Click on the pic to see a larger photo! 1.5 W Power Amplifier type class-C
This project will explain the basic function of a class-C transmitter.
I will explain how to dimension a transmitter and the purpose of the different components.
I will also explain how you can build a 1.5W PA transmitter.
The project will include PCB, components and instructions how to make coils, assembly and testing.
All contribution to this page are most welcome!

You can find thousands schematic of RF power amplifier on internet, but almost non explain how they were made and how they works. Beside that you will also find very strange instructions how to make coils.

So, let's once and for all dig into the subject of RF-Power Amplifier.
How does it work and how to calculate the different components values.
Please prepare yourself with espresso coffee, efedrin C15H10NO and guarana, because this will be a long page with lot of math and explaining :-)
If you don't want to read all this, you can click here to go straight down to the party.

Relationship between the two shapes square wave and sine wave signal. RMS (Root Mean Square)
First let's discuss and learn about the RMS value of a sine signal. Take a look at the picture at right.
The left figure shows a square wave signal (blue) and to the right you see a sine wave signal (red).
The coloured area show the time integration of the DC voltage.
The area will later on represent the power your will get from the specific signal shape.
The relationship between the two shapes are 1 to 1/rot(2). It gives about 1 to 0.707.

This means that if you put 1 W power from a square wave or pure DC signal into a load, you will only get 0.707 W from a sine wave signal with equal amplitude. From the figures you will easy see that the area of the sine wave is less than the area of the square wave. You can also think like this:
Let's say you want to heat up a resistor with the two signals. To achieve equal amount of power (heat) in the resistor, the sine wave must be rot(2) times larger in amplitude than the square wave or constant DC signal.

So how about the power. We know P = U * I and U = R*I
For the square wave signal or a constant DC signal we will get the formula P = U2/R.
When calculating the power from a sine wave signal we have to consider the relationship of 1/rot(2) to a square wave or a constant DC signal. This gives the formula P=U2/2R You can find the two formulas under each figure.

Amplifier stages
Basic amplifier block diagram. It is time to look at the class-C amplifier and try to identify the different stages.
You will find 3 stages.

  • Input (red)
  • Output (blue)
  • Driver (black)

  • Let's start focusing on the output stages.
    At RF out you find a 50 ohm resistor which represent an antenna.
    The DC voltage is 12V which also will be the peak voltage of a sine signal.
    Let's say we want our transmitter to deliver 5W. We can then use the formula to calculate the necessary impedance. R = U2/2P = 12*12/2*5 = 14.4 ohm.

    The driver (transistor) want to see a 14.4 ohm load to be able to deliver 5W of power. As you can see the antenna is 50 ohm and we wish to have 14.4 ohm load, so we must have some kind of impedance transformation.
    A commonly used impedance transformer is the pi-filter. It consist of 2 capacitors and an inductor. At the input stage we will have the opposite relation.
    The driver (transistor) has a low impedance (1-10 ohm) and we wants the input to be 50 ohm which is a kind of standard.
    Again we need a transforming stage to convert the input power with 50 ohm impedance to (1-10 ohm) impedance at the driver input. Most often the input impedance of the driver is not purely resistive, it also contain a reactive part. In this example the input impedance is (5-j5) ohm.

    Basic Amplifier
    Main components in a basic amplifier. The picture at right show you the same stage with the main components to form a class-C amplifier. The output pi-filter is shaped with C3, C4, and L4.
    An important note is that the transistor has a stray capacitance Cce from emitter to collector. You can find this data in the datasheet of the RF transistor.
    As you can see, Cce is parallel with C3.
    When you have calculated C3, you must subtract the stray capacitance Cce to get correct value of C3.
    Below I explain how to calculate pi-filters.

    L2 is a large inductor which is not critical in value (1-100uH). Sometimes a ferrite pearl will do the job. L2 act as high impedance (invisible) for RF signal, but it will give the base of the transistor a DC voltage grounding.

    C1, C2 and L1 forms an input transforming stage which will make the input 50 ohm purely resistive and convert all energy to the low impedance of the transistor.

    How to calculate a PI-filter (Output filter)
    Math is fun don't you think. Picture at right explain how to calculate a pi-filter.

    First you have to define the Q-value of the system.
    For an electrically resonant system, the Q factor represents the effect of electrical resistance. The Q factor is defined as the resonant frequency (center frequency) f0 divided by the bandwidth BW.

