# Transfer function for Small signal model of inductor series peaking [closed]

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

However, how does this transfer function contribute to peaking since there are no "zeroes" at the nominator ? • 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. Dec 17 '16 at 7:39
• Express the reactances in Laplace equivalent form. Then it's simple circuit analysis to give Vo/Iin
– Chu
Dec 17 '16 at 8:28
• Usually $H(s) = in/out = V_{in}/V_{out} or I_{in}/V_{out}$ I'll give you a hint $Z_L = L*s$ Dec 20 '16 at 19:13
• I'm voting to close this question as off-topic because Homework question with zero attempt to solve Dec 20 '16 at 19:16
• this is not homework question. It is derived from figure 2 of the paper "A Wideband Low Power Low-Noise Amplifier in CMOS Technology" 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: -

$\dfrac{R+sL}{1+sCR+s^2LC}$

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$\dfrac{R}{R+sL}$

So,

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

So,

$\dfrac{V_0}{I_{in}}=\dfrac{R}{1+sCR+s^2LC}$

Any help would be appreciated

I see little sign of that!