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It's a normal effect. in a), T1 is 'on', and you sweep the gate of T2. At say the midpoint of the sweep, T2 is partially on (say 5 V on the gate), and some current is flowing through it and T1. T1 being on doesn't mean it has zero resistance, and so T2's source is not at 0 V. Therefore its gate-source voltage is not the same as Uin. conversely, in b), T2 ...

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Bootstrapping is not really the correct term for this. MOSFETs all have capacitance between the gate and drain. While some is a parasitic due to the wires, most of it comes from the actual structure of the MOSFET itself. This capacitance has two effects: When driving the FET via the gate, as the FET turns on and off, the change in voltage across the ...

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The parameter that you should use in your design, is $V_{min}$, because this is the guaranteed value from the manufacturer. $V_{typ}$ is a typical value, means that most devices operate with this value, but the manufacturer can't guarantee for all devices. Then, $V_{min}$ is a little bit higher, and applies for all devices manufactured.

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This question is all about the required input levels into a logic device The minimum value for $V_{IH}$ is the guaranteed value for a "1". In other words, if you equal or exceed this level the input circuit will definitely recognize it as a logical 1. Typically it might work a bit lower than this. As for $V_{IL}$, the maximum value is that input level ...

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Your horrible noise could be grossly underdamped LC ringing .You cant up the C because you have to keep fast rise time .If you place Resistance in your circuit you should be able to damp out the oscillations .More resistance lowers the Q so why not start with 100 ohm.Keep increasing to see your waveform improve.Twisting all the loose wires will help too .The ...

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It probably refers to a non-folded cascode "telescoped" with a diffamp like this: But you should really ask at office hours (for your class) to be certain.

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As you commented, the 4e- in the valance shell of Phosphorous ion joins the bond leaving behind no free electron. But the substrate/wafer becomes positively charged because of the extra proton present in Phosphorous. So as implantation progresses, the substrate becomes more and more positive. If we don't provide a proper method to neutralize this charge, it ...

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If you don't mind adding a little control logic for the start-up sequence, you could do something like this: With no voltage applied to the gate of the P-channel JFET, it is a relatively low resistance. Once your circuit has biased, pull the JFET gate high to make it a high resistance. The value of the resistor to the MOSFET (if any) I will leave to you. ...

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As the input signal starts to rise from 0 V, the NMOS transistor in the inverter will turn on when it reaches a threshold voltage; as it rises further, eventually the PMOS transistor will turn off. In the initial portion of the rise, the PMOS is still (fully) on, and the NMOS has not started to turn on. However, there is gate-drain capacitance in that ...

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As you know, you use the upper part of the structure (from Vout to Vcc) to set the 1's and the lower part of the structure (from Vout to Vss) to set the 0's. You can do this directly from the function expression. For example, let say your function is F=A·B*+C. I've chosen a different (and shorter function) for a simpler illustration of the method. The ...

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