Connect with us

Differential to single ended

Discussion in 'Electronic Design' started by Rene Tschaggelar, Dec 31, 2003.

Scroll to continue with content
  1. I should convert a differential signal to a single ended one.
    This is usually done with an OpAmp in a subtracting setup.
    For speed reasons I only have a current feedback amp.
    That should work, shouldn't it ? With the same 4 resistors ?

    I somehow lost a paper from { analog ? linear ? national ? }
    named "me and the current feedback amp" or similar. And these
    parts are too tiny and too fast to just give them a try on
    a breadboard.

  2. Mac

    Mac Guest

    There are some pretty fast voltage feedback amps around nowadays, too.
    I've never done what you are trying to do. But I have used current
    feedback op-amps before and I have read their datasheets.

    The thing to watch out for with a current feedback amp is that the
    non-inverting input is NOT high impedance. Also, you don't have as much
    flexibility in setting the feedback resistor. Whatever value or values the
    datasheet recommends is what you should use.

    So, if you can live with the impedance issue, and the feedback resistor
    value is OK, it should work. Be sure to check the common-mode input range
    on the op-amp, too.

    You can probably breadboard it on perfboard. If the amp is sensitive to
    capacitive loading, you may want to put a series resistor immediately
    after the output, and before the load. Also, make the feedback path as
    simple and direct as possible.

    Good luck.

  3. Ben Bradley

    Ben Bradley Guest

    I dunno if this is what you lost (it's nothing like the title you
    gave), but it does address the issue of differential inputs using CFB
  4. Mac

    Mac Guest

    OOPS! I meant to say that the inverting input is NOT high impedance. The
    non-inverting input typically is high-impedance.


  5. I have little experience with current feedback opamps, but I think the
    two amp subtractor will allow a high impedance input with the resistor
    network values selected to be optimum for the inverting low impedance
    and amplifier speed. Then you can terminate the high impedance inputs
    as you desire. See second example on page 16 of:

    I haven't browsed through them, but there are several current feedback
    opamp application notes about half way down on:,2175,967,00.html
  6. I'll have to either do a simulation or an experiment to figure out
    the useability for the intended purpose.

  7. Thanks,
    The later reference was about like the one I remember.
    I'll have a look at the documents.

  8. Thanks all for the suggestions.
    Beside doing some reading, I guess I'll give it a try.
    I happen to have two amplifiers both in the MSOP8 case.
    One is a voltage feedback amp, the THS4275, the other is
    a current feedback amp, the THS3001.
    While the VFA has a rather low slewrate of 1V/ns causing
    a limited full power bandwidth, the CFA is absolutely stunning,
    its slewrate is supply dependent, but somewhat higher.
    A bit more than 1V swing would be appreciated.
    The noise voltages are 3nV/rtHz for the VFA and 1.6nV/rtHz
    for the CFA.
    And the CFA tends to be less prone to oscillations.

  9. If you come across specific info that relates to using current
    feedback opamps as subtractors, please come back and educate me.
  10. Bill Sloman

    Bill Sloman Guest

    I've done this - some 15 years ago - with Comlinear parts, and the
    National Semiconductor equivalents, which used somewhat higher
    resistance feedback resitor values, which saved a useful amount of

    Both parts are long obsolete, so I can't usefully give you part
    numbers, even if I could remember them.

    In the second go-around, we used a transmission line transformer (
    wound with a twisted pair of 36 swg enamelled transformer wire) to
    push up the common mode rejection at high frequencies. Nobody was more
    surprised than I was when the damn things worked straight off the
    drawing board, but I managed to keep a straight face.

    I think I might have schematics at home in Nijmegen - e-mail me if you
    want more detail.
  11. I read in that Bill Sloman <>
    I wouldn't bother, Bill. The weather is b. awful here, and we keep
    swapping the rain and fog back and forth with the Netherlands.

    Marginally happy new year, though. (;-)

  12. I found an article amongst the material mentioned that was
    convcerned about subtractors. As I understand it, the same
    4 resistors do the subtract with a fine detail though.
    The '+' input is highimpedance and the '-' input tries to follow
    the '+' input, but the '-' input is a low impedance current input
    Whatever current appears there due to different voltages or so,
    is amplified and appears at the output. So in a closed loop
    such as a subtractor, the input current at the '-' input is kept
    at zero, statically. Again the feedback resistor makes the
    speed, so it should be rather low, as suggested in the datasheet,
    in the few hundred ohms range.

