# Relationship between Frequency, Inductance and Resistance

Discussion in 'General Electronics Chat' started by abuhafss, Oct 30, 2014.

1. ### abuhafss Thread Starter Active Member

Aug 17, 2010
170
2
Hi

Kindly have a look at this article. It describes a cheap and easy method of finding inductance of a coil.

My question is how do we get the relationship:

Inductance = 50 x Time period of one Cycle

or

Inductance = 50 / Frequency of the Oscillation

A detailed explanation or an article/tutorial link shall be appreciated.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
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I would say that this expression is only approximate. It is going to be a function of the transistors, and the battery voltage. Look at this:

3. ### WBahn Moderator

Mar 31, 2012
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First, there '50' should be '50 Ω'. Inductance has units of henries, which is Ω·s, and time has units of sec, the coefficient has to have units of Ω.

Clearly it is related to the L/R time constant of the test inductor and the 68 Ω resistors. You essentially have something charging during one half cycle and then discharging during the other (or possibly a shorter charge/discharge cycle occurring during each half cycle depending on the specifics and how you look at things). Whatever happens is happening in the time of a half cycle which is T/2, where T is the period of one cycle. If we assume, at least for now, that in each half cycle we have the circuit changing from one level to another level via a first order response, then the two levels are separated by some multiple or fraction of the time constant. Thus

T/2 = kτ = kL/R

And therefore

L = T [R/(2k)]

therefore

R/(2k) = 50 Ω

and

k = R/(2·50 Ω) = 68 Ω /100 Ω = 0.68

This is a quite reasonable fraction for lots of circuits (probably anything from about 1/4 to 4 with 1/2 to 2 being most common, though that is certainly a crude approximation that I am in no way prepared to back up). In a first order system, a signal will traverse 50% of its way to the new level in 69.3% of a time constant. If the voltage is going from 0V to 1.25V (a NiCd cell), then 50% is right at one diode voltage drop.

What is happing is that the voltage across the inductor is going from +Vl to -Vl in each half cycle with Vl being the difference between the transistor base voltages corresponding to one transistor being nearly on and the other being nearly off. It is highly likely that neither transistor either fully turns off or fully turns on, and instead stays in the active region at all times. Now, in a differential pair, a differential input signal on the order of the thermal voltage (26 mV) is sufficient to essentially shift the current from one transistor to the other. That isn't quite what we have here, but still a differential base voltage of 60 mV will result in one transistor carrying 10x the current as the other. So we aren't talking about large voltage swings across the inductor.

That's certainly not an actual analysis an the physics in in the hand-waving, but maybe it gives you some ideas on how you might analyze the circuit yourself. If I get a chance I might do something similar and make a blog post out of it.

Sep 22, 2013
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