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Thread: 15kpps scanner signal, what are the values?

  1. #21
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    Bump. Still don't get what I need to connect to the negative pin (or positive?) of the output +-5V connector that goes to the scanner in the diagram above.

  2. #22
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    I believe there is a problem with the circuit that Andrew posted above. Granted, I'm not an electrical engineer (and he most definitely is), so I could be mistaken, but I think his circuit will only act as a buffer. Andrew may have simply misunderstood what you wanted to do, but if you input 0 to +5 VDC to his circuit, it will output the exact same voltage. (It's configured as a unity gain-follower, if I'm reading it correctly.) EDIT: My bad - That should have been an inverting gain follower... The output is the inverse of the input.

    I think you said you want to use a 0 to + 5 VDC input signal (thus, a single-ended input) to control a set of scanners. And the scanner amps are expecting a differential input. With that in mind, the following circuit should do exactly that. I admit that there are probably more clever ways to accomplish this task with fewer components, but I've built examples of this circuit before and played with it several times over the years, so I know it works.

    Since you already had the fixed 5 volt regulator, I just built a voltage divider to get the 2.5 volt reference. The pair of 200 ohm resistors should be higher tolerance if possible. (Ideally 1%, so Red, Black, Black, Black, Brown is the code you want) 1/4 watt is fine. And obviously, you'll need to build two of these circuits (one for X and one for Y), although you don't need to re-do the voltage regulator. You can use the same 2.5 volt reference for both X and Y. Also, pay particular attention to the feedback resistor on the second op-amp (in the middle of the schematic). It's the only one that is not a 10 K resistor. It needs to be 20 K. Otherwise you'll only have +/- 2.5 volt differential output instead of the +/- 5 volt that you want.

    To connect your input to this circuit, simply connect the positive signal lead to the 10 K resistor on the inverting input to the first op-amp. The negative (or ground) lead from your signal source should be connected to ground on the +/- 12 VDC power supply that is powering everything.

    Note that this circuit does not contain any protection against over-voltage... If your input exceeds the 0-5 volt range, your output will go beyond the normal +/- 5 Volt differential signal that the scanners are designed for. If you need to clip the outputs to protect downstream components, or if you need size and offset adjustments, then I can post the more complicated version of this circuit if you want. But this should get you started.

    Sorry for the large image file!

    Last edited by buffo; 06-17-2018 at 11:30.

  3. #23
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    Quote Originally Posted by buffo View Post
    I believe there is a problem with the circuit that Andrew posted above. Granted, I'm not an electrical engineer (and he most definitely is), so I could be mistaken, but I think his circuit will only act as a buffer. Andrew may have simply misunderstood what you wanted to do, but if you input 0 to +5 VDC to his circuit, it will output the exact same voltage. (It's configured as a unity gain-follower, if I'm reading it correctly.)
    It's actually configured as a 2x buffer with a -5V offset. If you replace the +5v connection with Gnd you'd have a regular 2x buffer (in a non-inverting configuration, gain is resistor ratio+1), but by moving that reference point the output is offset an equal amount in the opposite direction. So it converts 0 to +5V to -5V to +5V.

    I'm afraid your circuit won't quite work. The first stage inverts the signal about the 2.5V ref (so it changes 0 to +5v to +5v to 0), then you've got a -2x gain stage converting that to 0 to -10V, then with the complementary output you've got 0 to +10V. So the differential output winds up being 0 to -20V. If you change your first stage to the one that I showed, replace that 20k resistor with another 10k (since the input now has 2x gain), and swap the X+/X- outputs (because the input is no longer inverting) it will properly convert a single-ended 0-5V signal to +/-10V differential. (You could also use a 4-resistor diff amp configuration for the first stage, but that just needs more resistors and a +2.5V ref instead of a +5v ref.)

    So with our powers combined, we get the below:

    Click image for larger version. 

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  4. #24
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    Question Whoa... Now I'm confused!

    Quote Originally Posted by aberry View Post
    It's actually configured as a 2x buffer with a -5V offset.
    Well, that's what I was expecting. But just to be sure I plugged the values into a couple different op-amp calculators before I posted, and with 5 volts on the inverting input, every one of the calculators I tried produced exactly 0 to +5 volts on the output for a 0 to +5 volt input.

