anonymous
  • anonymous
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)
MIT 6.002 Circuits and Electronics, Spring 2007
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anonymous
  • anonymous
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)
MIT 6.002 Circuits and Electronics, Spring 2007
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
vIN = -0.5 VOLT & RL=8000 ohm
anonymous
  • anonymous
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
  • anonymous
0.093 FOR 3rd que.......

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anonymous
  • anonymous
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
  • benzo
the above answer are correct but i need all three formula
benzo
  • benzo
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
  • anonymous
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!!!!

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