Push-Pull Output Transformers - Part III, The Final Countdown:

Discussion in 'Amateur Radio Amplifiers' started by KD2NCU, Sep 28, 2017.

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  1. KD2NCU

    KD2NCU Ham Member QRZ Page

    Some follow up calculations demonstrating the transformer action of the bifilar feed coil.

    Below is some analysis demonstrating that the bifilar coil is acting as an impedance transformer in addition to its other functions. Also below is the answer to the question, “How can it possibly be acting as a transformer if I can replace it with two separate drain/collector chokes and the circuit still functions properly?”

    First some voltage/current/power calculations to demonstrate its function as an impedance transformer.
    Below is a circuit from an application note “Microsemi Application Note 1819: A 700W Broadband Amplifier Using VRF2944”. This amplifier produces 700 Watts across a 50 ohm load using a 65 volt supply. Let’s see how this is possible. Let’s work backwards from the load.

    Load Voltage & Current
    •Given 700 Watts into a 50 ohm load, the Load Voltage will be 187.1 Volts RMS, 264.6 Volts Peak.
    •The Load Current will be 3.74 Amps RMS, 5.29 Amps Peak.

    T3 Primary Voltage & Current:
    The final output transformer is a 1:4 Guanella Transmission Line Transformer (1:2 turns ratio)
    So the voltage at the primary of the final transformer will be ½ that of the secondary and the current will be double that of the secondary.
    •Primary voltage will be 93.5 Volts RMS, 132.3 Volts Peak
    •Primary current will be 7.48 Amps RMS, 10.58 Amps Peak

    DC Supply:
    We only have a 65 volt DC supply. How are we getting 132 Volts Peak or 264 volts peak to peak across the primary of the output transformer? The answer is, as shown earlier in this thread, the bifilar feed coil is acting as a transformer with a 1:2 turns ratio or 1:4 impedance ratio. It is doubling the DC source voltage of 65 volts resulting in the 130+ volts peak across the output transformer primary winding.

    As a quick review, at the peak of Q1’s conduction cycle at full power, Q1 causes the full 65 volt supply to appear across one winding of the bifilar coil. This produces another 65 volts across the other coil. The final transformer sees the voltage across the two coils in series or twice the voltage across the individual coils. IE; at full power, the bifilar coil is doubling whatever the DC source voltage is (at full load).

    If the bifilar feed coil were not doubling the DC supply voltage, the peak voltage across the final transformer primary winding would be ½ of what it is and the output power would be 1/4th of what it is.

    Relationship Between DC Source Voltage, and Circuit Impedance.

    So clearly, the application notes that use the formula above are assuming that the peak voltage across the final transformer primary winding will be twice the DC supply voltage by one means or another. When used, the bifilar feed coil is doubling the DC supply voltage via transformer action so that the peak voltage across the final transformer primary will be 2 x the DC supply voltage.

    Another method of doubling the DC supply voltage is through the flywheel action or inductive kick provided by two separate large inductors between the DC supply and the collectors or drains.

    Why can I replace T2 with two inductors and still get the same results?

    One person reasoned that if you can replace T2 with two inductors and the circuit still works essentially the same, then T2 cannot be acting as a transformer. Here’s the flaw in that logic. The transformer action of T2 is doubling the supply voltage Vdd. The inductors also double the supply voltage.

    In a nutshell, if selected correctly, the two drain inductors provide a flywheel action very similar to the drain or collector inductors in Class C, D, E, and F amplifiers and they end up doubling the voltage across the primary of the final transformer the same way the T2 bifilar coil does.

    If you look up analysis of a single ended class B amplifier you will see that if properly set up the peak to peak output voltage will be twice the DC supply voltage because of the flywheel action of the drain inductor. In a class B or AB linear amp, the outputs of the two single ended Class B amps are connected across the primary of the final transformer so the peak to peak voltage across the final transformer primary is four times the DC supply voltage as is the case with the bifilar feed coil.

    The class B or AB push pull circuits essentially bring together two single ended amplifiers 180 degrees out of phase.


    Direct Comparison Using Collector Chokes and Bifilar Feed Coil With Identical Results
    In both cases, the peak to peak voltage on the primary of the final transformer is 4 x Vcc.
    In one case the DC supply voltage is doubled by the flywheel action of the collector chokes.
    In the other case, the DC supply voltage is doubled by the bifilar feed coil.


    Example: W6PQL 1 KW SSPA for 1.8-54 MHz – Using Bifilar Coil and Using Drain Chokes
    This is an actual example of replacing a Bifilar Feed Coil with drain chokes.
    Shown below are the drain circuits for W6PQL’s 1 KW linear amplifier before and after replacing the bifilar feed coil with drain chokes for various technical reasons. Notice that this was the only change made to the output portion of the amplifier.

    In both cases, the power output is about 1000 Watts into a 50 ohm load using a 50 Volt DC supply. See the URL above for full schematic and details.
    T3 and T4 are both Ruthroff 1:4 transformers made using coax cable.
    T3 and T4 are connected to the drain circuit and the output balun such that they form a 1:9 impedance transformer (1:3 turns ratio).

    The power, voltage, and current calculations are the same for both cases.
    In both cases, to obtain 1 KW, we need about 100 volts peak at the input to the T3/T4 transformer but we only have 50 volts DC available.
    In one case, the bifilar coil doubles the DC supply voltage via transformer action. In the other case, the DC supply voltage is doubled via the flywheel effect of the inductive chokes on the drains.

    1. T2 is clearly acting as a transformer. The transistor causes a signal voltage to appear across one coil of the transformer. T2 causes a replica of this voltage to appear across the other coil of T2. The sum of these two signal voltages appears across the primary winding of the final transformer.
    2. This transformer action allows the peak voltage across the final transformer to be twice that of the DC supply. If T2 were not providing this function, the voltage across the output transformer would be one half of the desired value and the power output would be one fourth of the desired value.
    3. Replacing T2 with two correctly sized drain or collector inductors employs inductive reactance as a “flywheel” or “inductive kick” principle to essentially double the signal voltage resulting in the peak voltage across T3 primary being twice the DC supply voltage as in the case of using a bifilar coil.
    4. Several working examples have been provided showing that the same results can be achieved using a bifilar coil or correctly sized drain chokes.
    5. The worked examples show that to obtain the stated output power, both the use of drain chokes and the use of the bifilar coil have to have the ability to cause the final transformer primary winding to see a peak voltage of twice the DC supply voltage.

    Attached Files:

  2. ON4LDY

    ON4LDY QRZ Member

    And again, no indication of how to practically calculate the number of turns to give T2 and the size of the ferrite. So bad !!
    This discussion realy goes round in circles....
  3. KD2NCU

    KD2NCU Ham Member QRZ Page

    The post above was answering someone else's questions, not yours.
    Check your personal email. I will be sending you two documents via email that answer your questions.
    1. Why is T2 smaller than T3?
    2. Design of T2.

    Let me know if you have questions.
  4. WA1GFZ

    WA1GFZ Ham Member QRZ Page

    Designing T2. look at your peak voltage, then you need to select wire to handle the DCcurrent. Wind it on a core that is running at less than 200g at the lowest operating frequency' If it runs hot you need more turns or a bigger core. The reactance of the winding needs to be at least 4 times the device output impedance.
    Lost track of this, busy hot rodding an old EFJ Valiant with 4 finals.

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