RF Sampling circuit HELP!!

Discussion in 'Wireless & RF Design' started by scottshannon, Jul 6, 2012.

  1. scottshannon

    Thread Starter New Member

    Aug 4, 2009
    I would appreciate some help trying to understand this circuit:

    I am looking at a RF Sampling circuit on youtube. Here is the video:


    I was curious how he calculated the power dissipation in the resistors.

    In the circuit there is a 47KΩ & 690Ω resistor. The author told me the following:

    "The assumption made with this circuit is that it is inline between the transmitter and a properly tuned 50 ohm load/antenna. Therefore, the 100W is dissipated in the 50ohm load. That means the voltage on the line is 70.7Vrms. Nearly all of this appears across the 47K resistor, dissipating about an 1/8th of a watt. Then, I used a 4x safety factor."

    Would someone who knows more than I know be good enough to let me ask them some questions about this circuit? I would like to calculate the power dissipated in the resistors. It is driving me crazy!
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    It's already half explained.
    To recap...

    At 100W carrier power into a standard 50 ohms load the RMS voltage would be 70.7V.

    With 70.7V across the 47k plus 690 ohm resistor the power dissipation is V^2/R or 5000/47690=0.1048W. Most of this will be dissipated in the 47k resistor. To work it out exactly use the voltage divider rule to find the voltage across individual resistors.

    For example..

    V_47k=(47000/47690)*70.71=69.69V RMS.
  3. KL7AJ

    Senior Member

    Nov 4, 2008
    Power dissipated in a resistor is simply E squared over R.