learn the basics of LC filter design at the level necessary to follow the more advanced topics provided later. Includes a great Javascript calculator I've written to work out L or C from resonant frequency!

Ian Purdie's Amateur Radio Tutorial Pages

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Electronic Basics

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ESSENTIAL BASICS YOU MUST BEGIN WITH

Introduction

Please make sure you have thoroughly reviewed Radio Terminology and Basic Electronic Units and have preferably printed out a copy of each and have them both handy because I will continually refer to both those units. I don't want you to be unecessarily confused, that's not the object of the exercise but without that basic understanding you will go nowhere here.

You will need:-

You should have a decent calculator to do your work. It should be capable of:
(a) At least one memory location.
(b) an exp or exponential function.
(c) pi of course.
(d) square root function.
(e) a powers function - often denoted Yx where the first number is raised to the power of the second number.
(f) decimal to hex-decimal and vice versa. This proves very helpful for graphics and some programming.

Buy

A cheap exercise book as this will prove beneficial. Leave the first few pages vacant to enter commonly used formula and the next few pages vacant to record the L/C of the frequencies you might happen to refer to quite often.

Happy with that?.

What is this LC anyway?

Well the letter L stands for inductance and the letter C stands for capacitance - (I love this high tech talk) - therefore having grasped that fundamental you should read on.

If you don't know or understand inductance, capacitance and resistance you are now in deep trouble for ignoring my introduction above - no short cuts here.

Good, now any quantity of inductance and capacitance will resonate at a particular frequency. As an example assume we have on the bench an inductor which measures exactly 159.155 uH and a capacitor which measures exactly 159.155 pF.

In the real world we usually don't measure to that level of precision and in fact you wouldn't be that accurate anyway. Mostly, if you are within about 5% you are doing O.K.

Using our calculator you will find if we multiply L times C i.e (L * C) we get 25330.314. This very important figure you will now ensure becomes permanently embedded in your mind for evermore (and written in the book I suggested you buy) because it is the very foundation of my easy formulae for L/C circuits.

There are other ways and means of calculating resonance etc. but guys. this is the simplest yet. You will learn in the next section how I arrived at it but for the moment learn these formulae and also write them in the book. They are one and the same, just re-arranged by simple high school algebra.

LC= 25330.3 / Fo [Mhz]2

Fo [Mhz] = SQRT[25330.3 / LC]

Where obviously LC is the combination of inductance and capacitance, Fo is our frequency of resonance and of course 25330.3 is our constant we have just learnt. SQRT means "square root of".

One formula will give you LC for a known frequency and the other will give you a frequency for a known LC combination. It's just simple re-arrangement of an algebraic expression. And you thought maths at school did / does suck! and had no practical real-world value to you.

Now I've made it a lot easier for you now with my skilfully crafted Javascript LC Calculator below.

PROCEDURE:

1. Simply enter your desired frequency in Khz

2. Press "Solve LC for me"

3. Read LC value from the window. If you want a different frequency then Press "Clear" to start over.

4. If you are happy with the answer then enter a value of either capacitance or inductance as a possible component and Press "Find the other value"

5. Lo and behold there is the remaining component value.

LIMITATION - DESIGNED IN:

I deliberately wrote the Javascript program to give answers as whole integers. As an example, if you ask the LC Calculator for the LC of 5000 Khz you will receive an answer of 1013 instead of the more accurate 1013.211836. Similarly if you the enter a component of say 33 pF the other value will be shown as 30 instead of 30.696. The purpose of restricting ourselves to integers was (a) we don't need decimal places of precision because they are irrelevant (unless you get much higher in frequency such as 100 Mhz) and; (b) whatever the answers, component tolerances and strays will be a much bigger factor anyway.

LC CALCULATOR

Enter your frequency in Kilohertz to calculate LC

Enter your frequency in kHz:
Then LC =    inductance (uH) X capacitance (pF)
Enter a possible component [either (uH) or (pF)]:      (uH) or (pF)
Remaining component value =    uH OR pF

Links, Copyright and copying pages - If you want to link to this page or copy any of this page you should read this first.

