I need to measure the inductance(L) of a certain coil I have wound.How do I do it experimentally without using LCR meter?

How to add filters and lenses to Canon SX100is

by
Ruta9

How to Convert Scanned PDF to EPUB for iPhone 4/4S

by
Ericmac

Make a 48 LED Macro Ring Light for SLR for $10

by
Ruta9

How to Create a Java Unit Converter

by
eabrosius

A bidirectional logic level converter (for I2C)

by
janw

Convert any 2D image to a 3D object using OpenSCAD (and only free software)

by
R-I-H-A-M

Mintyboost on the Cheap!

by
blrplt1

Converting a Google Sketchup Design into DXF format

by
timmy954

You can access an existing collection each time you add an Instructable to it, or from the "You" menu and page.

forgot your password or username?

it happens.

it happens.

Enter the email associated with your account and we will send you your username and a temporary password.

Not a member? Sign Up »

We have sent you an email with a password reset code. Please enter it below.

Not a member? Sign Up »

Choose the R so that it is much larger than the actual copper resistance present in your inductor. That is so you can just sort of ignore the small copper resistance of your inductor.

Also if your inductor has some sort of core, like iron or ferrite, that can saturate, choosing big R can help to make I(t) small, hopefully smaller than the expected saturation current.

http://en.wikipedia.org/wiki/Saturation_%28magnetic%29

If it is an air core inductor, don't worry about saturation.

Vinput is the square wave you put across L and R in series, so:

Vinput = L*(dI/dt) + R*I

And the output is the voltage across the resistor, which is R*I

R*I = Vinput - L*(dI/dt)

In the limit of

frequencies much less than f0= (1/2*pi)*(R/L), the voltage across the resistor R*I is much bigger than the voltage across the inductor L*(dI/dt), and the output will look much like the input, like a square wave.In the limit of

frequencies much greater than f0, the voltage across the inductor L*(dI/dt) is much larger than R*I, and the output will look like a sawtooth, whose slopes are plus or minus (R/L)*(Vinmax-Vinmin)At

frequencies close to f0= (1/2*pi)*(R/L), you should actually be able to see the shape of the exponential. See:http://en.wikipedia.org/wiki/RL_circuit

The time constant τ, the amount of time it takes for the exponential to fall to (1/e) its previous value, is τ= (L/R)