Differential OP-AMP

Discussion in 'General Electronics Chat' started by khier, Jan 22, 2014.

  1. khier

    Thread Starter Member

    Jan 16, 2013
    30
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    Dear All,

    I am trying to amplify the difference coming from two identical sensors. The two inputs are fed into an OP-AMP configured as difference amplifier. Both signals are biased with 2.5 V midway between Vcc+ and Vcc-. A transformer is used as a power source.
    I tried to connect the signal directly to a single amplifier, to dual amplifier with the other amplifier used as voltage follower to generate ground, and finally to quad amplifier chip. In the last case the signals are fed to the additional amplifiers in voltage follower mode and their output is supplied to the difference amplifier. The last configuration is shown in the attached file - BTW, there is a mistake in the sketch: R4=R5=470 KOhm in reality. The difference between the two inputs is around 40 mV.

    Now to the problem. When S1 is greater than S2 I get a proper behaviour. The output voltage increases linarly with the difference S2-S1 until saturation at -2.5V. However, when the difference is positive, i.e. S2-S1 is greater than zero, the amplifier saturates faster at 1.25 V, exactly midway between 0 and Vcc+.

    I tried different gain ratings 20, 50 and 100. The behaviour is the same except it reaches the saturation level faster as the gain increases, as to be expected.

    Did I do something wrong or is there fundamental understanding problem?

    Thanks in advance.
     
  2. #12

    Expert

    Nov 30, 2010
    16,284
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    R4 is one ohm? Gain in the negative direction is 10,000? Look up the equations for a differential amplifier and make sure you aren't using an LM741 op-amp.
     
  3. ramancini8

    Member

    Jul 18, 2012
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    118
    What is the input signal swing? I don't like the ground scheme because of the high impedance. You operate single supply and configure two op amps as inverting amplifiers and sum them.
     
  4. BobTPH

    Active Member

    Jun 5, 2013
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    What opamp are you using? It must be a rail-to-rail to work properly. If if is an LM324, then that explains it because the output cannot go to the positive rail.

    Bob
     
  5. crutschow

    Expert

    Mar 14, 2008
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    3,233
    For a difference amp, R4 should equal R5.
     
  6. khier

    Thread Starter Member

    Jan 16, 2013
    30
    1
    The input signals are DC signals.

    I trield LM081 and LM082. The Quad chip is indeed an LM324. Does a TL084 suffer from the same limitaions?
     
  7. khier

    Thread Starter Member

    Jan 16, 2013
    30
    1
    The input signals are DC signals biased to 2.5 V. Their values vary between -40 and 40 mV (i.e. lie within the range of 2.46-2.54 V)

    I am not sure I got your point properly. How can I get the difference by summing them? Could you give me more details please.
     
    Last edited: Jan 22, 2014
  8. khier

    Thread Starter Member

    Jan 16, 2013
    30
    1
    The input signals are DC signals.

    I trield LM081 and LM082. The Quad chip is indeed an LM324. Does a TL084 suffer from the same limitaions?

    Walid
     
  9. #12

    Expert

    Nov 30, 2010
    16,284
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    The TL08x series has a minimum power supply of 6 volts.
     
  10. khier

    Thread Starter Member

    Jan 16, 2013
    30
    1
    I thought it can be used with anything betwen -18 and 18 V.
     
  11. #12

    Expert

    Nov 30, 2010
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    Those are the absolute maximum voltage supplies.
    Page 2, table 2, operating conditions.
     
    khier likes this.
  12. t06afre

    AAC Fanatic!

    May 11, 2009
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    khier likes this.
  13. khier

    Thread Starter Member

    Jan 16, 2013
    30
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    I was thinking about something like that, but always thought it is an overkill. I mean what is more trivial than estimation of the difference between two DC signals around 2.5 V using an op-amp? This must be possible with a common op-amp. All feedback statements indicate I made the wrong choice twice: once when I used the TL08X with 5V, and the second when I expected an LM324 to act like a rail-to-rail. This leads me to the following question: how would you get this differnce? I mean how the circuit should be configured and which op-amp shal I use?

    Thanks

    Walid
     
  14. khier

    Thread Starter Member

    Jan 16, 2013
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    1
    I see I misinterpreted the specifications.

    Thanks
     
  15. #12

    Expert

    Nov 30, 2010
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    Geeze! It took me about a week to look up every single thing on a datasheet, the first time I tried to understand it all. No surprise that beginners find them over complicated and confusing. (I used to be a beginner.) Lucky you have the internet now. I did it in 1970 :eek:
     
  16. khier

    Thread Starter Member

    Jan 16, 2013
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    Now I am back to position zero. Could anyone suggest an op-amp for my application? Sofar I came across the LM6582 and the TLV24XX family. Are they recommended for rail-to-rail output @5V?

    Thanks
     
  17. BobTPH

    Active Member

    Jun 5, 2013
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    Microchip has a large line of rail-to-rail input and output opamps that operate on a 5V single supply. The other choice is the use the LM324 but power it with a higher voltage. If you used 9V, a 5V output is not a problem.

    Bob
     
  18. khier

    Thread Starter Member

    Jan 16, 2013
    30
    1
    The voltage level is not the biggest problem with the 324. I could reduce the gain and be happy with the 5V. The biggest problem is the asymmetric behaviour. When the signal at the inverting input was stroger than the non-inverting signal I could get up to the negative saturation voltage (Vcc-) at the output, but only 0.5 Vcc+ for the opposite case.

    Now if I limit the gain to have Vcc/2 as the maximum voltagge, will the lm324 behave symmetrically or will it again go to -Vcc/2 and Vcc/4 this time?
     
  19. BobTPH

    Active Member

    Jun 5, 2013
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    The problem is that the output of the LM324 cannot go as high as Vcc, more like Vcc-2V is the max it can output, so if you keep the output to under 3V it should behave symmetrically. Try making the offset 1.5V and reducing the gain to get +- 1.5V from that as the output.

    Edited to add: Also, the load on the output limits the output voltage. What is it loaded with? For max output, the load should be no more than 10K Ohms.

    Bob
     
  20. ramancini8

    Member

    Jul 18, 2012
    442
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    On second thought this is the perfect application for an instrumentation amplifier. It will do everything I suggested with much higher precision. A diff amp made with 1% discrete resistors can be off by 10% considering purchase tolerances, tempco, age, etc. I am not going to suggest a diff amp because you can do the data sheet search as well as I can, but look for a single supply instrumentation amp.
     
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