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large inductance measurements of Leybold coils (from 1uH up to 1 kH)

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Dr. Slack

Wed Feb 01 2017, 10:26AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Apparent inductance will change with measurement frequency due to the self capacitance effects. The self capacitance appears in parallel, which will raise the apparent inductance, that is reduce the susceptance, as the frequency approaches the SRF (self resonant frequency). It will be worth measuring the SRF of all the coils, this is most conveniently done with a signal generator and an oscilloscope.

Physikfan

Wed Feb 01 2017, 12:27PM

Registered Member #60240 Joined: Mon May 16 2016, 07:01PM
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Posts: 304

Hi Dr.Slack

The self resonant frequency (2*PI*f)srf of an inductance means exactly the value where 2*PI*f*L = 1/(2*PI*f*C),
C corresponds to stray capacitance, winding capacitance etc.?
And the inductance at this frequency is purely resistive, therefore the current trough the coil as a function of the frequency has its maximum value, experiment with a sine oscillator and oscilloscope?

That means if you want to measure the "real" effective inductance the measuring frequency should be very low compared to the self resonant frequency?

Uspring

Wed Feb 01 2017, 02:38PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
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Physikfan: To the lowest order, the coil appears as a parallel tank. The current through the coil has its _minimum_ at resonance. At low frequency the stray capacitance has the lowest effects.

Plasma

Wed Feb 01 2017, 03:23PM

Registered Member #61406 Joined: Thu Jan 05 2017, 11:31PM
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Physician, the resistance to the inductor would have a large effect on the amps from the power supply. If you treat the resistance in parallel..
2*3.142*120*1024 = 772178 Xl
2*3.142*12*1130 = 85,211 Xl

Vin*(Xl^2/1100) is the amps from the source at high frequency more current will be drawn from the supply, showing a smaller inductor,
The resonance frequency would be 1/(6.284*sqrt(L*C)

Stray capacitance Screen = 1/1024 *(1/6.284*120)^2
If you can find the resonance frequncy, and use that frequency you should be able to get pretty accurate values

Just my 2 cents

Physikfan

Wed Feb 01 2017, 08:21PM

Registered Member #60240 Joined: Mon May 16 2016, 07:01PM
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Posts: 304

Hi Sulaiman, BjÃ¸rn, Patrick, Dr. Slack, Uspring and Plasma

Many thanks for your comments and hints.
I think I could learn a lot on this subject.
My next step will be to measure the self resonant frequencies of all these Leybold coils.

Ad Plasma
Are looking at the coil with 23 000 turns, right?

"If you treat the resistance in parallel..
2*3.142*120*1024 = 772178 Xl"

2*PI*120*1024, do you mean Omega*L, but, please, from where you get this 1024?

"2*3.142*12*1130 = 85,211 Xl" this is 2*Pi*f*L(23000 at 12 Hz) okay

"Vin*(Xl^2/1100)" please, from where you get this 1100? I do not understand this equation.

"Stray capacitance Screen = 1/1024 *(1/6.284*120)^2"

What I understand is:
If I measure the resonance frequency then I can calculate the stray capacitance of this special coil (23000 turns), right?

Regards

Physikfan

Plasma

Thu Feb 02 2017, 02:17AM

Registered Member #61406 Joined: Thu Jan 05 2017, 11:31PM
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Posts: 268

Sorry 1024 should be 1060. The 11000 is the resistance of the coil, you treat that with the power two of the inductor impedance. The resistance of the coil is treated as parallel to both the stray capacitance and induction. Three leg parallel. It gets added to the inductor, and that is times with capacitor.
Prf = (RsL+XL)+ XC. / (RsL+XL)*XC
It doesn't change the real resonance frequency by much

If the frequncy is 50000, and the voltage of the power supply is 10v, you measure 100mA
10v/.1=ans*11000=sqrt(ans)= inductor.

Dr. Slack

Thu Feb 02 2017, 07:22AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

The SRF is 1/2.pi.sqrt(L.C) in Hz, where L and C are the DC inductance and the coil effective self capacitance respectively.

If you measure the inductance well below this frequency, say a factor of 10 below, then the error from the self capacitance is of the order of this factor squared for factor of 10.

Resonance is most easily detected by connecting a high impedance scope to the coil, and using a high value resistor to a signal generator, then sweeping the frequency until you get a maximum AC voltage (minimum current). Don't forget that the scope will add some extra capacitance to ground.

Although at resonance you can model the loss to be due to either a series or a parallel resistor, a resistor in series with the L matches more closely what you can measure at DC as the copper resistance. The equivalent parallel resistor is rather different in value. The effective resistance at resonance will be a little different from the DC value due to skin and proximity effects reducing the area of the wire available for conduction.

A good alternative way to make your inductance measurements is to load the coil with a capacitor, and sweep for resonance. A plot of frequency and capacitance will allow you to estimate the self capacitance that is always in parallel with the added C. This allows you to measure at any frequency you have capacitors for.

You can estimate the losses at resonance by measuring the Q factor, which is the centre frequency / 3dB bandwidth. Don't forget that the 'high value resistor' that you use to connect to the sig-gen, and the scope input resistance, will also contribute to parallel loading of the LC.

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