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Stereo and mono out the same jack?

Discussion in 'Electronic Design' started by Jon, Feb 14, 2004.

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  1. Jon

    Jon Guest

    I have two piezo pickups. I can handle building the preamps, but then
    I want to send the line level signals through a regular 1/4"
    instrument/phone cable. It would be really nice if it would send them
    out as stereo tip left/ring right with a TRS plug in the jack, and
    then automatically mix the two and put the mix out with a TS plug in
    the jack (so it mixes if the ring is grounded). Anyone know of a
    circuit offhand to do this? I will think about it myself as well...
    I am envisioning something similar to a cross coupled output, like
    http://www.rane.com/n124fig5.gif
    but without the inverting section at the beginning (I can invert just
    by switching polarity), and two different signals into the inputs.

    Is this the right approach? Maybe there is something blatantly
    obvious that I am not thinking of. And cross coupled outputs are
    touchy, anyway, so maybe I should not be basing it off of those if
    possible.
     
  2. N. Thornton

    N. Thornton Guest

    I cant say I followed your cross coupled idea. Care to explain?

    One simple approach would be to put dc through the groundable ring so
    that when a mono plug is inserted a relay switches the tip output to
    mono. A cleaner way would be to monitor its Rout, and when it drops to
    near zero switch your opamps to mono.

    Regards, NT
     
  3. Jon

    Jon Guest

    Hmmm... Maybe I'm not thinking about this right. I was seeing the
    cross coupled output as a way to output the same signal level if
    output to a balanced or unbalanced cable, so each is putting out
    opposite and equal signals if the cable is balanced, but if you ground
    one of the output terminals the other opamp becomes a summing amp and
    puts out two times the signal to the other output. Whether that's how
    the cross coupled output behaves or not (I will think about it some
    more...) that is the idea I was thinking of using. Grounding one of
    the outputs automatically turns the other amp into a summing amp which
    outputs the summed mono signal out one pin.
    Yeah, kind of like that, but where 'monitoring' just consists of the
    ground altering the feedbacks of the opamps so that they sum the
    signals instead.
    I should have mentioned that this is inside an instrument, so the
    smaller the better. Relays probably not a good idea. The best method
    of all would be one that only uses the preamp opamps, I guess, but
    that doesn't seem very feasible. I am basically just using the
    circuit at http://www.oz.net/~walterh/oppiezobuf.htm times 2.
     
  4. I believe that Rane circuit is a Leunig stage to be a transformerless
    line driver. There was a thread last year about that driver having
    less than desired balance. I don't believe its what you expect.
    GG
     
  5. N. Thornton

    N. Thornton Guest

    As far as I can see the cross coupler feeds L into R and R into L, if
    in opposite phase, so youd never get straight stereo out, maybe more
    like a stereo wide signal out.

    A blend switch can use near zero power: the FET that connects the 2
    channels has essentially no power consumption, and you can just feed
    an extremely small current to one jack ring plus to an opamp input,
    plus a Vref to the other opamp input, and the opa will switch state
    and turn the fet on or off.

    If you used a separate micropower opamp it would eat nothing
    noticeable extra. If smallness is the prime issue, maybe use a quad
    opamp.


    Regards, NT
     
  6. Tony

    Tony Guest

    The Rane circuit has been used many times, both in discrete form (eg using two
    "B" type SIP resistor packs, so you can vary them to set the gain) and in IC
    form (from AD/SSM at least). It does its job well, but you need to watch what
    happens with no load (the DC common mode output can easily hit the supply
    rails).

    In a simple variation on the circuit, the two circuit nodes that are shown as
    grounded can be used as a second input, eg for the "-R" signal (where +L is
    applied to the main input). This then provides balanced-in, floating
    balanced-out - like a transformer.

    But not much use for the intended application, which needs something simpler,
    eg...

