Op Amp - Purpose

Discussion in 'General Electronics Chat' started by qwertyuiop23, Apr 30, 2014.

  1. qwertyuiop23

    Thread Starter New Member

    Aug 21, 2013
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    I have the attached op amp circuit, and I trying to figure what it is meant to do.

    Obviously removing the 100k resistor and +3V bias makes it into a straight forward inverting amplifier. With it there I am not sure what this circuit is meant to achieve.

    Any help would be greatly appreciated. In real life the function generator is a piezo crystal.

    The reason I am trying to understand this circuit is because I want to place a full bridge rectifier in front of the inputs but I need to know if the voltage drop due to the diodes will impact the system performance significantly.

    Regards,
    Lance
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    R2 100k to +3V applies a +3V bias to the non-inverting input.
    Hence the opamp will amplify signals that are just slightly below +3V.
     
  3. qwertyuiop23

    Thread Starter New Member

    Aug 21, 2013
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    Ie with the gain of 1000 2.97V between inputs would equate to 0V output while 3.00V would equate to 3V output? Assuming op amp if rail to rail?

    EDIT::

    The reason I ask is that this is not what happens when I simulate this. Using circuit attached in the first post (made with TINA TI) and varying the input voltage from 1V - 4V the trigger point occurs at 1.85V?

    EDIT2::
    Using an ideal op amp in the simulation moves it to triggering at zero volts.
     
    Last edited: May 1, 2014
  4. pwdixon

    Member

    Oct 11, 2012
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    Was this the circuit you meant it to be? Is the input voltage floating wrt GND?
     
  5. alfacliff

    Well-Known Member

    Dec 13, 2013
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    the gain is set by the resistors connected to the - leg. gain is 1
     
  6. pwdixon

    Member

    Oct 11, 2012
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    How did you calculate that gain?
     
  7. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Depends what you mean by 'significantly'. With only a 3V supply the diode drop is 40% of the supply! I'd call that significant. You can easily see the effect in a simulation. BTW, what impedance are you using in the sim model of the piezo? That impedance has a 'significant' effect on the opamp response too.
     
  8. crutschow

    Expert

    Mar 14, 2008
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    If you want to full-wave rectify a small signal, you should use an op amp in a precision rectifier circuit. With that, the diodes are in the feedback loop and their forward drop becomes negligible.

    What is the highest signal frequency.
     
  9. qwertyuiop23

    Thread Starter New Member

    Aug 21, 2013
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    To answer your questions:

    @pwdixon
    Yes the piezo is floating as described

    @alfacliff
    I thought it was an inverting op amp, therefore a gain of -1000?

    @crutschow
    Highest signal frequency is around 50kHz. I had thought about using a precision rectifier; however, this is a device that has a five year life span running of two AA batteries, so anymore active power draw would eliminate that option.

    @Alec_t
    Agreed 40% is a large voltage drop (Using two zeners conducting at anytime given a 0.24V forward voltage drop gives only a 16%).

    However, if the circuit is triggering at 3mV (I believe it is) then the difference between 3V and 1V is insignificant as the piezo is being hit with approximately 10,000G.

    I use a square wave input going through a capacitor (10u) and inductor (1m) in series. Then across the output I have a capacitor (10u) and a resistor (10M). Essentially an oscillator with resistance across the output. The input wave is a square wave of amplitude XXV (directly proportional to force hit with).


    I believe because of the 100K on the non inverting input to 3V is also attached to the non grounded unit it essentially acts as a grounded input.


    Currently the opamp is always putting out zero when I attach a diode full rectifier to it.


    - Regards
    Lance
     
  10. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Why don't you bias the input at 1.5V and rectify the output?
    BTW, the loop gain will be -1000 only if the source impedance is zero :(
     
  11. BobTPH

    Active Member

    Jun 5, 2013
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    And it will be -1000 at 50KHz only if the GBP of the opamp is > 50M.

    Edit: never mind, I see that an OPA300 is that fast.

    Bob
     
  12. qwertyuiop23

    Thread Starter New Member

    Aug 21, 2013
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    @Alec_t
    Biasing the input would require a voltage regulator or voltage divider to get the 1.5V correct? Therefore increasing the current draw during no signal input and during signal input - something that I need to keep minimised. That being said I have considered doing this; rectifying the signal after amplification.

    In regards to the open loop gain, agreed the open loop will only work if there is zero impedance output from the piezo system, and this is the reason why usually I would buffer the voltage coming from the piezo with a simple non-inverting voltage follower to utilise the zero (or near zero) output impedance of the opamp. As far as I know a piezo has a very large output impedance during no signal (dc output impedance in the range of 10M) - it can be modelled essentially as a high pass network. Therefore assuming at high enough frequencies it acts as zero impedance (using the impedance calculation for a capacitor and assuming 10u capacitance - unsure of true capacitance - the impedance is below one down to approx 5kHz), so for an oscillating signal the open loop gain should be -1000, for no signal it should see infinite impedance and therefore the output will be 3V.

    Comparing this to a diode which when conducting is proportional to the DC current flowing through it, which can be approximated to being very small due to thermal voltage. Therefore also presents a low impedance input (in parallel with the piezo capacitance impedance) to the op amp giving the -1000 open loop gain. Additionally a non-turned on diode has infinite impedance thus meaning the output from the opamp will be 3V.

    Does my logic follow?

    @BobTPH
    Thank-you for clarifying, I also checked this earlier to ensure that it would indeed suit.


    Regards,
    - Lance
     
  13. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Yes, though a Vreg would be current-hungry. Perhaps try 10meg instead of 100k from pin 3 to 3V, and another 10 meg to ground? (I haven't checked what bias current your opamp needs).
     
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