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Bipolar amp advice

Discussion in 'Radio and Communications' started by Mosaic, Oct 23, 2016.

  1. Mosaic

    Mosaic Well-Known Member

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    Hi all:
    Since RF power is (Vpk)^2/100 for a 50 ohm system.
    How does the (88-108Mhz) schematic attached manage 25 W output with just 28VDC supplied? bipolar amp.png
     
  2. JimB

    JimB Super Moderator Most Helpful Member

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    Without doing any calculations, my best guess is that the 2T inductor (and microstrip) in conjunction with C16 and C17 make for a nice impedance transformer.
    C18 thru C20, and the two 5T inductors make a "half wave" low pass filter, so the 50Ohm of the load (antenna) is presented to C17. The 50Ohm is than transformed to something lower by C17, C16 and the inductors so that the available voltage at the collector of Q1 is operating on a lower impedance load.

    JimB
     
  3. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    What mode do you think the amp is in? Class A, Class B, Class C?
     
  4. dave

    Dave New Member

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  5. Mosaic

    Mosaic Well-Known Member

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    It's a class C RF power amp.
     
  6. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    I asked about Class C because your power formula is not for class C. I think you used the formula from another type of amplifier.

    With your 24V supply you will have a collector voltage of 40 to 60 volts peak, depending on how you bias the amplifier.
     
  7. Mosaic

    Mosaic Well-Known Member

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    Can you clarify that 40-60Volts for me please?
     
  8. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    There is a inductor from supply to transistor collector. We can not have DC across an inductor. So the average collector voltage is close to the supply voltage.

    Another way to say it: If you turned the transistor on hard for 50% of the time so that 24 volts is across the inductor then for the other half of the time you should have a average voltage of about 48 volts. (think boost PWM power supply)

    Picture: Using a 13V supply. The voltage on the collector is not a sign wave but 1/2 sign. The filter fills in the rest of the sign wave. The transistor is on for the flat bottom time in the blue trace. This puts 13 volts across the inductor for more than 50% of the time. So the inductor must fly up above supply for the same amount of volt seconds. This easy causes the collector voltage to get to 2x the supply and 3x is possible.
    upload_2016-10-23_21-19-17.png
     
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  9. Mosaic

    Mosaic Well-Known Member

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    Kick back when the tranny switches open?
     
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  10. Nigel Goodwin

    Nigel Goodwin Super Moderator Most Helpful Member

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    'Sort of' - it's a class C amp, so it provides pulse to the tuned circuit, which then 'rings' - it's rather like a small hammer hitting a bell.

    And as already mentioned, the tuned circuits provide impedance matching, so power output isn't related to supply voltage.
     
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  11. Mosaic

    Mosaic Well-Known Member

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    Would it be reasonable to estimate about 70% efficiency for an RF CW Class C amp and then estimate the power output relative to the average power consumed?
    I found a clear explanation here that clarifies this thread for me:
    https://en.wikipedia.org/wiki/Amplifier#Class_C
     

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