In A series LC circuit , what is SQRT(L/C)?

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Is it the impedance which could critically damp the resonance?.
 

yes but any resistance will damp the tuned circuits response...why specifically sqrt(L/C)?
 

The expression SQRT(L/C) gives the magnitude of both the inductive and capacitive impedance at w=wo:

wo*L=1/(wo*C)=SQRT(L/C).

From this follows

wo=SQRT(1/LC)

Damping is caused by resistive losses.
More than that,
The quality factor - as a measure of "sharpness" of resonance - is expressedby (Rs=series loss R)

Q=SQRT(L/C)/Rs
 
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The expression SQRT(L/C) gives the magnitude of both the inductive and capacitive impedance at w=wo:
Click to expand...

that's definitely useful to know...but what is the significance of this?.....and is "w0" called the "characteristic impedance"?

When one is doing a high side fet gate driver in an smps, one has a series capacitor , C, in series with the primary of the gate drive transformer....this is also in series with the wretched leakage inductance of the gate drive transformer......and when one wishes to damp the ringing of C and L(leakage), one adds a series resistance of SQRT(L/C) ohms...but why does one have to pick this value?

If a step input is put into an LC series circuit, then is SQRT(L/C) the impedance seen by the step at the instant that the step is applied.?
 

treez,

at first, I have assumed that you know that "omega" (here abbreviated as w) is the angular frequency which is defined as w=2*Pi*f (f=frequency).
Consequently, the dimension of SQRT(L/C) is Ohm (because w*L and 1/w*C also is in Ohms)
Of course, wo is the resonant frequency wo=SQRT(1/LC) as defined in my first post (inductive and capacitive impedances equal).

As to your last question: Now you can see that SQRT(L/C) is not identical to the overall input impedance at t=0 (step input).
At t=0 the input impedance of the inductor - and, thus, the overall impedance - is infinite.
 
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woops, sorry, of course as you say "w0" is the resonant freq, I goofed up there.

I think SQRT(L/C) ohms is that resistance that gives "critical damping" to the LC series circuit......ie it damps the ringing, but not too much.
I wonder if SQRT(L/C) can be termed the "critical damping resistance for a LC series circuit"?

I believe that SQRT(L/C) makes the LC ringing decay with an exponential (1/e) envelope?
 

treez said:
I think SQRT(L/C) ohms is that resistance that gives "critical damping" to the LC series circuit......ie it damps the ringing, but not too much.
I wonder if SQRT(L/C) can be termed the "critical damping resistance for a LC series circuit"?
Click to expand...

No - just the opposite is true. If SQRT(L/C) is large in comparison to the damping resistor R, there is a weak damping of the oscillatory step response.
By the way - if you speak about "critical damping", what do you mean using this term?
treez said:
I believe that SQRT(L/C) makes the LC ringing decay with an exponential (1/e) envelope?
Click to expand...

Yes - the ratio L/C is responsible for the ringing - however, the damping is caused by the resistive losses.
 
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"critical damping"....just means the envelope of the decay in the ringing has gone down to 1/e times what it started out at after one second
 

treez said:
I think SQRT(L/C) ohms is that resistance that gives "critical damping" to the LC series circuit.....
Click to expand...
"Critical damping" is the minimum amount of damping that gives no overshoot or ringing. To achieve critical damping, we need Q = 0.5

For a series RLC circuit, Q = sqrt(L/C) / R.
Thus if you set R = sqrt(L/C) then Q=1 and the circuit is underdamped. For critical damping you need R = 2 * sqrt(L/C)

To see the effect of different amounts of damping on the step response, why not experiment in LTSpice?

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treez said:
"critical damping"....just means the envelope of the decay in the ringing has gone down to 1/e times what it started out at after one second
Click to expand...
If you think about that for a moment, you'll realise it's complete nonsense.

What if the resonant frequency is 100KHz? After 1 second, there will have been 100 000 cycles of oscillation, and the amplitude will only have reduced by a factor of e. Looking at it another way, after 100 cycles of oscilation, i.e. 1mS, the amplitude will still be 0.999 of it's starting value.
 
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Thus if you set R = sqrt(L/C) then Q=1 and the circuit is underdamped. For critical damping you need R = 2 * sqrt(L/C)
Click to expand...
thanks I see what you mean.....maybe the engineer meant that we should put R=SQRT(L/C) ohms in the primary of the gate drive transformer, and another R=SQRT(L/C) in the secondary of the (1:1) gate drive transformer....this gives 2*SQRT(L/C) total, and gives "critical damping"

SQRT(L/C) is called "characteristic impedance" with microwave transmission lines....do you know what its called for low frequency analog?....is it simply the "critical damping resistance/2"?..is that what its called?


"Critical damping" is the minimum amount of damping that gives no overshoot or ringing
Click to expand...
I thought LC circuits *always* ring?...even if they have series resistance....the resistance just makes the ring die off exponentially...it still rings though?
 
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Or maybe he simply meant what he said. Maybe he considers Q=1 to be ideal in that application, or maybe the exact amount of damping isn't important and Q = 1 is "good enough".

treez said:
SQRT(L/C) is called "characteristic impedance" with microwave transmission lines....do you know what its called for low frequency analog?....is it simply the "critical damping resistance/2"?..is that what its called?
Click to expand...
I don't know if there's a name for it.

treez said:
I thought LC circuits *always* ring?
Click to expand...
No. It depends on the amout of damping.
 
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godfreyl said:
To see the effect of different amounts of damping on the step response, why not experiment in LTSpice?
Click to expand...
Hi Godfrey

Good idea! I ran some simulations. The results are attached below.

Regards, Godfrey


 
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treez said:
I am sure that if sqrt(L/C) = 0.5, then if a 2A current source suddenly runs into a series LC circuit, then the peak ringing voltage on the capacitor is 2*0.5 = 1V?
Click to expand...
1) I assume you mean "sqrt(L/C) = 0.5Ω". That would make sense.
2) Again, why don't you run a simulation and find out?
 
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sorry I appreciate it would just charge the cap up and up.
There was some similar thing involving connecting the mains to a pfc boost converter, and estimating the ring voltage peak on the boost output cap.
 

The ratio of L to C has to do with the amount of current it takes to operate an LC tank circuit.

For a given center frequency, and resistance...

High Henry value, and low C, operates on less current.

Low Henry value, and high C, operates on greater current.

Whether this makes more or less current available, is a question of how you use the tank circuit.

L:C ratio also determines Q (although volt level is reduced with higher Q).

These simulations illustrate:

 
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