Find LC resonance hidden by inductance

Do you have an estimate of the equivalent circuit, and the expected range of L1, L2, C, R values present? Any nonlinearities? (That ringdown (or whatever it was) doesn't look very sinusoidal.)

An LLC (series L into parallel LC tank) is very common for resonant converters and some induction heaters, but the matching ratio and Q factor are intentionally reasonable.

Tim

-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Contract Design Website:

Hi

I have a black box with a large inductance and a LC circuit in series

I need to find a way to measure the LC circuit resonance

When I sweep the black box, and record the current it only records something like 15% change of the current at resonance

Red Curve: Current Blue Curve: Voltage into black box

See the red curve. It reduces with frequency, due to the large inductance in series. The resonance I need to detect is at 360Hz

So, I could detect the dip (very small, so that won't be a robust way) I could monitor the phase shift also. But this will also be small I could do a curve fit to the general curve, so the dip will stand better out

Any other way?

Thanks

Klaus

L1 L2 50mH 100uH ======= ====== .-.-.-. .-.-.-. X ----' ' ' '----+---' ' ' '---. '-||--' | C1 | Cp1 | 1,000uF | '------||-----+ | ===

1) You could couple a signal generator to L2-C1 inductively, with a loop, then monitor V(X) with an oscilloscope while sweeping the frequency. The resonance peak should be very pronounced. 2) Most physical realizations of L1 will have a large parasitic capacitance Cp1, so step-pulsing will make L1 self-ring, and L2-C1 ring too.

You could excite the two resonances by applying a d.c. current, then breaking the circuit. L1 and L2 will each ring, L1 at its self-resonance, L2 will parallel resonate with C1.

Cheers, James Arthur

The graph I linked to was the current into the black box, which shows the impedance did not go to infinite, but there is some extra parasitics in the model which results in only a small dip on the curve

Cheers

Klaus

The parasitics are not visible in the schamatic you posted.

The inductors have dots. Are they coupled?

On 9/25/2015 5:48 AM, Klaus Kragelund wrote

Stepping out on a limb here, but can you tune out the large inductor with a series capacitor, to see the impedance increase at 350Hz?

Mikek

Adding a series capacitor will create a series resonant circuit at some frequency. That will appear as a short at resonance. At all other frequencies it will appear as either inductance or capacitance which does little to help your measurement.

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Rick

One of your many problems is that it takes only a few tens of milliohms in the 1000uF capacitor to completely screw up your measurements. You've got a problem if they are not ideal components.

That certainly kills the Q, yep, which means there won't be sharp peaks anywhere.

As long as the two parallel resonances are widely-spaced, I like the idea of coupling the probe signal inductively, then monitoring the accessible node. Both the 50mH self-resonance and the L-C could be easily measured. I think.

But, not knowing the physical construction, that might or might not be feasible.

Cheers, James Arthur

That's my point, tune the large inductance to a short at the 350Hz. I may not understand the problem maybe he doesn't know the values, although his schematic is labeled. I guess as you go off frequency, the large LC would start to show XC or XL and throw off the peak. I don't know if it will alter frequency or just Q. Tell me.

Mikek

He is trying to measure the frequency of the peak at about 350 Hz. How can he set the series resonance to the same value until he measures it? I expect the peak from the series resonance will swamp the lower value (and likely lower Q) parallel resonance anyway.

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Rick

That's actually potentially a very accurate approach.

L1 L2 50mH 100uH ======= ====== .-.-.-. .-.-.-. Vx --||--' ' ' '----+---' ' ' '---. Cx '-||--' | C1 | (var) Cp1 | 1,000uF | '------||-----+ | ===

FOR f = start TO finish apply signal(f) to Vx adjust Cx to series-resonance with L1 (indicated by max current) Vx is now virtually connected to L2-C1. Measure current, phase. NEXT

This method eliminates the effect of Cp1. It's a bit interactive, but manageable, I think(tm)...

Cheers, James Arthur

I'm getting out of my comfort zone, (one L and C are plenty for me.) but I think the Cx, L1 resonance should be fairly broad, (large R in L1) and maybe one value of Cx (picked to be near the L2/C1 resonace) would be enough.

George H.

I haven't thought it all the way through, but I think the L1-Cx resonance shouldn't be a problem. L1 is quite large. That usually makes for high-Q, since L goes with turns^2, while resistance only increase with turns.

Ideally, if you series-resonate Cx-L1 (per Mike), L1 disappears, directly connecting Vx to L2-C1 for analysis.

Cheers, James Arthur

They are not coupled. The big inductance is a real inductance, the LC is actually a mechanical resonance

Regards

Klaus

Nice idea

The aim of my investigation is to find the 350Hz resonance, whatever frequency it resides at. So I cannot add a capacitor to tune out the large inductance, at least not straightforward

Cheers

Klaus

I think the Q is quite low, it seems that it is since the dip is pretty small

I only have access to the big inductance, LC resonance which is a mechanical resonance I have not access to.

Cheers

Klaus

Really nice approach, I can try to do it manually on monday back to work. It's a quite simple test :-)

Cheers

Klaus

Yikes, NOW you tell us!

Do you think that the schematic represents the actual equivalent?

Put an accelerometer on the box, sweep, use a lock-in algorithm.

For the actual implementation, it will be difficult to implement a tuned capacitance. Maybe I could use something like a capacitance multiplier, so I can do it primarily in HW?

Cheers

Klaus