    Bandwidth BW = f2 - f1, where f2 is the upper and f1 the lower cutoff frequency. On a graph of response versus frequency, the bandwidth is defined as the 3 dB change in level (voltage, current, or power) on either side of the center frequency.

    The example show you the values for our 5W example above.
    R1 represent the antenna and R2 is the "wanted" impedance that the transistor should see, to be able to deliver 5W.

    Formula 1, 2, 3 will give you the reactance's of the components C1, C2 and L of the pi-filter. With formula 4 you will be able to calculate the two capacitance's C1,C2 and with formula 5 you will be able to calculate the inductance L.

    A conventional PI filter which has one more capacitor than inductor has a characteristic of slow start voltage rise which we will discuss more later on.

    How to calculate Input filter
    Even more math. Picture at right explain how to calculate an input filter.
    First you have to define the Q-value of the system.

    R1 represent the input resistance of the transistor and R2 represent the "wanted" input resistance of the amplifier.
    In our example we had and input impedance of 5-j5 ohm.

    This means that the transistor has an input resistance of 5 ohm in serial with a capacitance (Cs) of 5 ohm (reactive) at 100MHz.
    The input impedance (R1 and Cs) is printed into the dotted square.

    First we will only focus on the resistive part (R1) which was 5 ohm.
    Formula 1, 2, 3 will give you the reactance's of the components C1, C2 and L of the filter.

    As you see in the calculation example the reactance of the inductance L is 25.00ohm. Since L is in serial with Cs the two of them will subtract to give the total reactance.

    The total reactive part will then be :
    +25.00 (inductive) - 5 (capacitive) = +20 ohm. The reactance is positive which will represent an inductance of 31.8nH

    You can think like this: Some part of the inductor reactance (L) will be eliminated by the internal serial capacitor reactance (Cs) of the transistor.

    The transistor is simulated with a switch. Colector choke inductance
    The last part we have to discuss is the inductor at the collector of the transistor.
    This inductor has nothing to do with filtering purpose. The inductor acts as a energy storage for the output filter during the time the transistor is OFF. Take a look at the picture at right and I will explain.

    The transistor works in two state, ON and OFF.
    When the transistor is ON, the transistor will sink current Ic to ground. Ic = I1 + I2.

    When a current flow through an inductor (L) a magnetic energy filed starts to built up around the windings.
    The amount of energy is dependent on the current I1 and the inductance of L.

    Take a look at the bottom figure which is the OFF phase.
    When the transistor goes into the next phase which is OFF the current Ic disappear.
    The magnetic filed in the inductor (L) will now start to collapse and induce a current I1 which will flow into the pi-filter. I2 flows in opposite direction and is equal to I1.
    As time goes on, the current drops and so will the voltage at the collector of the transistor.
    When the transistor is about to go back into ON phase, the energy level in the inductor (L) should be close to zero. This means the current I1 is close to zero and so will the voltage at the collector also be.

    What happens if we use a too large or too small collector inductor (L) ?
    If we use a very large inductor (L), the transistor current will charge it with to much energy during the ON phase. The conventional designs of PA often use large inductor (choke), because you will then have the inductor working as a energy "fly-wheel" delivering energy to the pi-filter while the transistor is OFF stage.
    This is why you often see the collector inductor labelled as choke, (no value of inductance).
    It will work and is simple, but it is actually not the best solution.


    Since the inductor (L) still have lot of spare energy when the transistor is about to go into ON phase, the current I1 will still be high and so will the voltage at the collector also be. When the transistor enters ON phase it will drag this voltage down to zero and this "discharge" will create high current in the transistor causing a spike of power dissipation. Remember, there is one ON transition, and one OFF transition every cycle of the RF wave.
    This will happened 100 million times/second so the transistor will have to work harder and become warmer.

    If we use a too small inductor (L), we will not have enough energy to the following pi-filter and the collector voltage will swing to a negative voltage - an undesirable state.

    If we choose the inductor (L) value carefully, the collector voltage can swing right down to zero at the moment of transistor turn-ON.
    This is exactly what is needed to practically eliminate the power dissipation spike at the ON transition. This single change can significantly boost efficiency.
    Take a look at the simulation below and you will see the difference more clearly.

