Calculating inductance using a RL circuit and a frequency generator

In summary, the conversation discusses the process of calculating the inductance and impedance of an unknown air cored inductor over a set frequency range using a series RL circuit. The tools required for the calculation include a scope, multimeter, frequency generator, and amplifier. The steps involved in the calculation are described, along with a sample table of data. The conversation also addresses potential issues with using a 100 watt resistor and suggests alternative methods for more accurate results.
  • #1
yoda2026
11
0
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
 
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  • #2
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
 
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  • #3
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
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
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
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
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 can't 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:
  • #8
hmmm ill try changing to a non inductive resistor and see if my values change
thanks for the heads up
 
  • #9
i have changed resistor and i am getting a much more sensible answer

thank youusing 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 doesn't make any sense
 
  • #10
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
and from the description of your hookup
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)

it seems that voltage across the inductor would be from POSITIVE terminal of amplifier to side b of coil...
instead of
voltage across the inductor (voltage from the negative terminal of the amplifier to the inductor terminal)
which sounds more like voltage across resistor.
 

What is an RL circuit?

An RL circuit is a type of electrical circuit that contains both a resistor (R) and an inductor (L). The inductor is a coil of wire that stores energy in the form of a magnetic field when an electric current passes through it. The resistor is a component that limits the flow of current in the circuit.

How does an RL circuit calculate inductance?

An RL circuit uses the formula L = V/Iω, where L is the inductance in henries, V is the voltage across the inductor, I is the current flowing through the inductor, and ω is the angular frequency of the alternating current. This formula takes into account the voltage, current, and frequency of the circuit to determine the inductance.

What is a frequency generator?

A frequency generator is a device that produces electrical signals of a specific frequency. It is commonly used in electronics and scientific experiments to generate alternating current at different frequencies. In the context of calculating inductance in an RL circuit, a frequency generator is used to provide a steady source of AC power at a specific frequency.

Why is it important to calculate inductance in an RL circuit?

Calculating inductance in an RL circuit is important because it helps determine the behavior of the circuit. Inductance affects the flow of current in a circuit and can cause changes in voltage and current over time. By knowing the inductance, scientists and engineers can design and analyze circuits more accurately.

What are the practical applications of calculating inductance in an RL circuit?

Inductance calculations in RL circuits have practical applications in various fields, including electronics, telecommunications, and power systems. They are used to design and analyze circuits, as well as to troubleshoot and repair faulty circuits. Inductance calculations can also be used to optimize the performance of electric motors, transformers, and other electrical devices.

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