SELF-TEST

Consider if we were designing any type of filter for say, 7.150 Mhz (which is exactly the same by the way as saying 7150 Khz).

Now 1000Khz = 1Mhz remember that. Using our formula above we would need an L/C combination totalling 495.5 - what combination would you need for some sort of filter at;

(a) 75Khz (b) 1750 Khz (c) 9Mhz and (d) 101.7 Mhz

if you're keen email your answers to me, in this way I can at least get some idea if anyone is actually doing this tutorial. The answers to these particular questions do NOT appear in the answers section as with the other self-tests as set out below.

QUIZ - No. 1

1. What is the L/C for the following frequencies;

(a) 455Khz (b) 7224 Khz and (c) 9 Mhz.

2. What capacitor will resonate at the above respective frequencies with the following respective inductors ;

(a) 679 uH (b) 22uH and (c) 3.81 uH.

3. What is the resonant frequency for the following L/C (inductor - capacitor) combinations to the nearest Khz;

(a) 2,533,030 (b) 116,716 (c) 312.72 and (d) 2.53303.

Yes Ian, I did complete the above quiz in my exercise book. Can I please now see the answers to check out my work.

What is Reactance?

Just don't react too badly to this.

inductive reactance is denoted as X_L whilst capacitive reactance is denoted X_C.

Reactance is somewhat similar to resistance, but don't take that statement too literally. See - inductance, capacitance and resistance - except where resistance applies to D.C., reactance applies to A.C. which also includes radio frequencies or as we say in the game, R.F. (radio-frequency) for short.

Reactance might be considered as a quantity of resistance to alternating or varying frequencies.

Inductive reactance is = (2 * pi * Fo * L) and;

Capacitive reactance is = 1/( 2 * pi * Fo * C ).

Where 2 * pi = 6.2832, Fo is Mhz, L = uH, C = pF, * means multiply and 1/(xxxx) means one divided by the result of the numbers in brackets.

Let's use our examples from the previous tutorial. If we use 159.155 uH inductance, 159.155 pF capacitance and using a frequency of 1.0 Mhz substituted into our reactance formulas you should get reactances of XL = 1000 ohms and XC = 1000 ohms respectively.

XL is dead easy or should be but XC might be difficult for you because you need to calculate as follows:-

1 / ( 2 * 3.1416 * 1,000,000 * .000000000159155 )

OR enter into your calculator as;

1 / [( 2 * 3.1416 * (1*¹⁰⁺⁶) * (159.155*^10-12)]

Where (1*¹⁰⁺⁶) = one million or a one followed by six zeros and;

(159.155*^10-12) = 159.155 with the decimal place moved twelve places to the left.

If you have a decent calculator then always enter both as follows:

(a) For frequency (always in Mhz) enter 1, then hit your exp key and always enter 6.
(b) For capacitance (always in pF) enter, in this example, 159.155, then hit your exp key and always enter 12 and then the change
sign key to - or minus.

And remember reactance is always called ohms just as we call resistance by the same unit and use the same symbol.

QUIZ - No. 2

1. What is the reactance of a 100pF capacitor at;
(a) 100Khz (b) 1.5 Mhz and (c) 22 Mhz.

2. What is the reactance of a 22uH inductor at;
(a) 100Khz (b) 1.5 Mhz and (c) 22 Mhz.

3. This is a revision of the previous tutorial:- What capacitor will resonate with the above inductor at;
(a) 7234Khz (b) 1224 Khz and (c) 3.5 Mhz.

Yes Ian, I did complete the above quiz in my exercise book. Can I please now see the answers to check out my work.

3.What is Impedance?

Now here comes one of the most confusing aspects of electronics - which I will de-mystify by taking my customary casual approach.

What is Impedance?

Now here comes one of the most confusing aspects of electronics - which I will de-mystify by taking an extremely casual approach, so what's new!. I have known electronic enthusiasts who still couldn't even mentally visualise the concept even after 25 years.