    Wait - actually, that's not right. It *does* invert the voltage... So +5 input gives zero out, and 0 input gives +5 out. I just edited my initial post to correct that. But even so, I don't see how you're getting 2X gain when the input and feedback resistors are the same value... What am I missing here?

    If you replace the +5v connection with Gnd you'd have a regular 2x buffer (in a non-inverting configuration, gain is resistor ratio+1), but by moving that reference point the output is offset an equal amount in the opposite direction. So it converts 0 to +5V to -5V to +5V.
    Now I'm really confused. Can we back up a bit?

    The first op amp is comparing the fixed voltage (+5) on the inverting input with the signal on the non-inverting input, right?

    If so, wouldn't the gain be equal to the ratio of the the feedback resistor (which goes to the inverting input) and the input resistor (on the non-inverting input)? So for a zero volt input, compared to +5 volts reference, that's +5 volts difference, times a gain of 1 (10K/10K), but the output is inverted, so the output is zero.? And for a +5 volt input, compared to +5 volts reference, that's zero difference, again times unity gain, but inverted to the reference voltage gives you +5 VDC out. No?

    In this scenario, with the reference voltage equal to the maximum input voltage, I don't see how the output voltage could ever swing negative. That's why I compared it with the middle point of 2.5 volts, so you'd have a swing of -2.5 volts (on a zero input) and +2.5 volts (on a +5 volt input).

    I'm afraid your circuit won't quite work. The first stage inverts the signal about the 2.5V ref (so it changes 0 to +5v to +5v to 0)
    This first stage is the only part of my circuit that I changed. When I was first drawing this, I almost eliminated that stage entirely, as I was pretty sure the rest of my circuit would (with the proper feedback resistor), be able to accept the 0 to +5 V input directly. But then I thought about needing the signal to swing above and below ground. I thought I could do that by injecting an offset, but then I thought it would be easier to add that first amp and just compare the input to the offset instead. I guess I made a mistake there, but I still don't see it.?.

    then you've got a -2x gain stage converting that to 0 to -10V,
    I thought it would be +5 to -5? (Assuming the output from the first stage was -2.5 to 2.5, you need to double it...)

    OK - if my first stage is already producing double gain, then I see why that feedback resistor on the second stage should be reduced. (And indeed, as you'll see below, the original circuit does have a reduced gain section here, but for other reasons.) I'm still lost as to why you say the output peaks on the second stage are zero and -10 V. I expected the signal to be varying above and below zero at this point in the circuit.?.

    This is really bugging me now; I'm obviously missing something big here, but I'll be damned if I can figure it out.

    then with the complementary output you've got 0 to +10V.
    That last stage should be the exact opposite of the output of the middle op-amp. So -5 to + 5...??? Again, why is the output not able to cross zero?

    So the differential output winds up being 0 to -20V.
    Do I have another conceptual error here? I thought "+/- 5 volt differential" means the two edge cases are as follows:

    1) positive terminal at +5 volts, negative terminal at -5 volts, net difference of +10 volts, peak positive signal
    2) positive terminal at - 5 volts, negative terminal at +5 volts, net difference of -10 volts, peak negative signal

    And yes, I agree that this is 20 volts peak-to-peak. Which is the ILDA standard, no? Either pin (X+ or X-) can be at a maximum of +5 or -5 volts... But they need to be able to swing above and below zero...

    Maybe this will help... Here is the original circuit I've been referencing. In this original (and actually constructed form) of my circuit, that first stage accepted a differential input and converted it to single-ended. So, yeah, a little different that what we're doing here, but close. The complete circuit as built (and tested) looked like this:



    I know this one works, because I built several of them and played around with a controller on the input and an oscilloscope on the output to verify that it worked as intended. In the end the project that this was for fell apart for other reasons, but this circuit *would* faithfully recreate the signal on the input while allowing for a little extra gain adjustment (to compensate for line losses on the input) and also included zeiners to protect against over-voltage on the output. The 5.6K feedback resistor in the middle knocks down the gain of the second stage to allow for more adjustment on the size pot (it won't clip until you get close to full scale).

