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measuring coil inductance with resonance

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IamSmooth

Mon Oct 05 2009, 12:49AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I took some 3/8" OD copper tubing today and made a 5 turn coil. ID is 2"; OD is 2.75"

I tried to measue its inductance by connecting it in series with a capacitor of 1uf. I also tried 10uf. I put this in series with a 1 ohm resistor and connected the circuit to a signal generator with an amp. I measured the voltage across the resistor.

Now, when I did this with a 125uH inductor everything worked perfectly, and I got a peak voltage at the resonant frequency. However, when I did it with the coil I could not get a peak. Is the inductance too low for me to measure this way? Do I need more current through the coil so I can make the measurements? Is it that the winding capacitance for so low an inductance value is affecting the circuit?

Using calculations for a solenoid I get an inductance of about 0.8uH. I would think I should be able to measure this.

Frosty90

Mon Oct 05 2009, 02:21AM

Registered Member #1617 Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139

Some types of resistor are quite inductive, like the wire-wound and carbon-film (?) ones. This may be effecting you result, and swamping out the small inductance of the coil, giving a much lower resonant frequency.

Cheers,
Jesse

IamSmooth

Mon Oct 05 2009, 02:57AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

A good thought, but the resistor I am using is non-inductive. The legs of the coil were close together and may have been adding some stray capacitance in parallel. I twisted the legs apart. Using a 1uf capacitor I can actually get a peak somewhere near 95kHz. This would give an inductance near 2.5uH.

The solenoid formula is not that accurate if the length is not >> radius if I recall, so my calculated result may be right.

Proud Mary

Mon Oct 05 2009, 03:21AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Hi Smoothie, here is a rough answer to your question.

According to my ancient ivory slide-rule, I reckon the inductance of your home-made coil is near as damnit 1.2uH.

Setting aside your 1R resistor, which makes little difference,

The resonant frequency of your first combination of 1.2uH and 1uF has a resonant frequency of about 150kHz, but a characteristic impedance of just 1R1.

Your second combination of 1.2uH and 10uF has a resonant frequency of about 47.56kHz, but a characteristic impedance of just a tiny 0R346.

Now when you changed the tiny inductance of your coil for the 125uH one, although the resonant frequency fell to 14.23kHz, the characteristic impedance increased to 11R2 in combination with the 1uF cap, and 3R53 for the 10uF.

You don't say what the output impedance of your sig gen is, so I will assume it is 50R.

Can you know see how serious is the mismatch between the sig gen's output impedance of 50R, and your 1.2uH/1uF combination of about IR1? The ratio of mismatch is almost 50:1, which will look pretty much like a short circuit to the sig gen. You will have almost no signal voltage to excite your resonant circuit.

With the 10uF, the situation is far worse, as you will now understand.

Now when you substituted the 125uH inductor for your home made 1.2uH coil, the characteristic impedance rose to 11R2, a mismatch ratio with the sig gen's 50R output impedance of about 4.5:1, which gave you enough exciting voltage to get your LC circuit to resonate. It no longer looked like a short circuit to the sig gen.

How's that? It's down to you now, Smoothie, to Google 'characteristic impedance' and find out what you can about it! It's a key concept in AC theory.

Harry.

Steve Conner

Mon Oct 05 2009, 10:04AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Harry is spot on. My own efforts at measuring low-impedance tank circuits failed for exactly that reason. With such a low impedance, you should shunt-feed the circuit and look for a voltage peak at resonance, instead of series feeding it as you're currently doing.

IamSmooth

Mon Oct 05 2009, 01:00PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I was using a signal amplifier, which I thought would have taken care of the source impedance. I will check and see how the open voltage drops with a 1R resistor and determine the actual value.

What is shunt feeding? Is this setting the circuit up as a parallel LC tank?

How did you come up with the characteristic impedance of 1R1? Does it have anything to do with the transmission line formula found here:

GeordieBoy

Mon Oct 05 2009, 02:33PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
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shunt feeding is connecting the source ACROSS the parallel resonant LC circuit. So you have the current source, the inductor and the capacitor all connected in parallel. You then tune the source and look for a peak in the voltage measured across the whole arrangement.

As Steve said it works well for things like Induction Heating tank circuits or TC primary circuits where the Inductance is relatively small and the capacitance is high.

-Richie,

teravolt

Mon Oct 05 2009, 04:17PM

Registered Member #195 Joined: Fri Feb 17 2006, 08:27PM
Location: Berkeley, ca.
Posts: 1111

another way to do this is to put the cap and inductor in parallel put a current toride in the LC ciruit like a DRSSTC and mesure the circulating current. As resonance happens current will be max like in a VTTC. If you use a 100 ohm carbon resistor in series with your circuit your generator wont be loaded down so bad

IamSmooth

Mon Oct 05 2009, 04:51PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

teravolt wrote ...

put the cap and inductor in parallel put a current toride in the LC ciruit... As resonance happens current will be max

For a parallel LC isn't impedance highest at resonance, and current would be lowest?

Steve Conner

Mon Oct 05 2009, 04:58PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

From the perspective of the current transformer, which is connected in series with the tank circuit, it's a series resonant circuit, but from the perspective of the generator shunt-feeding it, it's a parallel resonant circuit. At resonance, current drawn from the generator is at a minimum, but circulating tank current is a maximum.

There is really only one kind of resonant circuit, and "series" and "parallel" are just ways of looking at it. This duality may take some thinking to wrap your head around, but once you get it, you understand L-match circuits and impedance transformation, too.

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