"Phase delay" problem in current source amplifier

Discussion in 'General Electronics Chat' started by MisterNo, Oct 22, 2015.

  1. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    I have a problem with the presented current source amplifier. Two attached scope figures show voltage measured across the load for two different input sinusoidal signals of 100Hz and 1000Hz.
    The input signal is feed in from a signal generator without any offset, and voltage across the load has a "step" rise from zero instead raising smoothly zero no matter how small amlitudes are (see scope figures). The step is some kind of delay as it increases with frequency meaning that there is a constant phase delay that is more visible at higer frequencies and still present even at 1Hz. It totaly not visible when input signal is a DC rised up in steps of 0.1mV, that is the current can be very low.

    I cannot figure out how to get rid of the step and what is causing it. My firt guess is that is related to turn on time of a MOSFET and a sleeve rate of an op-amp, but AD848 has a sleeve rate of 300V/us which is more than enough. I tried also another op-am with a sleeve rate of 20V/us, different MOSFETs with different gate capacitance, but the step always stay the same so there has to be something else going on?
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
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    It's slew rate, not sleeve; and it isn't the problem because it would also affect the negative going signal.

    Why do you have the 10pF cap and 10 ohm resistor? They seem superfluous to me and I'd remove them.
     
  3. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    You need to post the complete circuit.
    What are the power supplies for the opamp?
    What is the supply for the load? What is the load impedance?
    What is the offset, if any that the sinusoid is riding on?
     
  4. ifixit

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    Nov 20, 2008
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    And what is the source impedance of the signal generator and the DC source you used to do the 0.1mV step test.
     
  5. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    Yes slew. 10pF cap and 10 ohm resistor are for compensation. Without them the circuit still works (not as "nicely"), but the problem with the step still remains.
     
  6. dl324

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    Mar 30, 2015
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    You shouldn't require compensation at low frequencies and the resistor is definitely unnecessary and will only slow down the MOSFET by introducing delay.
     
  7. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    power supply for the op-amp is +-12V, all nicelly filtered with bypass caps 0.1u + 10u and 5ohm small resistance in series with the op-amp to create LP filter
    The load impedance is 40ohms, (non- inductive resistors ), 50V supply voltage for the load
    No offset as aleady stated. So the negative part of the input sinusoidal signal is negative and that is where the problem is, as when the input signal has offset making it is always positive everything is fine, no turn on delay.
     
    Last edited: Oct 22, 2015
  8. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    I don't see a differenece in the "step" without or without compensation, but I do see high frequency (MHz) oscilations) and overshoot with square input signal without compensation.
     
  9. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    I'm using old wavetek 23 signal generator, don't know exactly what is the output impedance, but I also inserted buffer amplifier in between (for some other purpose) and as I remember that didn't solve the problem. I will re-test it later today with a buffer o-amp to confirm again.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Your input is a sinewave with no offset so the op amp is going into negative saturation when the input goes negative. It's the delay in coming out of saturation that you are seeing.
    You need to keep the input always slightly positive (even just a few mV) to prevent this.
     
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  11. dl324

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    Compensation would improve high frequency response which would exacerbate problems with high frequency oscillations. Are you sure you have supplies well bypassed? Do you have crosstalk between input/output?

    Overshoot is a side effect of having a high slew rate. You also have to worry about matching input impedances to minimize the effect of higher input bias currents which are also a side effect of having a high slew rate.
     
  12. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    Thanks, yes that was my understanding as well, but I got confused as I tried a different op-amp that has slower sleve rate by 10 times, different MOSFET with 10x less input gate capacitance and I also changed the negative rail op-amp poser supply to just -1V and I didn't not see any substantial difference with different combinations.
    Yes, with small offset everything is fine and just 1mA is enough, but ideally I would like not to have any current through the load all the time. So any sugestions how to do that without swaping the load with some permanently attached resistance in parallel with the load?

    I guess the gate of a MOSFET should held somewhere near the threshold value
     
    Last edited: Oct 22, 2015
  13. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    Well with higher output current of 500mA and without 10ohm resistor (10pF cap is stll in place ) I do see ocilations in the output current (see the two attached figures) with sine and square wave input signals. Oscilations are small, and dissapear when I place 10ohm resistor in series with the MOSFET gate.
    Op-amp is well bypassed as stated in one of the posts with 0.1+10uF caps and small 5 ohm resistor in series.
    The loas is supplied for a 50V power supply that is actually formed with 5x220uF +2x2000uF caps connected in parallel which are charged constantly up to 50V (the charging curcuitry doesn't not make a difference being connected all the time or disconencted for the purpose of a measurement). Also I tried some Mastek volage power supply that can make a difference in high freq response, but didn't solve the major problem with the step rise in the output current.
    Cross talk between input/output can only be if there is something wrong going on with the ground plane of a PCB and I'm pretty sure that is all ok, otherwise no.

    "Overshoot is a side effect of having a high slew rate. You also have to worry about matching input impedances to minimize the effect of higher input bias currents which are also a side effect of having a high slew rate."
    AD848 has 6.6uA bias current, yes pretty high. I tried another op-amp with lower slew rate that has 180nA bias current and yes that has an effect on high freq response, but the main problem stays, that is the step rise in the output current.
     
    Last edited: Oct 22, 2015
  14. crutschow

    Expert

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    It's not the sleve (slew) rate or other variables that's the problem, it's that all op amps have a significant delay when coming out of saturation.

    If you add a diode from the op amp output (cathode) to the "-" input it will prevent op amp saturation by clamping the output at about -0.7V, so then the main delay factor would how be long it takes the op amp to slew from the -0.7V negative clamp to the MOSFET threshold voltage.
     
    Last edited: Oct 22, 2015
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  15. MisterNo

    Thread Starter Member

    Jun 28, 2010
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    Ok slew :)
    GOT IT, thanks! So the "trick" is to avoid op-amp saturation. I completelly forgot that saturation will cause a substantial delay on a time scale that makes other factors invisible.
    Will try it and let you know
     
  16. dl324

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    Mar 30, 2015
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    I don't see how saturation can be the issue. The supplies are +/-12V and saturation would show up as distortion when the output signal approached the supplies.
     
  17. crutschow

    Expert

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    It's not the supply voltages that's the problem.
    It's that when the input goes negative, the current can't, thus the loop becomes open and the op amp output goes to the negative rail and becomes saturated, in a vain attempt to generate a negative current.
     
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  18. ronv

    AAC Fanatic!

    Nov 12, 2008
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    I agree. I would think the diode would help. But at least in simulation - not much. :(
     
  19. kubeek

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    Sep 20, 2005
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    What is this circuit supposed to do?
     
  20. dl324

    Distinguished Member

    Mar 30, 2015
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    Thanks. I wasn't thinking about what the current source was doing...

    A diode connected as you mentioned will help with the saturation, but it won't eliminate it. The OP needs to prevent the input from going negative.
     
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