VT=0.5V.
what formulas do i need to find Vin (while operating in saturation), RL, and max Vin (while still in saturation)

anonymous · Answer

vIN = -0.5 VOLT & RL=8000 ohm

anonymous · Answer

so i ask for formula's yet i get answers... great job... helps me get the right answer... doesn't that i would like to know the equation's...

anonymous · Answer

0.093 FOR 3rd que.......

anonymous · Answer

HI Raizial!!! Here's the way to get solution
Actually in finding valid Operating ranges 
$$V0 = [-1 + \sqrt{1+2*K*RL*Vs}  ] \div KRL$$
In the above formula Vs= applied voltage across RL and Drain to Source( i.e Vds)..
$$Vs=ids∗RL+Vds(or)Vs=ids∗RL+V0$$
But in given problem , by applying KVL to output loop we get 
$$Vds+idsRL=Vs+(−Vs−)$$   $$Vds+ids∗RL=1−(−1)=2$$
So in finding V0 substitute$$Vs=2 $$  (NOT 1)
And we know Valid maximum input range is $$Vgs = V0 + Vthershold$$
Applying KVL to input loop $$Vin=Vgs+Vs−$$
Substitue the value ...You will get it !!!!!!
Cheers!!!!

benzo · Answer

the above answer are correct but i need all three formula

benzo · Answer

Solve the previous two results for iDS. You will get a quadratic equation, and you will have to choose the correct root: make sure that iDS=0 when vIN=VT.

What is your solution for iDS?

anonymous · Answer

Actually in finding valid Operating ranges

$$V0 = [-1 + \sqrt{1+2*K*RL*Vs} ] \div KRL$$ 
In the above formula Vs= applied voltage across RL and Drain to Source( i.e Vds)..

$$Vs=ids∗RL+Vds(or)Vs=ids∗RL+V0$$

But in given problem , by applying KVL to output loop we get

$$Vds+idsRL=Vs+(−Vs−)$$ $$Vds+ids∗RL=1−(−1)=2$$

So in finding V0 substitute$$Vs=2 $$ (NOT 1)

And we know Valid maximum input range is $$Vgs = V0 + Vthershold$$

Applying KVL to input loop $$Vin=Vgs+Vs−$$

Substitue the value ...You will get it !!!!!!

Cheers!!!!