    I realize that you probably have better things to do on a Sunday than to teach remedial electronics theory to me, but I'd really like to understand this better, if you don't mind indulging me...

    Adam

  5. #25
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    Quote Originally Posted by piydadorto View Post
    Bump. Still don't get what I need to connect to the negative pin (or positive?) of the output +-5V connector that goes to the scanner in the diagram above.
    Let's get back to your original question! Try tying the negative scanner pins together to the +/- supply common. Make sure the low side of your 0-5v source and the +5V regulator are tied there too.

    You may or may not be able to achieve full scanner deflection with aberry's original single stage amp, but hook it up and see what you get- we can go from there.

  6. #26
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    Quote Originally Posted by buffo View Post
    I realize that you probably have better things to do on a Sunday than to teach remedial electronics theory to me, but I'd really like to understand this better, if you don't mind indulging me...
    You clearly overestimate my typical Sundays

    It sounds like you're looking at the original circuit I posted as an inverting amp, but it technically isn't because it's the non-inverting input where we're applying our input signal. But this nicely highlights the fact that the standard inverting and non-inverting circuits are really the same thing, and thus operate the same way, but when we don't see ground where we expect to the standard simplified formulas have to go out the window. Most of the formulas and calculators you see published for op amps are very textbook in that they always present the reference node as Gnd, which is certainly the conventional case, but it makes it hard to see some of the other useful possibilities. Since "Ground" is just an arbitrary reference point, we can change our frame of reference and make the circuit look like a more familiar configuration (please excuse the red squiggles, Altium gets cranky about duplicate designators):

    Click image for larger version. 

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    Hopefully that makes sense, but it's not always so easy to morph an unusual op-amp circuit into one of the familiar configurations. So generally, there are two key factors that you can use to help suss out the operation of an unusual op amp configuration and generally develop a better intuition for how those circuits work:

    1) Any op amp circuit with net negative feedback will always move its output such that the voltage at the two inputs are equal*. So if you can identify the voltage at one input of the op amp (either because it's a fixed reference point or you know what you're applying to it), you know that the voltage at the other input will be the same, and then start solving things from there.
    2) The current into the inputs of the op amp is zero**. So in the case where the input is connected to the middle of two resistors, the current through those resistors can be assumed to be equal--that means the ratio of the resistors' voltages will be equal to the ratio of their resistances.*** Just pay attention to the direction of current flow which determines the sign (polarity) of the voltage!

    Understanding those two key things will go a long way towards deciphering the vast majority of op-amp circuits, and we can apply them to the circuit I posted:

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    We can also use those principles on the input stage from your circuit:

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    The second and third stages of your circuit are conventional inverting setups, so the normal formula applies.

    The reason that the second circuit you posted works properly is because its input stage is a differential amplifier, the output of which is proportional (equal in this case since the resistors are the same) to the difference between its inputs, and that difference can be both positive or negative depending on whether the positive input is greater or lesser than the negative input. Now if you take the first circuit you posted and imagine twisting those two 200R resistors around, you might notice that you have exactly the same topology! This is really a coincidence of the fact that you used a divider to get the +2.5V reference, but still gives another way of understanding the circuit: it's equivalent to a differential amplifier where the positive input is tied to +5V, and its output is therefore equal to the difference between the positive input (+5V) and the negative input (0 to +5V). Since the positive input will always be greater or equal to than the negative input, the output will always be positive or zero!

    Whichever way you look at it, the input stage's output is 0 to +5V, so then stage two (with a gain of -2) will give 0 to -10V for X+, and the third stage (with a gain of -1) will give 0 to +10V for X-, which makes the differential output range 0 to -20V (or 0 to +20V if you swap X+ and X-).

    Hopefully that helps. Let me know if any of that was unclear, and I'll be happy to issue forth ever larger walls of text on the subject.



