Unity Gain current buffer help

[duffman268]

So I'm asked to find an expression for the input impedance of the circuit, but I'm not quite sure how to do this.

As a unity gain buffer, is the current just Ix? Would the Zin just be zero?

I'm also asked to find the equivalent circuit of Zin, which I cannot figure out since I don't know how to find Zin.


**broken link removed**

 

[steveB]

The usual way to derive the input impedance is to calculate the voltage response to a current input. This is simple using Laplace domain impedances with the coil treated as reactance sL and the cap treated as reactance 1/(sC). You have simple linear equations using circuit theory. In other words, Zin=V/I where V is the voltage response to a current I.

It's not clear to me what the "unity buffer" is, unless this is referring to the current source Ix that exactly provides the measured current in the resistance.

Once you find the Zin, the equivalent circuit will be very easy to determine. I'll give you a hint that the final impedance resembles a series RLC circuit.

 

[duffman268]

I think I figured it out.

Zin is simply jXl - jXc + |XcXl|/R and from there its equivalent circuit is not hard to find.

Thank you for your help!

If i were to design an implementation with an active device in mind, how would I do so?

 

[steveB]

If i were to design an implementation with an active device in mind, how would I do so?

Active devices are always nonlinear. Hence, you need to make an equivalent AC linear circuit, and then do the analysis on that. Generally, such a solution is valid only for small AC signals at a particular DC operating point (i.e. bias point, or Q-point).

 

[MrAl]

I think I figured it out.

Zin is simply jXl - jXc + |XcXl|/R and from there its equivalent circuit is not hard to find.

Thank you for your help!

If i were to design an implementation with an active device in mind, how would I do so?

Hi,

You dont really need an active element to emulate this circuit. The equivalent network is an RLC series circuit with the following values:
Lnew=L
Cnew=C
Rnew=L/(R*C)

So the new network is just a passive RLC circuit with L and C the same as the old circuit, and R is the reciprocal of the original R then multiplied by the ratio of L/C.

If you want to design an active circuit in the same manner of the original circuit you could design a current controlled current source or since R is just a resistance and I=V/R a voltage controlled current source with a settable gain equal to v(R)/R. An op amp could be used if the current is low enough and frequency within range.