I am having problem understanding the Transfer function for Small signal model of inductor series peaking as shown below.

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However, how does this transfer function contribute to peaking since there are no "zeroes" at the nominator ?

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Retrieved from A Wideband Low Power Low-Noise Amplifier in CMOS Technology

  • \$\begingroup\$ If you derive I(in) to somehow result in a voltage across C, you have Vo/Vin. Or you could use I(R), and then it's about currents. If V-V or I-I is what you're looking for. \$\endgroup\$ Dec 17 '16 at 7:39
  • \$\begingroup\$ Express the reactances in Laplace equivalent form. Then it's simple circuit analysis to give Vo/Iin \$\endgroup\$
    – Chu
    Dec 17 '16 at 8:28
  • \$\begingroup\$ Usually \$H(s) = in/out = V_{in}/V_{out} or I_{in}/V_{out}\$ I'll give you a hint \$ Z_L = L*s \$ \$\endgroup\$
    – Voltage Spike
    Dec 20 '16 at 19:13
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    \$\begingroup\$ I'm voting to close this question as off-topic because Homework question with zero attempt to solve \$\endgroup\$
    – Voltage Spike
    Dec 20 '16 at 19:16
  • \$\begingroup\$ this is not homework question. It is derived from figure 2 of the paper "A Wideband Low Power Low-Noise Amplifier in CMOS Technology" \$\endgroup\$
    – kevin
    Oct 5 '17 at 6:42

Using current divider rule: \$\small I_R=\large\frac{1/sC}{R+1/sC+sL}\small I_{in}=\large \frac{I_{in}}{s^2LC+sRC+1}\$

Hence: \$\small V_o=RI_R=\large \frac{RI_{in}}{s^2LC+sRC+1}\$

and: \$ \large\frac{V_o}{I_{in}}=\large \frac{R}{s^2LC+sRC+1}\$


Assuming no external load at Vo, you need to calculate the impedance seen by the current source. It's the parallel combination of C and (R in series with L): -

Z = \$\dfrac{\frac{R+sL}{sC}}{R+sL+\frac{1}{sC}}\$

Multiply through by sC to get: -


So, your current source x the impedance above produces a voltage across C (Vc). To find Vo you have a potential divider problem: -

Vo = Vc\$\dfrac{R}{R+sL}\$


\$I_{in}\times \dfrac{R+sL}{1+sCR+s^2LC}\times \dfrac{R}{R+sL} = V_o\$



Any help would be appreciated

I see little sign of that!


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