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cascode_input_resistance

Let's draw the small signal model.

Note that: $v_{gs2}$ and $v_{bs2}$ are grounded (both M1 and M2 bodies are grounded, but not shown in the schematic).

Therefore the two dependent current source from M2 are equal to 0.

Writing the nodal equation at $v_x$ yields:

\[\frac{v_x}{r_{o2}}+\frac{v_x}{r_{o1}}-\frac{v_t}{r_{o1}}=g_{m1}v_{gs1}+g_{mb1}v_{bs1}\]

\[\frac{v_x}{r_{o2}}+\frac{v_x}{r_{o1}}+g_{m1}v_{x}+g_{mb1}v_{x}=\frac{v_t}{r_{o1}}\]

\[v_x\left (\frac{1}{r_{o2}}+\frac{1}{r_{o1}}+g_{m1}+g_{mb1} \right )=\frac{v_t}{r_{o1}}\]

The TRICK to solving the problem without a lot of algebra is to notice that all of the test current flows through $r_{o2}$ and therefore:

\[v_{x}=i_t r_{o2}\]

\[i_t r_{o2}\left (\frac{1}{r_{o2}}+\frac{1}{r_{o1}}+g_{m1}+g_{mb1} \right )=\frac{v_t}{r_{o1}}\]

\[r_{in}=\frac{v_t}{i_t}=r_{o1} r_{o2}\left (\frac{1}{r_{o2}}+\frac{1}{r_{o1}}+g_{m1}+g_{mb1} \right )\]

\[r_{in}=\frac{v_t}{i_t}=\left [r_{o1} + r_{o2}+r_{o1} r_{o2}\left (g_{m1}+g_{mb1} \right ) \right ]\]

\[r_{in}=r_{o1} + r_{o2}\left [1+ r_{o1}\left (g_{m1}+g_{mb1} \right ) \right ]\]

cascode_input_resistance.txt · Last modified: 2015/06/17 20:10 by admin