# input impedance of RLC tank, and characteristic impedance

Discussion in 'General Electronics Chat' started by dxt78, Jul 14, 2011.

1. ### dxt78 Thread Starter New Member

May 4, 2011
2
0
Hello,
I have an RLC tank as shown in attachment (2R in parallel with an L and a C). I have two questions about it:
1) Why is the characteristic impedance of this LC tank sqrt(L/C) and what meaning does it have?
2) When computing the input resistance of this tank, what role should the characteristic impedance play?
Thanks.

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2. ### TBayBoy Member

May 25, 2011
148
19
I might be wrong, but isn't a tank circuit when L and C in parallel at resonance?

3. ### nuckollsr Member

Dec 17, 2009
16
1
You have illustrated a series-resonant circuit paralleled by 2000 ohms total resistance. At DC, the terminal impedance of this network is 2000 ohms purely resistive because the capacitor blocks all conduction at DC. At very high frequencies, the terminal impedance is again 2000 ohms because the inductor blocks conduction at high frequencies.

As the frequency goes up from DC, the circuit will show declining capacitive reactance until the such frequency that the inductive and capacitive reactances are equal (resonance) at this time, terminal impedance goes to zero. Continue increasing the frequency and the impedance rises when the circuit goes inductive.

4. ### kubeek AAC Fanatic!

Sep 20, 2005
4,686
805
For this circuit what nuckollsr said happens at about 5Ghz.

5. ### billbehen Active Member

May 10, 2006
39
1
With the connection shown, the two 1KOhm resistors will not damp the LC tank's resonance at all, if the L and C are ideal, as you show them.

Near resonance (about 5Ghz as mentioned above) the impedance of the LC is so low, that the 2KOhm in parallel has little effect. At resonance, the impedance of the tank is ideally zero, so the 2R in parallel has no effect at all....

In order to control the Q , the R's would have to be in series with the L and C, and also in series with the impedance meter (off to the left in the schematic?) Further, Ro = 31.6 Ohms for the values of L and C shown, so if they were in series with 2KOhms you wouldn't have any resonant peak at all, since Q = Ro/2R = 0.014 << 1/2.

Hope this helps!