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I am designing active inductor using OTAs (Transconductance amplifiers) and have a problem writing an equations. enter image description here I wrote two equations for nodes V2+ and V2- and input current Iin. $$ 0=sCV^+_{2}+g_1(V^+_{1}-V^-_{1})$$ $$ 0=sCV^-_{2}-g_1(V^+_{1}-V^-_{1})$$ $$ I_o=g_2(V^+_{2}-V^-_{2})$$

If we express voltages from first and second equations, we get our new equations: $$V^+_{2}=-\frac{g_1}{sC}(V^+_{1}-V^-_{1})$$ $$V^-_{2}=\frac{g_1}{sC}(V^+_{1}-V^-_{1})$$ So now if I put, my new equations into the third equation I get: $$I_o=g_2(-\frac{g_1}{sC}(V^+_{1}-V^-_{1})-\frac{g_1}{sC}(V^+_{1}-V^-_{1}))$$ $$I_o=-\frac{2g_1g_2}{sC}(V^+_{1}-V^-_{1})$$ We also know that Io is equal to -Iin. After that we put above equation in equation for calculating the impedance: $$Z=\frac{V^+_{1}-V^-_{1}}{I_{IN}}=\frac{V^+_{1}-V^-_{1}}{\frac{2g_1g_2}{sC}(V^+_{1}-V^-_{1})}=\frac{sC}{2g_1g_2}=sL$$ $$L=\frac{C}{2g_1g_2}$$ But when I try to simulate the circuit below, I get the inductance 4 times bigger and the equation should look like that: $$L=\frac{2C}{g_1g_2}$$ enter image description here Can somebody explain what am I doing wrong here?

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  • \$\begingroup\$ You have 2 common mode , CM, caps loading Vo-'s and 2 differential OTA's . That shows incorrect DM signals \$\endgroup\$
    – D.A.S.
    Commented May 21, 2022 at 15:19

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Just a guess, maybe you missed a factor of 2 in the gain of OTA because they are fully differential. If in each stage you miss a factor of 2, then the end result will be 4x smaller than it should.

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