    Simulation of inductor (L)
    The left simulation show you the collector voltage from a large inductor. The green line show the transistor phase ON/OFF.
    As you can see in the figure, the collector voltage is still high when the transistor is about to enter ON phase.
    When transistor enter ON phase it has to discharge this voltage, which will cause a current spike!

    The right simulation use a more optimised inductor. Here you will see that the collector voltage is almost zero when the transistor is about to enter ON phase.
    There will be no current spike and you will have a cooler system and higher efficiency. Spice simulation of an amplifier using two different values of the inductor L

    Upside / downside
    The upside of choosing correct inductor (L) is that you get higher efficiency of the transmitter and your transistor will run much cooler. The peak collector current will be larger too. Peak collector current is larger because while ON, the transistor builds extra current through the small inductor (L). This current is dumped into the pi-filter (during the OFF phase) and must be replenished during the next ON period. The result is that the filter sees a larger swing, which translates to slightly larger output RF power. So you can see that the extra voltage and current that the transistor encounters is not wasted.

    With a small choke, collector voltage has a higher, narrower peak. This tends to increase even-order harmonics presented to the filter.
    The filter has more difficulty rejecting the 2nd harmonic. Second harmonics can be reduced by designing the filter for a slightly lower cutoff frequency, and extending the OFF period at the expense of the ON period (this must be done at the base drive). You'll get slightly less power out as a result.
    You'll also need to more carefully address power supply filtering at the power supply end of the choke.
    The small choke results in large peak-to-peak currents drawn from the supply. The supply-end of the choke will tend to get yanked around: excellent, short-leaded bypass capacitors should be employed to ensure a steady voltage at this point. Larger current swings around the transistor/choke/bypass area could cause ground-loop problems for the amplifier that drives the base. Even more care must be taken with component placement and printed circuit layout issues when this design technique is employed. The higher drive levels needed for efficient operation mean that final amp gain is less than normal. More power is required of the driver stage. A compromise is reached where driver dissipation eats into good final amp efficiency.

    How to calculate the inductor (L) ?
    The energy needed is dependent on many factors. Higher Q filters will have an inherent voltage swing that is larger due to a current dump, so an inductor having less energy is required. Inductance likely scales with filter Q. You also have losses in the components to consider.
    The inductor should dump its entire current charge while the transistor is OFF into the following filter. At the filter's input (collector), the current dump will result in a voltage swing dependent on filter Q. Many filters will have low Q, to keep losses low, and to keep component tolerances non critical.
    For a Q of one, here is a very simple approximation:
    The inductance required is collector impedance Z divided by 4f, where f is transmitted frequency in Hz.

    L = Z/( 4f )

    Once you calculate a collector choke, you should test it on a computer circuit simulator.
    The testing goal attempts to swing collector voltage just down to zero at the moment of OFF-to-ON transition, without swinging negative.

    Alright, it is time to wake up now….. theory part is over :-)

    Click on the pic to see a larger schematic. Hardware and schematic
    Let's have look at the schematic at right. You can click on the picture to enlarge it. This amplifier is run on +12V.
    At the input you find two resistors R1 and R2 which gives 50 ohm input impedance.
    This will make a stable load for a previous stage as FM PLL controlled VCO unit (Part II)
    A RF transistor BFG193 will pre-amplify the RF signal. The resistor R3 and R4 and R5 will set the gain on this transistor.
    If you wish to have high gain you should use R3, R4 as 100 ohm and R5 from 5 to 10k.
    With this configuration you will only need 5-20mW as input driving power.
    If R5 = 10k and removing R4, you will lower the gain and need 50-100mW as input driving power.

    C13, C14, C10, C19 and L3 forms the input matching filter for the main transistor Q2. The filter can be tuned by the capacitor C19. The power to drive the transistor Q2 comes from L4 which is an inductor. The value L4 is not critical at all. 10-100uH will do the job. The purpose of L4 is to keep the power line free from RF.

    Q2 is the main transistor and in this project I use MRF313, but 2N3866 or 2N4427 will work good to. Since this amplifier is at 1.5W the output filter is calculated for 50 ohm. I use a 5 pole filter element to give a good filter shape and simple solution. The filter can be tuned by the capacitor C20.

    Click on the pic to see a larger format. PCB of the 1.5 W Power Amplifier
    15w_313.pdf 1.5 W Power Amplifier type class-C (pdf).

