The basics of a step by step, easy to follow and duplicate tutorial on the design of LC Filters ranging from the simple to the complex together with required learning aids.

Authored by Ian C. Purdie VK2TIP

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L/C FILTERS - ESSENTIAL BASICS TO BEGIN WITH


you can have this page translated /vous pouvez faire traduire ces pages /Sie können lassen diese Seiten übersetzen /potete fare queste tradurre pagine /você pode ter estas páginas traduzido /usted puede hacer estas paginaciones traducir


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NOW CONTINUING WITH THE TUTORIAL

1. 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 you should read on.

If you don't know or understandinductance, capacitance and resistance  jump ahead now and then get back to us.

2. What is Reactance?

Just don't react too badly to this.

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.

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

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

5. You should have a decent calculator to do your work.

You will need:-

(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.

6. 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.

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LC BASICS

1. What is this LC jazz anyway?

Well the letter L stands for inductance and the letter C stands for capacitance. Ah! I still love this high-tech talk.

If you still don't know or understand inductance, capacitance and resistance  or you cheated then jump out now and then get back to us.

Happy with that?. 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 (L * C) we get 25330.314. This figure you will now ensure becomes permanently embedded in your mind for evermore because it is the very basis of my formula for L/C.

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 formula. They are one and the same, just re-arranged by simple algebra.

LC= 25330.3 / Fo [Mhz]2

or

Fo [Mhz] =  Ö[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.

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.

Consider if we were designing any type of filter for say, 7.150 Mhz (which is 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

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.

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INDUCTORS, CAPACITORS AND RESISTORS

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These are the principal passive components which make up most circuits we are discussing here. A capacitor is a capacitor, an inductor is an inductor and of course a resistor is a resistor or an impedance at its rated value.

Don't you believe it because it ain't necessarily so.

We shall soon see that a capacitor can at certain frequencies look like an inductor and an inductor can look like a capacitor and our resistor may look a bit like both of them.

1. What is an Inductor

Well the letter L stands for inductance. The simplest inductor consists of a piece of wire. A piece of #22 wire which is approximately 100 mm ( 4" ) long has a measured inductance of 0.14 uH. Well mine did because I measured it on an inductance meter (see kits) operating at about 740 Khz. - [but compare that value with formula later].

Now this 100 mm length of conductor is about 25 mils (i.e. 25/1000") or 0.644 mm in diameter but does NOT carry rf throughout its whole cross-sectional area of 0.326 mm2. ( [Pi * Dia2] / 4 ) - another useful formula as you will soon see because I then use the traditional pi * r2 which is a pain if you talk ( as we mostly do ) in diameters.

At low frequencies this would be the case,  but as the frequency is increased so is the magnetic field at the centre of the conductor and this leads to an increasing impedance to the charge carriers. This decreases the current density at the centre of the conductor and increases the current density around the perimeter of the conductor. We call this increase in current density around the perimeter "skin effect".

where A1 = pi * r1 and  A2 = pi * r22
Skin depth area = A2 - A1
or
=  [ pi * ( R2 2 - R1 2 ) ]

The net result of skin effect is a net decrease in the cross sectional area of the conductor and a consequent increase in the rf resistance. Consult the references I have suggested you read for a more detailed and informed discussion on this very important topic.

In particular I would recommend RF Circuit Design - Chris Bowick - Sams.

This skin effect we will see later on has a very significant impact on the Q of our inductors.

Throughout these tutorials I will incorporate as many relevant formulae as possible. Why?.

Because you will have under one roof all the formulae necessary to tackle any design goal you may have in mind.

Let's start out with the formula for the inductance for a straight piece of wire.

L = (.002 * length)[ 2.3 log {((4 * length)/ dia) - 0.75}] uH

where:
L = inductance in uH.
length = length in cm. (1" = 2.54 cm).
dia = diameter of the wire in cm also.

plugging in the details of our piece of wire which are, length = 10 and dia = 0.064, all dimensions being in cms.

