wheat stone bridge and peak detectors, not getting the results i thought

Discussion in 'Homework Help' started by ninjaman, Jan 31, 2015.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    hello

    I have a circuit
    upload_2015-1-31_19-5-13.png

    I thought I would be getting nothing but DC from the peak detectors. I have followed a schematic of a manufactured antenna analyser. when I use the oscilloscopes I get one DC and one AC. R5 is the unknown impedance. this is in place of an antenna. I have used the watt meter to find out if the circuit works. it seems to be ok. but to measure the voltages I want DC from the peak detectors to go into an mcp6002 op amp. this is a general op amp with 1MHz bandwidth. so it would have to be DC. high frequency (3-30MHz) would be going into the wheatstone bridge, antenna, from the antenna to the peak detectors that would put a DC signal into the amplifiers, then into an arduino to do some calculations and show results.
    I get two DC and two AC. not sure what im doing wrong here. the rest of the circuit is an non-inverting amp. though I didn't think this would do much. each op amp circuit is exactly the same, connected after the capacitors shown. each peak detector has an amplifier.
    so yeah, how do I get DC instead of AC?

    any help would be great!


    although I have just tried adding some more of the circuit that includes another capacitor and that seems to work?
     
  2. crutschow

    Expert

    Mar 14, 2008
    13,019
    3,235
    So show us what "some more of the circuit" you added.

    What is the signal amplitude?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    I don't see the point of including D2 & C2.
    If the diodes were ideal then one would see virtually AC / ripple free DC voltages at C3 & C4 (as marked) positive terminals. With the antenna load exactly 50 ohms the two voltages should be identical with the bridge in balance.
    What about the voltage at D1 cathode if it was also ideal? The difference between the balanced bridge centers would be zero - if the antenna load is exactly 50 ohms. Hence D1 would not be conducting at any point. So D1 cathode must be at essentially the same potential as the LHS bridge center and therefore appear as a pure AC signal with zero DC offset.
     
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