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enter image description here In the book i=2 ma is given I can get this ans by solving equations but how to intutively prove that the current would be same?

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Let's say that initially the gate voltages are at 0V. As the left transistor is off, all that current will be used to charge the gate capacitors. This means that the gate voltage will rise.

After a while, the gate voltage will cross \$V_T\$, the threshold voltage. From that point on, the left transistor will start conducting some current to ground. So from the initial 2mA, a part will go to ground through the drain of the left transistor, and the rest will go to further charging the gate capacitance. As the gate voltage continues to rise, the drain current keeps rising as well. After a while, all of the 2mA will be conducted to ground through the left transistor, and none will be left for charging the gate capacitors.
Let's say that - for whatever reason - the gate voltage where to increase even further, then the left transistor would conduct more than 2mA to ground, causing the gate capacitances to be discharged again. So this structure will self-compensate both ways! It will always try to converge to a steady-state.

It is then useful to look at what that steady-state is. The left transistor has regulated itself to a point where the gate voltage is exactly the gate voltage you would need to have 2mA through the drain. This voltage can then be passed to one or more other transistors, to turn it back into a current.

Adding a source resistor does not really influence this behavior. The current through both branches are equal, so that means the voltage across the resistors are also equal.

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If parts are all perfectly matched: Voltage on Rs on the left is R x 2 milliamps. Vgs on the left is the voltage that corresponds to Id of 2 milliamps. The voltages on both gates are the same (they are tied together). Vgs on the right is the same as Vgs on the left. Id is the same.

In real life there would be some mismatch.

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  • \$\begingroup\$ How did you prove that vgs on right is equal to vgs on left considering the fact that only gate is same between left and right \$\endgroup\$
    – Rac
    Commented Jul 20, 2018 at 3:52

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