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The Evolution of a Transmitter By Wayne McFee NB6M Salmoncon 2010 Designing a Simple QRP Transmitter • Let’s say we have a circuit we like for a 40 Meter QRP Transmitter, but would really like to put it on a different frequency band. • What do we do? The Answer………. • Modify the existing circuit so that it will perform well on the band of choice. Resource Material • The transmitter circuit we will use as an example is the “First Transmitter” circuit from Experimental Methods in RF Design, which was explored in yesterday’ Ham Radio Coloring Book lesson. • This circuit, from Wes Hayward, W7ZOI, is discussed in detail in the first chapter of EMRFD, pages 1.17 through 1.22. • EMRFD, published by the ARRL, is truly a “Must Buy” for anyone wanting to learn about radio and electronic Circuitry. What is the objective of this lesson? • To make the necessary changes to this 40 Meter Transmitter to make it work well on the band of choice. • We will first look at putting it on 20 Meters. How do we go about this? • Every transmitter has certain parts with values selected to optimize performance on the desired frequency band. • The first thing we need to do is to identify those parts. • The second part of the process is replacing those parts with part values that will optimize performance on the band of choice. “First Transmitter” Stages • Although relatively simple, this transmitter is thoughtfully designed for good performance. • It contains a Variable Crystal Controlled Oscillator (VXO), Buffer Amplifier with Power Output control, a Driver, and a Power Amplifier. • It also contains a Keying Circuit and a manual T/R Switch. • Let’s take a look at a block diagram of the circuit. • Now let’s take a look at the circuit. • You will note that there is a mistake in the drawings you have. Q5, 2N5321, is an NPN transistor, not a PNP as is shown. • The schematics shown on the screen are correct. Now let’s take a look at the circuit with the individual stages outlined. Using your colored pencils from yesterday, outline the stages on your own diagram. Let’s identify the parts that are specific to 40 Meter operation. Frequency Sensitive Parts • Crystal • Feedback Capacitors in the Oscillator • Coupling Capacitor to Buffer Amp • Impedance transforming “L” network between Driver and PA • Impedance transforming “PI” network between PA and 7 Element Filter • 7 Element Output Filter ffer • The Crystal is really easy, we just remove it and replace with a 20 Meter Crystal. • The rest of the parts we are concerned with are either capacitors or inductors. • What do we need to know in order to make the appropriate changes to these parts? • In order to answer that question, we need first to understand why these capacitor and inductor values were chosen. The answer is Reactance • Capacitors and Inductors have Reactance to radio frequencies. • These component values were chosen because they exhibit the needed amount of reactance at 7 MHz. Now let’s look at the easy part! • Do we have to know advanced electronics theory, or be able to work with advanced mathematics to be able to put this transmitter on the 20 Meter Band? NOT AT ALL • Since all of that advanced work has already been done by Wes, W7ZOI, we need only the following: • The ability to multiply and divide simple numbers • The ability to determine the number of turns needed on toroids to arrive at a desired amount of inductance. Coils - 2010 • I know that we can all multiply and divide simple numbers • And we can either work out the necessary number of turns on toroids ourselves, using published formula, or use a computer program such as “Coils – 2010”, by Wes Hayward, W7ZOI. • It is contained in a zip file, coils–zoi.zip and is available here: http://w7zoi.net/tech.html Stage by Stage changes • Let’s take this stage by stage, starting with the oscillator VXO Circuit Changes • Change the crystal to a 14 MHz unit. • Change the value of the feedback caps. • Change the value of the VXO coupling cap. KISS • Everyone knows this means: “Keep It Simple, Stupid”. • How do we apply that theory here? Use the Available Info • Wes has already worked out the part values required to provide the necessary amount of capacitive reactance at 7 MHz. • Do we need to do any of that work? • Do we even need to know what the reactance of a particular part is? NO • All we really need to know is the simple numeric relationship between 7 and 14. Firmly Grasp the Obvious • OK, so 14 is two times 7, or twice the original number. • How do we use that fact? Capacitive Reactance is Inversely Proportional to Frequency 1 Xc = ____________ 2πFC What is Xc of 390pF at 7 MHz? 6.28 X 7 X .00039uF (390pF) = .0171444 1 _________ = 58.32 Ohms .0171444 What Capacitance Reactance does 390pF have at 14 MHz? 6.28 X 14 X .00039 = .0342888 1 _________ = 29 Ohms .0342888 What does this mean to us? • This means simply that if we want a capacitor value that will provide the same amount of Xc at 14 Mhz that a specific capacitor value did at 7 MHz, all we have to do is divide the capacitor value by 2. • 390pF divided by 2 = 195pF Xc of 195pF at 14 MHz 6.28 X 14 X .000195 = .0171444 (look familiar?) 1 _____________ = 58.32 Ohms .0171444 • To optimize the oscillator performance at 14 MHz, we change the feedback cap values to 195pF, or the closest standard value that provides quick and reliable starting and running of the oscillator. This could be either a 180pF or 220pF cap. • The 7 MHz coupling capacitor between the oscillator and buffer amp is now a 100pF value, which works out to have an Xc at 7 MHz of 227.47 Ohms. • What capacitance value will provide that amount of Xc at 14 MHz? KISS • Yes! All we do is halve the current 7 MHz value to arrive at the 14 MHz value. • ½ of 100pF is 50pF, the closest standard value is 47pF. Moving On • Let’s apply our KISS method to the next set of components that need to be changed. This L Network transforms 200 Ohms to 50 Ohms at 7 MHz • What new Capacitance value is needed for 14 MHz operation if the 7 MHz value is 200 pF? Piece of cake, right? • 200pF divided by 2 is 100pF • What new inductance value is needed if we use 2.028 uH at 7 MHz? Inductive Reactance Xl = 2πFL 6.28 X 7 X 2.028 = 89.15 Ohms 6.28 X 14 X 2.028 = 178.30 Ohms 6.28 X 14 X 1.014 = 89.15 Ohms Amazing! Once again, all we need to do is divide the current inductance value by two. What’s Next, this is FUN! • Let’s look at the PA Circuit, specifically at the impedance transforming PI network and 7 Element Output Filter. Test Time • Here is the circuit for the PA, Impedance Matching PI Network, and 7 Element Output Filter. • Your test is to correctly identify the parts which need to be changed, and the new capacitor and inductance values for 20 Meters. Final Challenge • If you have a 1000pF cap in a circuit designed for 7 MHz, and you want to change it to a value suitable for 10.1 MHz, what would the new value be? • Hint - use the KISS method……. OK, a little help here • What percentage of 10.1 is 7? • Let’s see, 7 divided by 10.1 = 69.3 • So what’s 70% of 1000? • New value 700pF - Closest standard value is 680pF Turn the Question Around • What capacitance value would be needed for 3.5 MHz if the 7 MHz value is 1000 pF? • Think about it - the original frequency is double the desired frequency. You Got it • That’s right, the value for 3.5 MHz would be 2000 pF. • We could use a pair of 1000pF in parallel if we needed that exact value. See How Simple This Is? • To determine the new value, use the percentage relationship between the old and new frequencies. Rule of Thumb • For New L or C value, ALWAYS divide the original frequency by the new frequency. 7 ÷ 3.5 = 2 multiply old value by 2 7 ÷ 10.1 = ± .70 multiply old value by .7 7 ÷ 14 = .5 multiply old value by .5 7 ÷ 21 = .33 multiply old value by .33 7 ÷ 18 = .39 multiply old value by .39 Questions or comments? Thank you all for your kind attention! Happy Building! Wayne NB6M