    Above you can download a (pdf) filer which is the black PCB.
    The PCB is mirrored because the printed side should be faced down the board during UV exposure.
    To the right you will find a pic showing the assembly of all components on the same board.
    This is how the real board should look when you are going to solder the components.
    It is a board made for surface mounted components, so the copper is on the top layer.

    Grey area is copper and each component is draw in different colours all to make it easy to identify for you.
    The scale of the pdf is 1:1 and the picture at right is magnified with 4 times.

    Good grounding is very important in a RF system.
    I use bottom layer as Ground and I connect it with the top layer at four places (via-holes) to get a good grounding.
    Drill a small hole through the PCB and solder a wire in each via-hole to connect the top layer with the bottom layer which is the ground layer.
    The two four via holes can easy be found on the PCB and are labelled "GND" and marked with red colour.

    The construction is pretty basic to build and the only part I need to explain a bit is the making of the coils.
    In this construction I use four hand wounded coils, L3, L5, L6 and L7. The wires of the coils should be Cu or silver-plated wire with a diameter of 0.8-1.2mm.
    I really like the silver plated wire because you don't need to scratch of any enamel of the wire. The downside is that silver plated wire is a bit more expensive.
    The easiest way to make a coil is to use a drill. For L3 I simply use 5 mm and makes 2 turns. With L5, L6, and L7 I use 6.5 mm drill.
    Space the inductor a bit. You can click on the top photo to see my coils.

    Component support
    This project has be constructed to use standard (and easy to find) components.
    People often write to me and ask for components, PCB or kits for my projects.
    All component for 1.5W Power amplifier (Click here to download component partl313.txt).
    The kit cost 30 Euro (41 USD) and include these components:
    1 pcs
  • PCB (etched and drilled)
  • 1 pcs
  • RF transistor MRF313 (Q2)
  • 1 pcs
  • RF transistor BFG193 smd (Q1)
  • 1 pcs
  • 220uF hole mounted (C1)
  • 6 pcs
  • 100nF smd (C2, C3, C4, C5, C6, C7)
  • 2 pcs
  • 1nF smd (C8, C9)
  • 2 pcs
  • Variable capacitor smd (C19, C20)
  • 3 pcs
  • 22pF smd (C10, C11, C12)
  • 4 pcs
  • 15pF smd (C13, C14, C15, C16)
  • 2 pcs
  • 10pF smd (C17, C18)
  • 1 pcs
  • Resistors 10k smd (R5)
  • 4 pcs
  • Resistors 100 ohm smd (R1, R2, R3, R4)
  • 1 pcs
  • Inductor 27uH (L4)
  • 2 pcs
  • Inductor 10uH smd (L1, L2)
  • 4 pcs
  • Wire for Inductor (L3, L5, L6, L7)
  • Order/Question
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    Tuning and testing
    Simple testing unit which needs no power. There is only two capacitor C19 and C20 you have to tune for best performance.
    A simple way to test the amplifier is to build an extra dipole antenna and use it as a receiver.
    Take a look at the schematic at right. I use a dipole antenna as receiving antenna and the signal is then rectified by the germanium diode and the 10nF cap. An 100uA -meter will then show the signal strength. A very easy unit to build.

    I place the receiving antenna a bit away from the transmitting antenna and tune (C19 and C20) until I reach strongest reading from the 100uA meter. If you get too strong reading you can add a serial resistor to the ampermeter and if you get to low signal you can simply add a (MSA-0636 Cascadable Silicon Bipolar MMIC Amplifiers).

    Of course you can tune your system with a dummy load or wattmeter, but I prefer to tune my system with the real antenna connected. In that way I tune the true system and get the best performance.

    How far will it transmit?
    This question is very hard to answer. The transmitting distance is very dependent on the environment around you. If you live in a big city with lot of concrete and iron, the transmitter will probably reach about 400m. If you live in smaller city with more open space and not so much concrete and iron your transmitter will reach much longer distance, up to 3km. If you have very open space you will transmit up to 10km.
    One basic rule is to place the antenna at a high and open position. That will improve your transmitting distance quit a lot.
    Very ruff estimation of transmitting distances.

    Final word
    In this part, I describes the complete function of a class-C amplifier.
    I hope you enjoyed the page and feel inspired to build your own power amplifier.
    if you miss something or want to add some info, do please mail me.
    According to the law it is legal to build them, but not to use them.

    I wish you good luck with your projects and thanks for visit my page.

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    Copyright © Last modified on 22 mars 2006.