    Take the R signal via a buffer and 1k resistor to the Ring terminal.
    Feed the Tip terminal with the sum of L+R-Ring. Since R-Ring is zero while Ring
    is unloaded, this reverts to the L signal (ie, it's a proper stereo signal), but
    if the Ring is grounded, the Tip gets L+R.

    Tony (remove the "_" to reply by email)
     
  7. Tony

    Tony Guest

    Sorry, edit malfunction; it should have been...
    Tony (remove the "_" to reply by email)
     
  8. Jon

    Jon Guest

    Yes that is what that circuit is. It is not very near to what I was
    thinking after all. I thought about it a bit and here is a much
    better description of the concept I originally had:


    ___ ___
    Vleft-------------|___|---o----|___|----.
    100k | 100k |
    | |
    ___ | |\ |
    .-------|___|---o----|-\ | ___
    | 100k | | >-----o----|___|----Vtip
    | | .-|+/ 1k
    | ___ | | |/
    | .--|___|---' |
    | | 100k ===
    | | GND
    | |
    | '------------------------------------.
    | |
    | ___ |
    | .----|___|----. |
    | | 10k | |
    | | | |
    | ___ | |\ | |
    Vright----o-------|___|---o----|-\ | ___ |
    10k | >-----o----|___|--o-Vring
    .-|+/ 1k
    | |/
    |
    ===
    GND

    Note that this is just an IDEA. Brainstorming. But if I figure
    correctly (and I certainly don't sometimes), connect instrument amps
    or something to both outputs, and Vleft from the left preamp, Vright
    from the right preamp. Vtip will be -(Vleft+Vright-Vright) => -Vleft,
    and Vring will just be -Vright. (Overall inversion doesn't matter.
    Only bad if one is opposite the other.) If the ring is then grounded,
    by a TS plug, the right channel op amp is still driving 1k (is that
    big enough?) to ground and Vtip will now be -(Vleft+Vright+0) =>
    -(Vleft+Vright), a 50/50 mix of the two, although twice as loud.

    Does this look right? What problems would it have? Is there a
    simpler way? Could this be merged with the preamp circuit itself, so
    there are only two opamps total?

    There is a lot I don't know about practical op-amp design. (Which I
    am making a list of, to be asked in another post...)

    Also, ideally, the mixed output would be (Vleft+Vright)/2, so a mono
    or stereo output is at the same overall level, but it's not life or
    death.

    Yeah, I was trying to get the basic idea across, of one channel
    feeding back into the other and being grounded out. See above.
    That sounds like an even better idea.
    Yes, smallness and low power consumption are most important. I don't
    understand your blend circuit though. Is the quad op-amp an
    alternative to the FET? Or both are being used together?

    So two op-amps for the preamps, and one to mix? And one as a
    comparator to turn on or off the FET, which is used to switch the
    mixed or unmixed signal? Wouldn't that need 2 FETs? I am imagining
    an FET transmission gate. This is all very fuzzy...
     
  9. I read in sci.electronics.design that Jon <>
    No, it's the way, or at least a good way, to do what you want. It's
    virtually the same as the solution proposed by Tony Roe Whether 1 k to
    ground is OK or not depends on the op-amp. If you use NE5534 (or any one
    of several more modern types) it's OK, but 741 and 324 aren't so keen on
    1 k (for different reasons). I wouldn't use a 324 for this job anyway.
     
  10. N. Thornton

    N. Thornton Guest


    OK, I like your circuit. The only slight disadvantage is that it will
    double its distortion level when running in mono, but thats very small
    point if you use a reasonable quality opamp. The other thing is if you
    have a patchy connection on the shorted ring this would feed into your
    mono sound channel.

    What I had in mind, lets see...


    |\
    left --|-\ ___
    | >---|___|----+------- Vtip
    --|+/ 5k |
    |/ |_|
    -------|_
    |\ | |
    right -|-\ ___ |
    | >---|___|----+------- Vring
    -|+/ 5k
    |/


    There we go. With the fet off it works in stereo, with the fet on it
    outputs mono. To switch the fet youd need another opamp which detects
    the dc level on the jack ring, plus an RZR etc to put a small dc
    offset onto the ring. When a mono plug is inserted, this dc offset is
    shorted and the opamp switches the fet on, resulting in mono ouput.