I'll keep it dead simple, very inelegant but dead simple and give all the purists heart palpitations. I bet you walk away with a better understanding though.

If you need to know the technical answer for impedance and you should, then consult one of the must read texts I have suggested elsewhere.

Assume you have available these 4 items on your bench:

     (a) A series of eight fresh AA type 1.5 volt cells to create a total of 12 volts supply.
     (b) A 12 volt heavy duty automotive battery - fully charged.
     (c) a small 12v bulb (globe) of very, very low wattage. and;
     (d) a very high wattage automotive high-beam headlight.

Now if we connect the extremely low wattage bulb to the series string of AA cells we would expect all to work well. Similarly if we connect the high wattage, high-beam headlight to the heavy duty automotive battery all will be well. Well for a time anyway. Both of these sets are "sort" of matched together. Light duty to light duty and heavy duty to heavy duty.

Now what do you think would happen if we connect the high beam headlight to the series AA cells and conversely the low wattage bulb to the automotive battery?.

In the first case we could imagine the high beam headlight would quickly trash our little tiny AA cells. In the second case our min-wattage bulb would glow quite happily at its rated wattage for quite a long time. Why?, therein lies my expanation of impedance.

Consider it!

The heavy duty battery is capable of delivering relatively large amounts of power but the series string is capable of delivering only relatively minimal power. The first is a low impedance source and the other, in comparison is a relatively high impedance source.

On the other hand the high beam headlight is capable of consuming relatively large amounts of power but the minature bulb is capable of consuming only minimal amounts of power.

Again the first is a low impedance load and the other is high impedance load. If you're keen to apply ohms law you will discover why, research it through the text books.

Meanwhile take a well deserved coffee or tea break now and think it over. Me?, I'll just have another beer while I'm waiting for you.

Good break?. If you were paying attention you would now be able to understand an analogy - a particularly rough but effective one;

Imagine a tiny caterpiller chewing on a large blade of grass - no problem plenty to eat there. Now on the other hand imagine a poor cow stuck in a desert with only one similar blade of grass available to eat. I hope you have some better understanding now.

The term impedance is a general expression which can be applied to any electrical entity which impedes the flow of current. Thus this expression could be used to denote a resistance, a pure reactance (as above in this tutorial), or as is most likely in the real world, a complex combination of both reactance and resistance.

Don't get overly concerned if you're a bit confused by that statement at the moment. However this does then lead us on to "Q".

QUIZ

The good news is there isn't one here.

4. What is Q and why is it so important?

Another aspect which confuses some people no end. But it is absolutely critical to your understanding of filters later.

Well what is Q and why is it so important?

This is another aspect of radio which confuses some people no end. Q can be considered to mean Factor - 0f - Quality (that's my expression - not an official one).

It has no dimension as such. Usually when discussing Q in resonant circuits we can look at either the capacitor or inductor. Capacitors do have a Q many times greater than that of an inductor so we can generally ignore the capacitor and concentrate only on the Q of the inductor.

Throughout these tutorials we are dealing with resonant circuits and primary interest to us here is the response of our circuits. Now we're heading toward the pointy end of the stick.

The standard we will observe is the response at the 3db points. You have previously learnt that a combination of x - capacitance with y - inductance will resonate at a z - frequency. Now a very common fallacy - and a huge inexcusible one with some casual designers - is, if we arbtrarily select an L with a C then all is Okey Dokey. Many constructors building say a receiver for 40 metres will blindly follow this approach.

This constructor also blindly assumes his front end filter will accept all frequencies of interest and reject those which are of no interest to him or her (notice the political correctness here - I have a wife and 7 daughters, that's why!).

Have you always thought that was the case - I did for a number of years until I bit the bullet and learnt the theory just as you are now beginning to learn right here.

