Amplifier issues?

Discussion in 'General Electronics Chat' started by Twerpling, Apr 1, 2009.

  1. Twerpling

    Thread Starter New Member

    Apr 1, 2009
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    Hello All,
    I seem to be having an issue with a high voltage op amp circuit I have made. I used a LM675T (http://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?langId=-1&storeId=10001&catalogId=10001&productId=120926)
    to take a 0V to ~5V sinewave and make it into a 0V to ~60V sinewave of various frequencies. I did this by making a simple non-inverting amplifier with a compensating resistor where the R1 = 1K and R2 = 11k (making a gain of 12). I also added a compensating resistor on the positive port of the Op Amp of R1 || R2 which is ~930ohms. The rails are connected to 60V and 0V.

    The issue I am having is that I have a pretty severely capacitive load. I would imagine to the Op Amp it would look like two capacitors connected to the output with one going to 0V and one going to 60V. Without any load the current draw of the circuit is around 4 milliamps yet with the load attached the current draw becomes 500 to 600 milliamps (at 60V!). How can I decrease this current draw? It heats the amplifier far to much for my tastes.

    Another issue is that the output of the amplifier is somewhat noisy even without a load. I was hoping that I could somehow smooth it out.

    Additionally, If anyone can recommend a better amplifier to use in this sort of application I would love to hear it.

    Any thoughts?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    What is the nature of your load? It sounds like you have a DC load that is referenced to 0V without a decoupling capacitor. Do you have a decoupling capacitor in series with the load?

    hgmjr
     
  3. Twerpling

    Thread Starter New Member

    Apr 1, 2009
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    The load is quite literally two capacitors, one connected to output and 0V and the other to output and 60V. I do not have a decoupling capacitor anywhere in the circuit.

    If you were to look at the non inverting circuit here (http://www.allaboutcircuits.com/vol_3/chpt_8/13.html) and add two capacitors to the output, that would pretty much be the circuit I have.
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    What are the values of the two caps and what purpose do they serve on the output of the opamp?

    hgmjr
     
  5. Twerpling

    Thread Starter New Member

    Apr 1, 2009
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    The 'caps' are a simple contraction device I stuck on the end. The spec sheet for it says to treat it as two 2.4 uF capacitors when dealing with it electrically. It also states that it needs to be fed 60V for full motion. The idea is to get this thing to contract at the frequency we want.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    Ok. So what is the maximum frequency you plan to use to excite this contractor?

    hgmjr
     
  7. Darren Holdstock

    Active Member

    Feb 10, 2009
    262
    11
    The contraction device sounds like a piezo bar, and loading an op-amp with the output of one of those will pretty much guarantee instability. It's not just the capacitive load; they kick back, and the frequency vs. impedance response looks like the Alps.

    An old but trusted method of driving capacitive loads is to insert a 100 Ω resistor between the op-amp output and the load/feedback tap, and place the compensation cap directly between the output pin and the inverting input. It's not a cure-all, but it is effective at isolating most nasty cap loads.

    A more energy-efficient method would be to drive the piezo bar with an LC-filtered PWM signal, and then its intrinsic capacitance becomes an advantage.
     
  8. FaithPreacher

    New Member

    Apr 1, 2009
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    Darren is correct.

    you not expect a linear amplifier to drive a pure capacitance. It will add a huge lag to the feddback loop and cause intsbility as wellas output stage current limiting at high frequency. Isolate the load with a resistor to keeep the output stage from overloading. The r value will depend on the amplifier output drive capabily. For loop stability a value as low as 10 Ohms may be good enough.
     
  9. Twerpling

    Thread Starter New Member

    Apr 1, 2009
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    How would I go about doing this? I would imagine you would need to adjust the capacitance and inductance to generate a specific frequency rather then a range (which I was hoping to keep). Also how would this amplify the signal from 5 to 60?
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    If you didn't make the circuit in the datasheet with power supply decoupling capacitors and an RC network on its output then it is oscillating at a very high frequency which causes a high power supply current and lots of heat.

    Driving a 2uF capacitive load or more is explained in detail in the datasheet. Simply do what they tell you to do.
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    The power required to charge and discharge a capacitor is
    P=f*C*V^2. That power will be dissipated by your op amp.
     
  12. italo

    New Member

    Nov 20, 2005
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    obviously the amp saturate at those currents output. YOU MAY TRY TO INSTALL A LIMITING RESISTOR IN SERIES WITH THE OUTPUT BUT INSIDE THE LOOP SUCH THAT THE AMPLIFIER CAN BECOME SORT OF A LIMITED CURRENT SOURCE.
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    An output resistor inside the loop will shift the pole to an even lower frequency, probably guaranteeing oscillation.
     
  14. Twerpling

    Thread Starter New Member

    Apr 1, 2009
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    It looks like seems that this isn't the best way to go about doing this. Is there a better way to generate a signal with a greater stability and lower current draw (or either) using another method?

    I am looking to generate a range of frequencies here, somewhere in the area of 0 to 2KHz, though less will be acceptable.

    Any one have an idea of what sort of circuit topology to use? I would guess some sort of oscillators, I am trying to stay away from transformers since they may be too large for this project.
     
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