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Acquiring negative voltages in real life circuitry?

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data

    I'm working on a project for my circuit analysis class in which we designed a circuit that involves a wheatstone bridge, takes the voltage difference and then amplifies it to a useable level. I am past that phase of the design, I have a working circuit for that on a schematic. I also have build the "prototype" on a breadboard, which needs testing (tomorrow actually during lab). I am looking for help getting negative voltages for opamp Vcc+/- rails

    2. Relevant equations

    None really, it's a method question more than mathematics

    3. The attempt at a solution

    I understand the concept of getting negative voltages just fine on paper. You have a voltage source with one resistor at the positive, and another at the negative, the two meet up and go to ground, and bang, you got negative voltages on your negative terminal to ground. That part is fine for me.

    However, the practical application has me scratching my head, because I do not understand what is my "ground" on the physical circuit. All of my circuits in the past never dealt with grounds so to speak other than positive terminal gives voltage in, voltage out goes to negative terminal.

    My only thought so far on this, is this concept:
    Capture.JPG

    Oh, and the real life circuit will be using two 9v batteries (for the sole purpose of having a 9v and -9v supply for each opamp in my circuit)
     
  2. jcsd
  3. May 10, 2013 #2

    NascentOxygen

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    Had you stopped there, then I could have answered your question. There are basically two ways to provide an IC with both + and - voltage rails. You can use a pair of batteries, just as you indicate you are. Another way is to use just one battery, and to derive a negative rail from that positive one by means of a switching IC, a large capacitor, and a bit of circuit wizardry.

    But after reading the rest of your post, I can't see what you are asking. Maybe someone else can.

    It is sometimes feasible to operate an OP-AMP using just one supply rail and earth.
     
  4. May 10, 2013 #3
    Assuming you want a V+ and V- centered around ground, you just stick the ground between the 2 supplies in your picture. I'm not sure what those resistors are for, are they supposed to represent the internal resistance of the batteries?

    So I assume your circuit is:
    Wheatstone bridge -> Inverting/Noninverting amplifier -> output

    Also, why so much voltage? Are you using 741 op amps? Are they unity gain stable? If not, you should check to make sure you have enough gain to prevent oscillation.
     
  5. May 10, 2013 #4
    Yes we are using ua741 opamps. We want 9v as the initial voltage before the Wheatstone bridge, between each branch, one branch will have 240 ohms of resistance between two resistors, and the other branch has a rheostat rated for 120ohms and then another 120 ohms. The bridge is sent to an instrumentation amplifier and then the difference is amplified by a gain of 100 (as the differences we have are in the microvolt ranges before amplification. I'm assuming we cannot simply use the negative rail as ground because of the possibility of negative differences in the device we are designing (its a strain gauge for my professors strength of materials class for next semester).
     
  6. May 10, 2013 #5
    Sorry, but I'm still having some trouble following your setup. Let me give my interpretation of what I think you're saying, and you tell me if I have it right or wrong:

    Photoon5-10-13at219PM_zps0c4c40af.jpg

    So you adjust the rheostat until there is zero voltage drop across Rsense. That calibrates the strain gauge for 0 strain. Then any change in Rstrain will produce a voltage drop, which is amplified by the ua741 and sent to the output.

    So I guess the most important thing is: how big of a +/- change are you expecting to see in the strain gauge (which is assume is basically a strain-variable resistor). Try to be reasonably specific - are you expecting +/- 1 mohms, 10 mohms, etc.

    The reason this matters is that it tells you how big the other resistors in the Wheatstone bridge can be. That matters in this case, since you're dropping 9V across each one and that's going to produce a lot of current -> high wattage resistors.
     
  7. May 10, 2013 #6

    rude man

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    Your formula for Vout in terms of V1 and V2 is incorrect. And the impedance at V1 is essentially infinite whereas the input impedance at V2 is not only finite but there will be some voltage feedthru from V2 to V1.


    To amplify the voltage across the middle resistor of your bridge you ideally want an instrumentation amplifier. Such an amplifier does not load down either V1 or V2 input. You can make an IA from 3 op amps. Check the literature. But in this case it's not necessary, probably.

    If you connect your one (floating!) 18V power supply across the bridge the way you've shown, your + and - "9V" bridge excitation voltages will automatically adjust to close to + and - 9V thanks to the small-value resistances of the bridge. You then hook up the +/-9V to your op amp power pins also. Your op amp circuit needs fixing, as I said. Hint: adding 2 resistors will give you a circuit where Vout = K(V1 - V2), k a constant. Your op amp resitors should be > 10K thruout.
     
  8. May 10, 2013 #7
    Thanks for calling me out, I did have it wrong. The correct equation is:

    Vout = V1+(V2-V1)(1+R5/R4)

    Not sure what you mean by voltage feedthrough from V2 to V1. If you're referring to input offset biasing, that should be fixed by simply putting a resistor on V2 equal to R5||R4.

    Yes, using a differencing amplifier would be better, but I was under the impression he didn't want to do it that way. Plus I would point out that differencing amps usually have more noise, and the 741 already is very noisy. If they are trying to measure a non-DC signal, it might get buried.

    I agree that running the op amp off the 9V supplies probably would mess things up. Probably the best thing to do would be to buffer the voltages going into the Wheatstone bridge to isolate it from the op amps. Voltage references and low pass filters are easy enough.

    Honestly, I don't know why you would do a strain gauge like this anyway - it would be a lot more fun and interesting to stretch a wire and find the resonance frequency as a function of tension. Plus there's the benefit of that method allowing you to use a lock-in amp.
     
  9. May 11, 2013 #8

    rude man

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    If V1 ≠ V2 then a change in V1 generates a changing current into V2's input resistor, disturbing the bridge. No, this is not a voltage or current offset consideration.
    That would be negligible. You are talkig nanovolts/rt Hz. Besides, you are measuring dc, not ac.
    I don't believe I said that, quite the opposite. I would connect the op amp power pins to the +/-9V supplies directly. Op amps are not sensitive to even moderate power supply imbalances, and have good PSRR numbers, especially at low noise frequencies. Even the granddaddy of all fully compensated op amps, the 741.

    Wires are temperature-sensitive! How would you excite the wire and measure the frequency?
     
  10. May 11, 2013 #9
    Interesting, I'll have to look closer at it.

    Fair enough. I've been working on low noise high bandwidth stuff for a while, so I probably overreacted.

    You misunderstand. I was saying that given the batteries have internal resistances comparable to the bridge resistances, the op amp drawing power might change the rail voltages enough to disturb the bridge. So I was thinking it might be a good idea to power the bridge using a buffered reference.

    You could have a solenoid or something twang the wire with a pulse, then let it settle for a second to the resonance frequency. And you could easily measure the temperature with a diode or something.
     
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