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Impedance matching...

Discussion in 'Electronic Design' started by Chris Husser, Jul 17, 2003.

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  1. Chris Husser

    Chris Husser Guest

    I'm not sure how I would do this, but I need to take a 5 volt power
    source and drop it down to 0.7 volts and match the impedance to 75
  2. That sounds like a directive from the PHB.
  3. A E

    A E Guest

    That's a strange thing to ask for. Are we talking 5V DC here? That's my usual
    assumption when talking 5V power. Do you need .7 volts open load, or looking
    into a 75 ohm load?
    And what kind of specs do you need for accuracy?
    I'd say swing two silicon diodes (1N4148 style) with a 200R resistor in series,
    you get 1.4V at the junction of the diode and resistor. Now, take a 75 ohm
    resistor in series with the load and when you attach it to the load you'll get
    0.7V across the load, for a power dissipation of 7mW in each resistor. Slap a
    100nF cap across the diodes.
    There you go. Not the most efficient or elegant thing, but cheap.
    | 75R
    +----V^V^V^---- to LOAD
    +--- cap to ground
    V two forward biased diodes

    Oh dear that looks like crap. But you get the idea. Get rid of one of the diodes
    if you want 0.7V open load. Play with the type of diode and read the datasheet
    to figure out the resistor value for getting closer to 0.7V. But this should get
    you started, and should be pretty close to .7v at the 75 ohm load.
  4. Chris Husser

    Chris Husser Guest

    I'm sending the signal to a vga monitor, which has to be terminated at
    75 ohms.
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