    Actually :
    AN-597, OA-07 of NationalSemi mentione something about the subject
    OA-25 has a model of the CFA.

  13. Fred Bloggs

    Fred Bloggs Guest

    That TI app note is their typically low-quality stuff, and the Fig 6 for
    the differential amplifier shows the wrong equation, and why not, this
    is in keeping with the incoherence of the rest of the note, using
    undefined variables and loop equations that make no sense. The "author"
    must be a cooperative student on training assignment- confused individual.
    The CFB differential amplifier is exactly the same as in the voltage
    feedback case, the only difference is that you are constrained in the
    choice of feedback resistor from OUT to IN(-) to a small range of
    recommended values -selected for performance and stability purposes. I
    suppose that some confusion can arise because the IN(-) is actually a
    buffered output, BUT, since VOUT is a very large transimpedance gain
    function of the IN(-) output current, a negative feedback configuration
    using voltage-to-current converting resistors, will drive that IN(-)
    current to zero, AND, since the voltage at IN(-) is a unity gain
    buffered version of the voltage at IN(+), you have zero net differential
    input voltage. This means that for frequencies where the transimpedance
    gain can be taken as infinitely large, the circuit equations are
    identical to those of the ideal opamp:no input currents, zero input
    differential voltage, and zero output impedance.
  14. Mac

    Mac Guest

    Actually, I think it will work. The input impedance thing is largely moot
    because of feedback. The only catch is that the feedback resistor has to
    be as specified in the datasheet. This will probably force other resistors
    to particular values, depending on the gain you are looking for.

    If you are looking for unity-gain, then all four resistors will be the
    same as the feedback one. The input impedance for differential signals
    will be 2R. The input impedance for common-mode signals will be R. If that
    is acceptable, then I don't think you have any problem.


    If you
    want, draw the circuit, post it somewhere (you can post in here in
    ascii-art) and I'll be happy to look at it.
  15. Fred Bloggs

    Fred Bloggs Guest

    Fred Bloggs wrote:

    If wideband differential to single-ended conversion is required with
    minimal input loading, then you have this option in CFB- trim the R's in-:
    Please view in a fixed-width font such as Courier.

    R R R R
    | | | | |
    --- | | | |
    COM | |\ | | |\ |
    +--|-\ | +--|-\ |
    | >-------+ | >------+-> 2x(V2-V1)
    +--|+/ +--|+/
    | |/ | |/
    | |
    | |
    V1 >--------+ |
    V2 >--------------------------------+

  16. Why would you use that nastly asymmetrical configuration instead of
    the below?

    Vo = (Vinv- Vnon-inv) * 2 * Rf/Rg
    V(inv) |\|
    | >--+ _R_
    +-|-/ | +-|___|-+
    | |/| | | |
    | | | |
    | ___ | _R_ | |
    +-|___|-+-|___|----| |
    | Rf | |\| |
    .-. +--|-\ | Vo
    | | | >-+---o
    Rg | | +--|+/
    '-' Rf | |/|
    | ___ R__ |
    | | |
    | |\| | .-.
    +-|-\ | | |
    | >--+ | | R
    o---------|+/ '-'
    V(non-inv) |/| |

    Best regards,
    Spehro Pefhany
  17. I read in that Spehro Pefhany <[email protected]
    Because the simpler circuit is quite satisfactory, AIUI.
  18. Fred Bloggs

    Fred Bloggs Guest

    Actually there is a very elegant way to make gain adjust symmetrical for
    that circuit-and my apologies to JP who already pointed this ckt out on
    p16 of National AN-31 OpAmp circuit collection:

    Please view in a fixed-width font such as Courier.

    +----------/\/\ --------+
    | / |
    | |
    R | R R | R
    | | | | |
    --- | | | |
    COM | |\ | | |\ |
    +--|-\ | +--|-\ |
    | >-------+ | >------+-> Gx(V2-V1)
    +--|+/ +--|+/
    | |/ | |/
    | |
    | |
    V1 >--------+ |
    V2 >--------------------------------+

  19. Thanks both,

    I wasn't specific about the application.
    The application is an output amplifier for a 300MHz clocked DDS,
    with an output signal up to 150MHz or a bit less. It has
    differential outputs, great for differential input mixers,
    but less usefull for something else.
    So the only choice to go single ended is the 4 resistor
    subtractor with a damn fast amp such as the THS3001 CFA with
    a BW of 420MHz(@gain=1), slewrate 6.5V/ns and 100mA output
    There are very few parts available and building my own is
    not very rewarding either.

Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day