Help in understanding bypassing capacitor array.

Discussion in 'General Electronics Chat' started by moot, Jan 6, 2010.

  1. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    Hopefully this should be a quick question.

    I'm looking at a schematic for a lock-in amplifier (hat-tip: JILA), found here:

    http://jila.colorado.edu/bec/BEC_for_everyone/lockincircuit.htm

    Question: Do I assemble just one of the bypassing capacitor arrays (lower left corner) and then string it to all of the op-amps? Practically speaking, this seems problematic. I'm used to bypassing with a 0.1uF ceramic cap with short leads from power pins to ground on every IC. Here, it would seem that I would need lots of long wires (I'm working on a breadboard).

    Or...is this just something that sits in distant corner of my breadboard and filters out high-freq noise from the +/- 15V power supply to ground? The op-amps powered by the rails now reap the benefits?

    I can do without the Lumberg connector + Molex connector + Power LED, right?

    Thanks!
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The rail bypass caps were simply put in one place to reduce clutter in the rest of the schematic.

    The bypass caps actually belong right next to the opamps.
    U10 gets C100, C101
    U40 gets C400, C401
    etc.

    Sure, you can do without the connectors and power LED - but it might be nice to know if your supply rails are working without having to resort to a DMM to check them.
     
  3. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    Thanks!

    Let me make sure I understand, though. I redrew the bypass capacitors and attached it to this post. On each op-amp, I've connected the V+ and V- pins to their respective rails. I then connect the pins to ground with capacitors depending on that particular op-amp, right? So it's no different than what I usually do. It's just shorthand.

    (What's the deal with C1000 and C1001? Are they supposed to go with U10 or J9?..)

    (Also, I'm not going to use the power LED and connectors, since I don't have those parts lying around and I need to begin testing this circuit this afternoon.)
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    On your breadboard, try to keep the leads short. Having long leads means more inductance. This can result in resonant LC circuits that will give you odd problems.

    Leave C1000 and C1001 where they are.
     
  5. moot

    Thread Starter Member

    Sep 20, 2009
    46
    1
    Alright, got it all sorted out.

    Thanks for the help!
     
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