    The advantage is avoiding distortion, and avoiding possible problems
    with dirty ring contacts. But it takes an opamp and a fet. If you
    already have 2 opamps in place though this would be less parts than
    another 2 opamp cross coupled circuit. If youre designing the whole
    lot I'd probably prefer your circuit for lower parts count.


    Regards, NT
     
  11. Jon

    Jon Guest

    Ok. When you say distortion you are talking about the overall signal
    level and clipping? I don't think that is a problem. I intend to
    drop the signal down a bit anyway so that it is more like a guitar
    level signal, and so it can handle the battery dropping in voltage
    somewhat. (That is how it will behave, right? As the 9V battery dies
    it will just reduce the possible output swing of the op-amp (to a
    point, of course)?) The shoddy ring connection is a valid concern,
    but wouldn't it do the same thing with the FET circuit?

    That's a valid way of mixing? Whatever happened to "use a summing amp
    to sum voltages! never connect two outputs together!"? clearly
    something else I need to learn...

    So there are some different possibilities as I see it:

    1. Build my type of circuit with the preamp included so it is just two
    opamps total. But the current preamp circuit requires high input
    impedances (see below), to match the piezo, and my above switching
    circuit has 100k. If it is possible to use 10M or so resistors for
    the switching circuit, I guess that would work. (but I always see
    op-amp circuits with resistors from 1k to 100k and never more or less,
    so it must be bad to use those values, right?)

    2. Use the preamps and switching circuit separately, which requires 4
    op-amps and either two duals or one quad chip.

    3. Use the FET switching circuit, which requires 3 op-amps, and hence,
    two duals or one quad chip, same as 2.

    4. Can the FET be triggered with the dc voltage directly, without the
    op-amp? It seems like it could be, with the right voltages in the
    right places and the right FET.

    Here is my preamp circuit, exactly. I intend to drive either guitar
    amps, around 50k inputs, i believe, or sound cards, about 600ohm
    inputs (and certainly feel free to tell me if I did something wrong or
    need something else to make it better.):


    ___ |\
    .---|___|----o------------|+\ #| ___
    | 6.6M | | >---o-----#|----|___|----o
    | | .--|-/ | #| 1k
    --- Piezo .-. | |/ |
    [===]6 Vpp | |2.2M | | 4.7 uF
    --- at max | | '---------' Vout
    | '-'
    | |
    | |
    '------------o----------------------------------------o
    |
    | ---(Top section x2)---
    |
    9V |
    + |
    | |
    o------------|-------o-------.
    | | | |
    .-. | | |
    | |4.7k | | |
    | | | | |
    '-' | |\| |\|
    | | |-\ |-\
    o---------o--' | > | >
    | | |+/ |+/
    .-. | |/| |/|
    | |4.7k --- | |
    | | --- | |
    '-' | 103 | |
    | | | |
    o---------o----------o-------'
    |
    ===
    GND

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    I am using a tl062 right now because that is what i had in my parts
    box. I need to order an unrelated chip (and probably this circuit's
    output jack) from a catalog, so i should get some other things while
    I'm at it. ne5532s are pretty good quality? I see them used in audio
    a lot. Are the more expensive opa134 burr brown type chips really
    worth it?

    Oh wait... *Looks through box* I guess i do have some 5532s, and
    072s, 741s, and another 062, and one each lmc60 and lf412. I remember
    having problems with the lf412 and lmc60 in the older version of the
    circuit. Which would you use?
     
