# Calculating inductance using a RL circuit and a frequency generator

1. Jan 4, 2012

### yoda2026

any help would be much appreciated

i am trying to calculate the inductance and impedance of a unknown air cored inductor. over a set frequency range (1600hz - 2700hz in 100hz increments) using a series RL circuit

tools
scope
multimeter
frequence generator
Amplifier
1ohm 100W metal clad resistor(1.07ohms under test)

what i have done so far is

frequency generator is set at 1600hz
the output of a frequency generator is connected to the input of the amplifier
the output of the amplifier is used as the power-source

the positive terminal of the amplifier goes to side a of the inductor, side b of the inductor is connected to side a of the resistor, then side b of the resistor is connected to the negative terminal of the amplifier (if this is unclear please see the attached photo)

as a table of data i have (for this i am using the 1600hz information)

frequency = 1600hz
resistor = 1.07Ohms
voltage across the resistor = 0.84V
voltage across the inductor (voltage from the negative terminal of the amplifier to the inductor terminal) = V1-V2 = 0.19V

using this information how do i calculate the inductance

2. Jan 4, 2012

### yoda2026

http://img839.imageshack.us/img839/9320/20120105163452.png [Broken]

please not i did not calculate the table above and i am led to believe that it is incorrect
as the inductance decreases as the frequency increases.

i am trying to replicate this test using two unknown inductors to compare the results to see if one is a suitable replacement for the other

Last edited by a moderator: May 5, 2017
3. Jan 4, 2012

### Mike_In_Plano

The signal generator's internal impedance (typically 50 ohms) can make it difficult to get good excitation. This why I typically use a capacitor in series and sweep the frequency until I get a strong resonance.
Resonance can be detected by high voltage across across either the L or C. You can also place a small resistor in series with the ground lead and see the current peak at resonance. Then, find Xc from:

Xc=1/(2*pi*f*C)

At resonance, Xl = Xc

Xl = 2*pi*f*L

L=1/(4*pi^2*f^2*C)

4. Jan 4, 2012

### yoda2026

yes the frequency generator has an internal impedance of 50 ohms but then its is fed into an amplifier which i was hoping would counter the impedance

i am more than happy to add a capacitor if that will make the calculations eaiser. but i am not trying to find the inductance/impedance at resonance.... i am trying to find the inductance at a set frequency ie 1600hz

sorry if i have missed something

5. Jan 5, 2012

### Carl Pugh

Don't use 100 watt resistor. Some 100 watt resistors are incredibly inductive.

The Q of the inductor probbly won't change with any reasonable power, so test with signal generator and 1% metal film resistor.

I=Eresistor/Rresistor=0.84/1.07=0.785 amp

Inductor Impedance = 2*pi*f*Linductor= Einductor/I Solve for Linductor

Linductor= Einductor/(2*pi*f*I)= 0.19/(2*3.14*1600*0.785) = 24.08 microhenry

6. Jan 5, 2012

### Carl Pugh

Didn't check voltages before posting preceding.
You may have to use an amplifier to get reasonable voltages.
However it is probably best to use ten 10 ohm 1% metal film resistors in parallel for 1 ohm resistor. 100 watt resistors are likely to give you incorrect results.

7. Jan 5, 2012

### technician

I agree with Carl Pugh and I got the same value using your figures.
The voltages you have measured are across the 2 components.... R and L and therefore what goes on inside the power supply is irrelevant.
As the frequency increases the REACTANCE increases (2pifL) I cant think of any reason why the inductance (L) should increase
One other thing occurs to me.... have you measured the resistance of the coil itself? We have assumed it has zero resistance.
The actual resistor in the circuit is only 1 ohm so any resistance in the coil would be significant.

Last edited: Jan 5, 2012
8. Jan 5, 2012

### yoda2026

hmmm ill try changing to a non inductive resistor and see if my values change

9. Jan 5, 2012

### yoda2026

i have changed resistor and i am getting a much more sensible answer

thank you

using XL=2PI*F*L on the table i have supplied i get

2PI*1600*2.41*10^-5 = 0.2422

where there table gives 5.826615 :s which doesnt make any sense

10. Jan 5, 2012

### jim hardy

something's not right with the rightmost, XsubL, column in that table.

look at 1600 hz row

volts across inductor = 0.19

amps through inductor = 0.78

XsubL = 5.826615

ohms = volts/amps = 0.19/0.78 = 0.2420238+(use all the digits) , that should be XsubL not 5.something.
intuitively if you have way less than a volt at amost an amp you have less than an ohm.
With .8 amp through something and only 0.2 volt across it it's way less than an ohm.

now what's curious is if i divide their XsubL number by my XsubL number

i get the number that's in their L column.

5.826615 / 0.242038 = 24.0745

i think somebody did something that's not intuitive when making that spreadsheet.

11. Jan 5, 2012

### jim hardy

and from the description of your hookup
it seems that voltage across the inductor would be from POSITIVE terminal of amplifier to side b of coil...