L = (.002 * 4)[ 2.3 log {((4 * 4) / 0.064) - 0.75}] uH

L = (.008)[ 2.3 log {((16 ) / 0.064) - 0.75}] uH

L = (.008)[ 2.3 log {249.25}] uH

L = (.008)[ 2.3 {2.3966}]  = 0.044 uH

The reason I have included as many steps as possible is to help those people who do not necessarily feel comfortable with maths. Including as many steps as possible means they should be able to follow how we arrive at an answer. [now compare that answer with the measurement I made earlier???]

Now from the above equation you should begin to understand any piece of wire exhibits inductance. This of course includes capacitor and resistor leads as well as any  inter-connecting wiring in your circuit. Circuit board traces exhibit inductance and in fact this property is often taken advantage of at UHF. We can draw inductors on the PCB's.

What if we decide to wind our 100 mm piece of wire into a coil? Here is another formula which is referred to as "Wheelers' Formula" - * Proc IRE, p 1,398, Oct 1928., although here the formula has been metricated. If you are more comfortable with imperial inches then delete the 0.394 and use inches in lieu of cm.

L = (0.394 *  r2 * n2 ) /  ( 9r + 10b ) uH

where:
L = inductance in uH.
r = radius of the coil (in cm).
n2 = number of turns in the coil squared and;
b = the length of the coil in cm.

Our length of 100 mm wire can easily become 2.5 turns wound on a 12.7 mm dia. former or even 5 turns wound on a 6.35 mm dia. former. If for convenience we make both coils equal in length to 12.7 mm then using the above formula and not forgetting to convert to cm and NOT confusing radius with diameter.

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 Ö L     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?

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:

The only disadvantage I can think of

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 ) are:

TYPE            COLOR        AL                      Freq. Range

T50-26        Yel-Wh        320uH                  power freq.
T50-3          Gray           175                      50 Khz to 500 Khz
T50-1          Blue            100                      500 Khz to 5 Mhz
T50-2          Red              57                      2 Mhz to 30 Mhz
T50-6          Yellow          47                       10 to 50 Mhz
T50-10        Black            32                        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.

What is a Capacitor

Well the letter C stands for capacitance. A capacitor is formed by two or more conductive parallel plates separated by an insulating material or dielectric. Typical dielectrics you will encounter are air, mica, ceramic or plastic.

Consider an air variable capacitor of the type used either as a trimmer or even the tuning element in a radio. It is merely two or more conductive plates separated by air - is this not so?.

Go back to our inductors above, could not a similar description apply to them?.

That is why inductors also have capacitance!. We have also learnt a small piece of wire has inductance. This is why capacitors with leads etc. exhibit inductance.

At some point our inductor with its inherent capacitance (called stray) will resonate (you will learn about that later) and this is called its self resonant frequency. The same rule applies to capacitors. Mostly it is only the VHF and UHF aficionado who has to be greatly concerned about these properties. Stray capacitance is everywhere. Sometimes it can be used to advantage, usually you take it into account (that is another function of trimmers) but often it's a monumental pain.

I said mostly - which means if you forget about it, then you can surely guarantee one day this property will bite you. That will be the day when you're working on your pet project, which of course won't work as expected and you don't know why your theory doesn't work out in the real world - another "gotcha".

Stray capacitance in sloppy layouts can account for unexpected oscillations, no oscillations, different circuit responses and generally cause you to "become a victim of the bottle, eternally caught in the grip of the grape and, if you become like me, occasionally ruin an otherwise good keyboard with an even much better claret - well actually 'Tyrrel's Pinot Noir' when the budget allows".

Even the best of designs and careful layouts are affected. Not only stray capacitance but stray inductance can affect you. Try some high speed, double sided, digital PCB designs.

Back to capacitors. Apart from their uses in resonant circuits they are used as:

(a) dc blocking devices - a capacitor will pass ac or rf but not dc. That is often the function of coupling capacitors in circuits. The coupling capacitor will pass our required signal but block the dc supply from the previous stage e.g. .001 uF (1000 pF or 1N0).