  12. I read in sci.electronics.design that Jon <>
    wrote (in <>) about
    'Stereo and mono out the same jack?', on Mon, 16 Feb 2004:

    [big snip]
    Look at the 5 kohm resistors: they prevent the two outputs 'seeing' each
    other.
    Yes, that's what I recommend.
    It can be. Even your 2.2 Mohm is pushing your luck on offset current.
    Well, now, you've got 6.6 Mohm, which isn't a standard value (6.8 Mohm
    is). In that position, you can have as many megohms as you like. But the
    2.2 Mohm is too big. The way to overcome that is to use a 'bootstrap'
    connection. You apply the output signal of the op-amp, or a fraction of
    it, to the ground end of the resistor, which greatly increases its
    effective value at signal frequencies.
    .---|___|----o------------|+\ #| ___
    [rest snipped]

    The 220 k in series with the - input is to keep the impedances at the
    two inputs nearly equal, minimising current offset. The 220 k at the +
    input is bootstrapped with 0.9 times the output voltage, which is also
    0.9 times the input voltage, so appears at signal frequencies to be 2.2
    Mohm.
    It's a rather low-power device, not capable of driving loads below
    several kohms.
    Yes, but they are not unity-gain stable; you can't use them as
    followers.
    IMHO, not for your application.
    TL072s are quite good. I don't know the LMC60 at all. The LF 412 is
    good, and, although the data sheet doesn't seem to say so explicitly, is
    unity-gain stable, as shown by its use as a follower in the 'sample and
    hold' application circuit.
     
  13. Jon

    Jon Guest

    Ok. First, what problems would the offset current cause?

    Second, if i am combining the two circuits they will need to be
    inverting amplifiers so i can sum them and also to get -Vright.
    That's what i meant by input impedance. The inverting amps as i drew
    them had 100k inputs, which is not high enough for the piezos.

    as described here: http://www.amp.com/products/technology/articles/interface.stm

    so i need an inverting (summing) amp configuration that will interface
    with piezo. does such a thing exist? it looks like the inverting amp
    can be configured as in figure 9 on that site, but can it sum?

    here is the theoretical combined circuit with super big resistors,
    which i was talking about when i said that resistors this big are not
    "allowed":


    ___ ___
    .-----o---------|___|--o----|___|--.
    | | 10M | 10M |
    | .-.10M | |
    --- | | ___ | |\ | 4.7uF
    |===| | | .----|___|--o----|-\ | ___ #| Tip
    --- '-' | 10M | | >---o--|___|---#|-----.
    | | | | .-|+/ 1k #| |
    | | | ___ | | |/ |
    === === | .-|___|--' | |
    GND GND | | 10M === |__ .-.
    | | GND V | |
    | '-----------------------------------. | |
    | ___ | ____A | |
    | .----|___|--. | | '-'
    | | 10M | | | |
    | | | | | ===
    | ___ | |\ | 4.7uF | | GND
    .----------o----|___|--o----|-\ | ___ #| | |
    | 10M | >---o--|___|---#|--o--'
    | .-|+/ 1k #| Ring
    --- | |/
    |===| |
    --- ===
    | GND
    |
    ===
    GND


    maybe this can be bootstrapped, too? i don't see how...
     
  14. I read in sci.electronics.design that Jon <>
    If the d.c. impedances between the inputs and ground are not equal, both
    bias and offset currents will cause d.c. offset voltage at the output,
    perhaps affecting coupling to the next stage and restricting output
    voltage swing.
    Ah, well, you've now changed the question.
    Yes: the normal summing amp IS inverting. Indeed, the virtual earth
    point (the - input) is also called 'the summing junction'.

    Figure 9 shows a 'capacitive feedback' inverter, or 'charge amplifier'.
    It is particularly useful for capacitive sources, and has much better
    noise performance than a solution using high-value resistors.
    You wouldn't NEED to bootstrap 10 Mohm resistors. But you would have big
    problems with d.c. offset.
     
