Class A PA - question

Discussion in 'General Electronics Chat' started by RFeng, Jun 20, 2012.

  1. RFeng

    Thread Starter New Member

    May 30, 2012

    I'm trying to understand the device operation.

    It is said that its ideal-case efficiency is 50%, i.e. the RF power delivered to the load is half the DC power wasted on the transistor.

    My question is, the "load" here is the pull-up resistor connected to the collector? or is it actually the load itself? (which is not shown in the scheme below).

    If it's the load itself, how can we promise that the entire RF power will be delivered to the load and will not go to the transistor?

    Thank you.

  2. chuckey

    Well-Known Member

    Jun 4, 2007
    For a RF PA Rl is actually a tuned circuit. Also there is no Re. For class A, let the transistor take Iq standing current, so the minimum current it can take is 0 A. So the negative peak RF current is Iq, so the positive RF peak is 2Iq. so 2.8 X Irf = 2Iq, or the rf output current is .7 Iq, the same calculation can be made for the voltage, so the output power is .7 Iq X .7 Vc= .5 DC power in.
    If you use a resistor this will drop voltage across it so the output voltage will be lower and it will limit the current as well. So a resistor load is only used for voltage amplifiers NOT power amplifiers.
    #12 likes this.
  3. RFeng

    Thread Starter New Member

    May 30, 2012
    Hi, Thanks mate!

    I'm trying to understand your math.

    You mean that I_RF(min) = 0A (i.e. RF negative peak = 0A),
    and I_RF(max) = 2IQ?
    I mean, doesn't I_RF move around IQ? i.e. 0 < I_RF < 2IQ ?

    How did you conclude that 2.8 X I_RF = 2IQ?

    Well, in the article I'm reading, it is said that R_LOAD = [ V_RF(max) - V_RF(min) ] / I_RF(max) .
    Any idea how they reached it?

    Thank you very much.
  4. RFeng

    Thread Starter New Member

    May 30, 2012
    Perhaps you missed my reply :)