What is the 3db point anyway?. Well firstly let us consider the principal factor affecting Q. It's resistance. A length of wire wound into a coil to provide an inductor will have a certain D.C. resistance which can be measured with an ohmeter (actually you would need a milli-ohmeter because the D.C. resistance most likely would be well below 1 ohm).

However the resistance to a.c. or r.f. is altogether different, especially at low frquencies such as 455Khz.

It is this R.F. resistance which will affect or rather limit our Q to sometimes below useful values. You may define Q as the ratio of inductive reactance to R.F. resistance. Thus;

Q = ( 2 * pi * Fo * L )/ R

A typical practical toroidal inductor suitable for 40 metres may well be 23 turns of #20 wire wound on a T68-6 core. This inductor has an inductance of 2.42 uH and a measured Q of 295 at 7.0 Mhz. As an aside the Q at 5.0 Mhz is about 270 whilst at 9.0 Mhz the Q has already peaked at about 302.

What have we learnt here?. Firstly for a given inductance the Q is frequency dependent. It varies with frequency!. Do some more sums here based on what we have learnt so far. Go on, fire up the calculator, don't now or ever take my word for it. I've been known to be a blatant liar before. Use your own initiative and always follow me on your calculator - you learn by actually doing it.

If you are alert you should have established;

(a) that the reactance of 2.42 uH at 7.0 Mhz is 106.44 ohms and;
(b) it will resonate at 7.0 Mhz with a 213.6 capacitor because your L/C was calculated to be 516.94 and further;
(c) if our Q is 295 then the r.f. resistance must be 0.36 ohms.

Now I didn't just dream all that up. I took the toroidal coil inductance and Q from the Amidon Data Book. If it's wrong - and it won't be, blame Bill Amidon.

Where does this leave us? Simple - without any load connected to our inductor/capacitor (resonator) the 3db bandwidth is ( Fo / Q ) or ( 7000Khz / 295 ) or 23.729 Khz. Notice I said without any load connected to our resonator, this is referred to as Qu or the unloaded Q when the resonator is hanging around in thin air and is actually doing nothing of particular value.

Our resonator hasn't met either the caterpiller or cow yet. This resonator impedance may be described as ( Q * 2 * pi * Fo * L ) or as it will in future be referred to as ( Z = Q * XL ). This equates to 31,400 ohms or 31K4.

Why the big K?. Because with small type, commas and decimal places often get lost or obliterated. A very good example was a circuit requiring 4.7 ohms as an emitter resistance. The typesetter, this was in the olden days of course, had for so often saw 4.7K ohms or 47K ohms repeated everywhere and assumed your poor author was a dill and had merely omitted the K.

In an amplifier with emitter degeneration the difference between 4.7 ohms ( or as we will use it, 4R7) and 4.7K ( or as we will use it, 4K7) is infinitely profound. In short it don't work!. So if I say 31K4 then you should know beyond any doubt that I mean 31,400 ohms.

Circuit designers and publishers who cling to the old system should be summarily executed for inexcusible stupidity. Tell 'em I said so, that should have same the effect as hitting 'em with a wet lettuce leaf.

QUIZ - No.3

1. In our previous tutorial we had calculated the reactance of a 22uH inductor at;

(a) 100Khz (b) 1.5 Mhz and (c) 22 Mhz. - therefore:

What are the impedances if the unloaded Q's are respectively (a) 32, (b) 170, and (c) 110

2. What is the effective r.f. resistance of the inductor above at the respective frequencies and Q as follows;

(a) 100Khz / 32, (b) 1.5 Mhz / 170, and (c) 22 Mhz / 110

3. What is the correct method of describing the following resistances and capacitances.

(a) 0.68 ohms, (b) 10 ohms, (c) 220 ohms, (d) 1,800 ohms, (e) 82,000 ohms, (f)

470,000 ohms, (g) 5,600,000 ohms, and;

(h) 6.8 Pf, (i) 180 Pf, and (j) 4.7 uF

Yes Ian, I did complete the above quiz in my exercise book. Can I please now see the answers to check out my work.

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