(b) supply by-pass capacitor - a capacitor used to pass ac or rf. Examples include by-pass of a dc supply to an emitter bypass capacitor. When used on a dc supply line the capacitor shunts (shorts) to ground any unwanted ac or rf to avoid contamination of our supply e.g. 0.1 uF (100,000 pF or 100N)

(c) reservoir by-pass capacitor - Used in the output of a dc rectifier to smooth out the power line ac pulses and as a reservoir between the charging pulses. Think of it as a big water storage tank. This is where you have the high value capacitors e.g. 10,000 uF / 63V. (10,000U)

(d) emitter by-pass capacitor - In the case of an emitter by-pass our transistor may, as one example, be considered as two distinct models. One under dc conditions which sets up how we want the transistor to operate and another under ac or rf conditions e.g. as an amplifier. In this case our emitter by-pass capacitor merely makes any emitter resistor invisible for ac or rf purposes typical values might range from 2.2 uF down to 0.1 uF. (2U2 down to 100N)

Parallel plate capacitors have a formula for calculating capacitance:

C = 0.22248 * K * (A/t)

where:
C = capacitance in pF
K = dielectric constant (Air = 1)
A = area of plates or dielectric in inches2
t = thickness of dielectric or spacing in the case  of air variables.

Series or Parallel Capacitors:

Series and parallel combinations calculate in an opposite fashion to inductors, i.e. parallel = C1 + C2 + Cn and series are:

C total =  {1/ [(1 / C1) + (1 / C2) + (1 / Cn)]}

Now here come some very very useful formula. Using the one immediately above you should determine that a 22 pF capacitor in series with a 15 pF capacitor gives a total of 8.92 pF. Go ahead do it! check me out.

What if we had say 105 pF already in circuit and needed to reduce it to say 56 pF. A real world example might be say the only air variable capacitor available is one with a maximum of 105 pF. What do you put in series to reduce the maximum to 56 pF?.

Did you know the formula above, for two capacitors in series, can become:

C total =  [( C1 * C2 ) / ( C1 + C2 )] ** AND:

C series =  [( Cavail * Cwanted ) / ( Cavail - Cwanted )] THEREFORE:

C series =  [( 105 * 56 ) / ( 105 - 56 )] = 120 pF

If you disbelieve me then check it out using ** above as a double check. Nifty eh!. Write it in your exercise book. If that's flash try this red-hot formula.

Suppose we wanted an oscillator to tune from 7.0 Mhz to 7.2 Mhz. Now if you don't quite know what an oscillator is yet don't worry. For reasons we will explore later on in the tutorials you will find that your capacitance MUST vary, for tuning purposes, in the direct ratio of the frequency ratio squared or Cmax to Cmin =  (Fhigh / Flow)2 . Eh What!.

In this example our frequency ratio is Fhigh / Flow = 7.2 / 7.0 = 1.02857. Our ratio must be squared so then it becomes (1.02857 * 1.02857) = 1.05796. This must be our capacitance ratio i.e. Cmax to Cmin = 1.05796.

Using the 105 pF air variable capacitor is no help because it usually has a minimum of 10.5 pF and that's a ratio of 10:1. Assuming the 105 pF, for the moment ***, is acceptable then apply this formula (which is quite difficult to do in HTML language the way I wanted to do it - so bear with me):

Cpad + 105 / Cpad + 10.5 = 1.05796 / 1

What needs to be done is re-write that formula on a piece of paper using two lines and writing the division sign in the usual way, i.e. as an underlining of both Cpad +105 and 1.05796. I hope everybody knows what I mean. Next we perform a simple algebraic function, we cross multiply so you should end up with:

Cpad + 105  = (1.05796 * Cpad ) + ( 1.05796 * 10.5)

 Cpad + 105  = (1.05796 Cpad ) + ( 11.10858)

then subtracting 1 times Cpad from each side as well as subtracting 11.10858 from each side (in algebra what is done to one side of the equal sign must be done to the other side - I hope that's clear also) we get:

 93.89142  = 0.05796 Cpad  and  dividing both sides by 0.05796

Cpad  = 1619.93 pF

Now firstly I accept the fact that the formula may be obscure for many people and if that is the case then I will re-write it as a graphic but including too many graphics can cause other problems. It is MOST important (to me anyway) that I make this as easy as possible for you.

