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• Time out for tirade - I spent many years in the construction industry and cm is absolutely prohibitedas it should be everywhere else - unless you re a total
Message 1 of 1 , Sep 26, 2001
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I spent many years in the construction industry and cm is absolutely prohibitedas it should be everywhere else - unless you're a total idjut you can't mix up 1000 mm with 1000 metres but you can easily confuse 1000 mm with 1000 cm on a construction site. I have some recollection that at law, under the relevant original Metrication Act, cm is NOT officially recognized in Australia but is unhappily well beloved by retailers as well as my wife - boy we fight over that one.

Proceeding further you should have calculated inductances of (a) 0.054 uH and (b) 0.064 uH.

Our straight piece of wire has gone from 0.044 uH to 0.054 uH and 0.064 uH respectively.

A much flasher formula: (where coil length = diameter (which is best for optimum Q).

N = Ö[29L / 0.394r]  OR  L = [(0.394 * r * N2) / 29]

where:

N = number of turns
L = inductance required in uH.
r = radius of the coil (in cm).
All others (29 & 0.394) are constants

Using example (a) above we would get:

L = [(0.394 * 0.635 * (2.5)2) / 29] = 0.054 uH

Permeability:- So far our inductors have been air wound where the permeability of air is 1. If we introduce iron or ferrites into our core then we find that for a given number of turns, the inductance will increase in proportion to the permeability of the core. This means for a given inductance less turns will be required.

e.g. Neosid have a fairly popular former (part 52-021-67 page 279) designed to accept a core of 4 mm dia. This former has an external diameter of 5.23 mm and a typical core at 20 metres (15 Mhz) would be an F29 type.

If we conveniently wanted an inductance of say, and of course I obviously cheated, 3.4937 uH and an F29 slug gives the formula:

N = 10.7 ÖL     or    N = 10.7 Ö 3.4937

we very conveniently get exactly 20 turns (remarkable wasn't it)

On the other hand to achieve the same inductance on a 5.23 mm former we would need to wind?

Well you tell me  Ian C. Purdie  your answer. Hullo is anybody out there in cyberspace doing this tutorial?

If nobody is doing this then why would I bother continue typing reams and reams of interesting repartee?. I need feedback now and then.

Therefore introducing permeability means reduced turns therefore less rf resistance and hopefully greater Q.

Using the above former but with a different core and bobbin Mr. Neosid in his cattle-dog (australian for catalogue) page 264, tells me 150 turns of 3 x .06 EnCu wire, wave wound, will yield an inductance of about 670 uH. This would not be achievable without the permeability of core and bobbin, especially the claimed unloaded Q (Qu) of 150. Use the above formula and substitute a factor of 'X' for the 10.7 - what is the X?. If you need help, email me.

Toroids

These come in two types. Powdered Iron or Ferrites. Both introduce permeability. Toroids look exactly like doughnuts and come in various diameters, thicknesses, permeabilities and types depending upon the requency range of interest. Some of their advantages are:

• high inductance for the physical space occupied
• no interaction or coupling with adjacent components (unlike air wound and other inductors)
• various permeabilities available
• exceptional Q values when wound correctly and optimum core and windings selected
• wide range of diameters and thicknesses.
• relatively low cost
• often simple to mount or secure mechanically

The only disadvantage I can think of

• nearly impossible to introduce variable tuning of the inductance

Amidon Toroids available from  - www.bytemark.com/amidon/

A typically popular type is made available by Amidon Associates and a representative example is the T50-2. This core is lacquered red (so you know the type) and has the following main properties.

Being T50 it's outside diameter is 0.5", the ID is 0.3" and the thickness is 0.19"

The permeabilities or in this case AL factors i.e. ( inductance per 100 turns2 ) for all T-50 types are:

 TYPE COLOR AL Frequency Range T50-26 Yel-Wh 320 uH power frequencies T50-3 Gray 175 uH 50 Khz to 500 Khz T50-1 Blue 100 uH 500 Khz to 5 Mhz T50-2 Red 57 uH 2 Mhz to 30 Mhz T50-6 Yellow 47 uH 10 to 50 Mhz T50-10 Black 32 uH 30 to 100 Mhz

This is only a small sample to give you an idea. Your turns required to give a certain inductance based on the above AL  is as follows:

N = 100 * Ö[ L / AL ]

Therefore to obtain an inductance of 3.685 uH using a T50-6 toroid would require 28 turns (of course I cheated again) but check it out on your calculator as I may have left in a deliberate mistake to see if you're awake.

By the way don't get too paranoid about the exact number of turns because cores do vary in value anyway and particularly with temperature changes.

Series or Parallel Inductors

Just as is the same with resistors, if two or more inductors are in a circuit connected in series, then the inductances add together, i.e. L total =  L1 + L2 + Ln etc. If they are in parallel then they reduce to less then the lowest value in the set by the formula:

L total =  {1/ [(1/L1) + (1/L2) + (1/Ln)]}

Enter all of those formulae along with every other one above and below into your exercise book.

Now for the downer:

I said in the second paragraph in the beginning that an Inductor can look like a capacitor.

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