    * Assuming the op amp's output can swing far enough. An op amp with no or net positive feedback is a comparator. Note that feedback may not be direct, and the total feedback path may incorporate additional inverting stages (as in a conventional low-side current driver) or multiple feedback paths, hence it's the *net* feedback that's important.

    ** A JFET input op amp like the TL084 has typical input bias current of 30pA (that's 3x10-11), which is effectively zero for most purposes.

    *** Configurations with more resistors meeting at an input (IE, summing amps) are a little more complicated, but remember that the net current at the junction of all of those resistors must be zero.

  7. #27
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    Thanks to both of you.

    Could you please explain the purpose of the unused op amp U2D? And also U2E?


  8. #28
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    Quote Originally Posted by piydadorto View Post
    Could you please explain the purpose of the unused op amp U2D? And also U2E?
    My circuit uses the TL084 chip, which is basically 4 individual op-amps in a single IC package with a total of 14 pins. Each op-amp needs 3 pins (inverting input, non-inverting input, and output), which makes 12 pins total, plus you need a positive and negative voltage supply to the whole package, giving you 14 total pins. Power is distributed internally to all 4 op-amps from those two power supply pins.

    My circuit only uses 3 op-amps. The 4th one on the chip isn't needed. But to be complete, I included it on the drawing and simply labeled it as not used. (But I should have connected the output to the inverting input and then grounded the non-inverting input on that unused op-amp, as Andrew did on his drawing.)

    Regarding power - in my drawing the first op-amp also shows the power connections (pin 4 is the positive power rail, and pin 11 is the negative power rail). The rest of the op-amps don't show any connections, because the TL084 chip only has two power pins; the internal connections that supply power to the op-amps doesn't need to be shown; it's just inferred because we're using a quad op-amp IC.

    But when Andrew drew his version of the circuit, he separated the power connections to the TL-084 from the op-amps to make the drawing easier to follow. U2 is the TL084 chip and the numbers that follow are the individual parts inside. Thus, he has the 4 op-amps (Labeled U2-1 through U2-4) plus the power connection section (U2-5). It's just another way of drawing the circuit. If you count up the pins and match the numbers, you'll see that they're the same on both drawings.

    Quote Originally Posted by aberry
    Any op amp circuit with net negative feedback will always move its output such that the voltage at the two inputs are equal

    Whoaaaaaa! I never looked at it like this before... Damn dude - that helps a *lot*!

    I knew that the inputs were basically isolated and didn't draw any real current (barring a few trillionths of an amp), but realizing that the feedback is going to try to drive the inverting input to the same voltage as the non-inverting input is a huge insight! Thanks for that!

    OK, I'm going to play with this a bit more and make sure I understand it... But I already see my mistake on the first stage, and now I'm also able to follow your examples. (You would have made a good teacher, dude!)

    Adam

  9. #29
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    Quote Originally Posted by piydadorto View Post
    Could you please explain the purpose of the unused op amp U2D? And also U2E?
    The TL084 is a quad op amp, containing four fully independent op amps in a single IC, but the circuit only requires three op amps, so the fourth one is extraneous and should be connected as shown. (Op amps that are left unconnected can oscillate due to parasitic coupling or RF and at best consume additional power, or worse can disrupt the rest of the circuit, or worst of all, actually damage the op amp.) Actually, you could eliminate the third stage of the circuit and take X+ from the output of the first stage. Then you need only two op-amps and could either use a TL082 (same as the TL084 but with two op amps instead of four) or a single TL084 for both X and Y. You can swap the op amps around either within the IC or even between ICs, this won't impact the circuit and can really help make it easier to wire.

    U2E just has the power pins for the quad op amp, which are shared for all four op amps in the package. So U2A through U2E are all just different portions of the same IC that are simply split apart in the schematic to make it easier to read. buffo has shown the power pins on one of his op-amp symbols, both are standard ways of showing power pins on multi-part packages. His way is a bit more traditional, but my way can make dense schematics a little bit cleaner.

  10. #30
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    OK makes perfect sense. I have gotten the parts but haven't had time to build a proto board and test yet, will keep updated.
    Just to be clear, the final board schem you posted was for -5 - +5V scanners, not -10 - +10, right?

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