Secondly the figure of 1619.93 pF above is most likely too high to use in the real world. Having said that let's check our sums. We have 105 pF max and 10.5 pF min in parallel with 1619.93 pF which is the same as ( 105 + 1620 ) / (10.5 + 1620 ). This calculates out to 1.057977 : 1 which when applied to our tuning range would tune from 7.0 Mhz to 7.19999 Mhz or what we set out to do. Remember our strays earlier?. They would play havoc with this, so the 1620 pF would become in the real world 1000 pF + 560 pF + 100 pF trimmer ( i.e. 10 - 100 pF ).

I said earlier the 105 pF was possibly acceptable ***

Let's go back and reduce it to say 33 pF and redo our sums to get another Cpad value. Let me know your answer.

Some home brewers (hi-tech code word for building-it-yourself) advocate using a starting goal of 1 pF per wavelength of frequency. e.g. 7.0 Mhz = 40 metres (approx.) = 40 pF.

I must say I am not especially in love with that particular logic for many reasons which will become apparent as we become more deeply involved with filter and oscillator theory.

This is as far as I want to go on inductors and capacitors for the moment.

What is a Resistor

Resistance by technical definition "is the property of a material that determines the rate at which electrical energy is converted into heat energy for a given electrical current".

Mouthful?. Well look through the books I recommend to get a better insight. In particular if you don't already know Ohm's law backwards then learn it now!. It is as fundamental to your development as the first breath you took the moment you were born. Believe it.

It should be fairly obvious if you have managed to stay awake so far, that a resistor must exhibit some inductance. Depending on how it is mounted on a board or elsewhere then it will exhibit some capacitance also.

On a totally different tack, many people may not appreciate what is a cunning trick. In relatively low frequency circuits e.g. below 15 Mhz, surplus high value resistors make excellent stand-offs when building rat's nests. I use a heap of 1 Meg ohm or more resistors for this very purpose.

Of special interest to me is just what do you guys overseas use for resistors. Here in Australia our hobbyist suppliers (see Links) offer E24 2% metal film resistors for a reasonable price and I tend to use these. The 5% are becoming so small they usually become embedded in the feet of my grand-children. I would like to know what you use outside of N.Z., the U.S., Canada or U.K.

Now if we put two or more resistors in series (R1 and R2) then the total resistances will add together. If one were say 1K, or 1000 ohms and the other was 1K5 or 1500 ohms then the total is 2K5 or 2500 ohms.

On the other hand  if we put two or more resistors in parallel (R1 and R2) then the total resistances will be less than the lowest value. If one were again say 1K, or 1000 ohms and the other was again 1K5 or 1500 ohms then the total net resistance is 600R or 600 ohms.

How is this so? Well because we have now provided two paths for the current flow and that means less overall resistance. The formula is:

(R1 X R2) / (R1 + R2)  OR

(1000 X 1500) / (1000 + 1500)

Try it out for yourself on the calculator.
 

If I have omitted anything important let me know - but remember I am NOT endeavouring to be a substitute for basic text books.

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REACTANCE - ESSENTIAL BASICS

 What is Reactance

Because I don't want to have herds of graphic files or complicate programming in HTML, we will throughout these Tutorials denote inductive reactance as XL and capacitive reactance as XC.

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 we say in the game r.f. 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 * 6.2832 * 1,000,000 * .000000000159155 ) OR enter into your calculator as;

1 / [( 2 * 6.2832 * (1 X 10+6 ) * (159.155 * X 10-12 )]

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

(159.155 * X 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.

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ESSENTIAL IMPEDANCE BASICS

 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 in previous 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.

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ESSENTIAL Q BASICS

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 of 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 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 many years until I bit the bullet and learnt 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 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 what I mean.

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 dishcloth.

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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Created:26th December, 1998 and Revised: 26th July, 2000

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