For the circuit in the diagram, observe that Vgs is around 3V and so is Vds on the breadboard. I do not understand if it is necessary to keep the drain at 1/2 of V1 as it is done when using a drain resistor ( even in the case of using a current source). I am observing that the output is clipped for even around 6V p-p, there is another 10V to use. Basically not clear on how the current source works as a better drain resistor. Appreciate any help. Thanks
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