circuit intuitions: Example 1

Let us analyze the following circuit for resistance looking into the output node.

Let us identify the small blocks to analyze the whole circuit.

Block 1: Common source amplifier with load resistance.

Block 2: Voltage divider.

Gain of block 2 is A   = 1/2

Now for calculating the output resistance we use the concept discussed in the circuit intuitions article #2. The concept is shown in the above figure. The effective resistance seen at drain node is affected by the feedback gain A. In our example this feedback gain block is voltage divider.

R_out = 1/AG_m || R_o || R_L || (R₁ + R₂)*

= 1MΩ || 1MΩ || 1MΩ || 1MΩ = 250 k Ω   
* In our case the feedback gain block has input resistance which appears parallel to output node.