  15. N. Thornton

    N. Thornton Guest

    no, distortion, THD. No amp is perfect and the cross coupled circuit
    will increase it. But I dont think you need worry about it, it will
    still behave well.

    yes, down to a point. As long as the opamps still in spec, and the
    battery can deliver the required i_out.

    no, a small cap would be used to prevent switching at a.f.


    easiest option to build.
    easy conceptually
    yes, the only downside is going to be a large voltage thump when you
    plug stuff in, so that wouldnt be great practice, but it should work.
    You can do better though.

    John has covered the impedances. I'd add something else: a 1M resistor
    from output terminal to ground. It bleeds the dc off your output cap,
    so you dont get an ear bashing woofer bending thump when you plug
    things in.


    Regards, NT
     
  16. Jon

    Jon Guest

    yeah. that is a negligible problem for this circuit.
    ok. that's what i thought.
    ok. i will add that. in fact, i will use that for other appropriate
    circuits from now on...


    John Woodgate wrote in message
    ah. it's not going to be a lot, though, is it? and putting an equal
    resistor in the feedback loop (of the voltage follower) would make it
    somewhat better? i don't think a small dc offset is going to affect
    this type of audio signal in an important way. It gets a little bit
    of slowly changing DC offsets from the piezo already.
    of course. :)
    well, piezo sensors are modeled as purely capacitive sources, so that
    would theoretically be the best, however, i don't see a summing
    function being built with one. especially one that will sum from two
    piezos *and* an op-amp output.
    Is there a better way to get the same functionality and the same input
    impedances?

    what's that you say? a toggle switch? hmm... that would solve the
    dirty ring problem too. and could also have a 'panic' setting to
    output the piezos directly into whatever, in case the battery dies. i
    used to play it like that. it doesn't sound great, but it would be
    useful in a tight spot. hmm...
     
  17. I read in sci.electronics.design that Jon <>

    It can be: it depends on a number of factors, so the simplest thing is
    not to let it happen.
    A lot better, if the resistor is hundreds of kohms.
    It might NOT be small.
    I don't see how: the piezo is a capacitor.
    No problem whatsoever! A conventional inverting summing amplifier has a
    resistor in series with each input and a feedback resistor. A summing
    amplifier for capacitive sources has capacitors at the inputs and a
    feedback capacitor. You make the op-amp output look like a capacitor by
    simply putting a capacitor in series with it (you may need 100 ohms as
    well, to improve op-amp stability). For the same sensitivity, use the
    same value capacitance as that of your piezos. You can adjust the
    sensitivity by varying the capacitor value.
     
  18. Jon

    Jon Guest

    "slowly changing" being the key words. if i plug the piezo directly
    into the op amp input, noninverting, it will create a dc offset when a
    constant pressure is applied. sure, it leaks through the input
    impedance slowly, and drops back to zero, so it is technically just
    very low frequency, but it stays around for quite a while, and since
    this is under a violin bridge, which has different constant pressures
    at different times, it was initially causing problems because the
    signal would float up and down past the opamp's rails, without the 10M
    resistor to ground.
    ooh! that is how i will do it, then. i have never tried the charge
    amplifier before but i will try it out soon and report on my success.
    since i don't know the piezo's capacitance, just play with the values
    until i get a voltage i like? (could i reliably measure it with an
    LCR meter?) and the feedback resistor will now determine how quickly
    the dc offset mentioned above drops off?
     
  19. I read in sci.electronics.design that Jon <>
    Yes. Don't make a noise, though, because it's a microphone, of course.
    Probably best measured when detached from the instrument.
    Yes. The ratio of feedback cap to piezo capacitance determines the gain.
    Then the low-frequency -3 dB frequency is given by 1/[2 pi{(feedback cap
    x gain) + piezo capacitance} x feedback R}].

    If you have Spice, it's worth simulating your circuit after you've
    designed it, because it does interesting things at high frequencies
    (usually well above 20 kHz).
     
  20. N. Thornton

    N. Thornton Guest

    Suggest setting your low end response to whatever the minimum a guitar
    can put out is, so the dc offset cant float around.


    Regards, NT
     
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