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Please see images of question, my working out and the correct solution. I seem to be incorrect for some reason. This is without resorting to the small-signal model.

Question - Find Rout My Working Out Correct Solution

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    \$\begingroup\$ 1) You cannot determine Rout without using the small signal equivalent circuit. 2) the right circuit in the picture with Rg grounded and the gate of M2 grounded makes no sense 3) Start with Rg and M2 and determine the impedance (to ground) of Rg and M2. Then add M1 and recalculate. \$\endgroup\$ Commented May 3, 2016 at 17:04
  • \$\begingroup\$ Oh. The reason why I grounded them, is that I assumed infinite impedance at the gate terminal hence no current flow. \$\endgroup\$
    – Arsenal123
    Commented May 3, 2016 at 17:26
  • \$\begingroup\$ Also, why has Rg been totally neglected in the final answer? \$\endgroup\$
    – Arsenal123
    Commented May 3, 2016 at 17:27
  • \$\begingroup\$ Rg can be neglected because no current can flow through it since the gate is basically an open (infinite impedance). But that does not mean Rg does nothing ! The gate of M2 has the same voltage as the drain of M2 because.... Rg provides a connection. So if you ground Rg (like you did) that connection is no longer present. Changing the behaviour of M2 ! \$\endgroup\$ Commented May 3, 2016 at 18:47
  • \$\begingroup\$ Oh. So would that mean the Vgate(M2) = Vdrain(M2) = Vsource(M1) ? \$\endgroup\$
    – Arsenal123
    Commented May 4, 2016 at 9:10

1 Answer 1

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The answer is simple: That depends on the voltage applied to Vb1. Remember those transistors act like variable resistors function of applied voltage at Gate. For specific values of Rgs refer to the data sheets from manufacturers, they usually have all values you need to solve this problem.

Remember that different transistors have different Drain to Source impedance depending on applied voltage and other conditions.

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    \$\begingroup\$ "Remember those transistors act like variable resistors function of applied voltage at Gate" Only when in linear (or triode) mode. Not when they're in saturation mode which is the usual case in these scenarios. Besides, the solution for linear mode gets a lot more complex, also the circuit is useless when the transistors operate in triode mode. \$\endgroup\$ Commented May 4, 2016 at 9:16

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