RLC circuit problem / Are inductors polarised?

Forum 2005-2010 (read only) Hardware Interfacing
1

I am trying to make an oscillating circuit, but so far luck had been avoiding me. Here is the circuit design I am trying to achieve and corresponding fading away oscillation.

I am supposed to get some oscillation, at 2.32 KHz, that is gradually fading away. But all I get are two jinks, like this:

The components I am using are:

  • 10 Ohm resistor,
  • 100 nF ceramic capacitor,
  • 47 mF standard radial lead inductor.

Is it possible that this type of inductor is polarized? I am saying that because one of its legs is significantly longer than the other. That is usually a sign of polarization.

2

Is it possible that this type of inductor is polarized?

No.

I am saying that because one of its legs is significantly longer than the other.

It tells you the wind direction. This might be important when you have two inductors close to each other.

3

I arrived at that conclusion by running this online Java simulation here:

RLC circuit

Are you saying that this whole circuit is wrong? I just want to do a little experiment to satisfy myself that I understand how inductors work.

4

I would make the 10R resistor 1K and the 100R 100K. The problem is that the simulator assumes perfect components that do not match what you have in the real world.
You are getting the right response but it is damping down far greater than the simulator due to losses in the components.

5

thanks for the suggestion, I'll try that and report back.

6

There are two things I am trying to learn:

  1. I want to see inductor in action. So any circuit that will manifest inductance.

  2. I want to create some sinusoidal oscillations and measure their period and amplitude.

thanks.

7

Are you using the passive probe that came with the scope? If so, check the probe's loading characteristics. Most of those passive probes introduce at least 10pF of capacitance. Whatever the amount is, make sure you model your circuit with that load attached to it.

The probe will have a capacitance and a resistive value (usually 1Mohm). Put those two components in series to create a 'probe model'. Then add the 'probe model' into your simulation, where you are probing.

8

CMiYK, you are 100% right, with such a small capacitor the probe's own capacitance will count, wouldn't it!

I replaced 10 Ohm resistor with 1000 Ohm and got a three jingles. No sinusoidal waves yet.

9

This is an extremely good advice for everybody who wants to compare expectations/simulation model results to measurements..
It had fooled me many times....

10

The "ring-ability" of an LC circuit is dependant on the circuit Q which is a function of the circuit resistance. Might I suggest that, rather than increasing the 10ohm resistor that you reduce it, so increasing the circuit Q. Try reducing it to zero (remove it entirely). That will give you the lower limit ! I'd also suggest that, for demonstration purposes (to prove that you can actually see a ring) you increase the value of C so say 1microfarad so's it'll store more energy during the charge stage. The ring will occur when you open the switch, since when you close the switch the 100ohm charge resistor and battery effectively dampens the circuit.

You don't need any extra "active" components to produce a ring, you already have one - it's called the on/off switch. The original telegraphy transmitter radios didn't use any - simply working a morse key into an LC circuit was enough to generate RF rings.