Help with op-amp issue

Discussion in 'General Electronics Chat' started by DC_Kid, May 8, 2008.

  1. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    i am using a BurrBrown INA114 8pin dip op-amp (see attched pdf) to sense a current pulse across a current sense resistor. the problem i have now is that i dont see the oa's output as expected. i do see something, but i'll explain that later.

    power comes from a single positive +5v regulator (lm7805) and its filtered well. pin 7 has two caps on it, a 10kpF poly and a smaller ceramic. the gain is set by a 1% 510ohm connected from pin 1 to pin 8. pins 4&5 tied to gnd. pins 2&3 are biased to gnd with 360k ohm resistors. pin 3 (V+in) is tied to the input of the resistor, and pin 2 (V-in) is tied to the output of the resistor. the load being switched on/off is basically a 5.6mH inductor. the flyback voltage is clamped at 34v via zener diodes.

    the current flow is approx 1amp and is pulsed on/off 0-60Hz between 20-80% duty cycle. the gain of oa is set right at 100, so with 1amp flowing across the "curent sense" resistor i was expecting the oa output to almost reach supply voltage. the voltage drop across the resistor (as seen across pins 2&3) should be 1amp x 0.05ohm = 0.05v, and with gain of 100 the output should be 5v.

    i was also expecting to see the shape of the output signal to follow along with the current flow. in other words, as long as current is flowing the oa output should stay high, etc.

    what i observed look like noise. when the inductor current is switched off i was seeing very very short spikes with varying voltage up to about 3.4v.

    its a very simple design, but is not working as i expected...

    any suggestions or questions to lead me in the right direction.

    thanks.
     
    Last edited: May 8, 2008
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    Without the complete schematic, it's hard to be certain. But, this is not a device that is intended for unipolar voltage supply. I wouldn't be at all surprised if it didn't help to give the INA +/- 10 volts. It's also not a rail-to-rail device, so it probably can't swing the output beyond 3.5 volts or so. You need headroom for the output to be able to go to 5 volts.

    I guess you have the current sense resistor across the inputs so the voltage drop gives the differential to amplify, and the driving voltage is common-mode, and thus not visible.
     
  3. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    the pdf does note "It operates with power supplies as low as ±2.25V, allowing use in battery operated and single 5V supply systems". i assumed this meant a single +5v supply (battery), is this what it means? i can use a +9v regulator to power oa.

    can you explain what you mean by "thus not visible".

    thanks
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    A common-mode voltage should not appear in the output, and so will not be visible as an offset.

    Perhaps it will go on a single supply. I've never tried it. I'm pretty sure you can't swing the output closer than 1.5 volts to the positive supply rail. Don't know how close to 0 volts it will go, either. Try two batteries for +/- 9 volts.
     
    Last edited: May 8, 2008
  5. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    here is a schematic of my amp circuit.

    i was expecting to see the oa output waveform follow that of the current, but i didnt see anything at all.

    is there a easy way to use a voltage regulator and a +14v supply to obtain some -V that i can attach to pin 4? i see how to use two 9v batteries to get +/- 9v to the oa (by having one bat + and the other - as the common), but how can i achieve this using a regulator?

    thanks
     
    Last edited: May 9, 2008
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Check your PMs.

    You're going to have to use a negative supply. Check the INPUT COMMON MODE RANGE paragraph on the bottom of page 9 of the datasheet. To stay in the linear region of the instrumentation amp, you have to have > 1.25 between your input signal and the rails. Your negative rail is at 0, which when Q1 is conducting, places both inputs too close to the negative rail.
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    You posted your schematic over at Imageshack instead of posting it here. It is "forbidden". It seems to be extremely dark in the thumbnail pic so cannot be seen there either.

    The inputs of the instrumentation amp don't work when they are near 0V or near +5V in your circuit. Look at the input common mode voltage range on the datasheet.

    The output doesn't go to 0V nor to +5V. It goes to +1.5V and to +3.5V without a load.
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    delete the .th before the .jpg in the address line for the large image address.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    Now the link works.
    The input voltages are way higher than the supply voltage of the instrumentation amp.
    The inputs won't work at such a high voltage but won't be damaged because they have over-voltage protection circuits.

    If each input has a voltage divider to reduce the voltage to within the input common mode range then they will work with your single 5V supply. But maybe the output going between +1.5V and +3.5V is not enough for what you are doing.
     
  10. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    when Q1 turns on there is 1amp flowing through the 0.05 ohm resistor. the voltage diff across this resistor should only be 0.05v when Q1 conducts, and the oa inputs will also be very close to 0.05v and gnd. when Q1 is off the voltage on both sides of the resistor is 14.4v, no diff so no diff to amplify (but shouldnt the oa work if i had 14v on +in and 13v on -in ??). i'm only interested in oa output when Q1 conducts current. the "be careful" part for the oa inputs is the flyback voltage which climbs to 34v clamped by zener.

    as noted the inputs are too close to -rail voltage (gnd). if i manage to get -5V on pin4 (-V) would this solve the problem?

    thanks for the help...., learning something new.
     
    Last edited: May 9, 2008
  11. SgtWookie

    Expert

    Jul 17, 2007
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    If you get pin 4 to -5v, you will then be able to see more like what you expect on the output of the INA114 while Q1 is conducting. When Q1 turns off, what you'll see at the output of the INA114 won't be predictable, because the input voltages will be swinging outside both of the rails.
     
  12. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    i'm actually looking for a 5v pulse from oa output, so i guess i need +-9v supply to the oa.

    as for when inputs are higher than the rails, the oa only amplifies the diff of the inputs, so even if they are both at 14v the oa output should be zero, no?
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    The INA114 output V swing with a 5mA load and +/-15v rails is +/-13.5v. If you used less voltage for the rails, the 5mA load vs output voltage would likely decrease somewhat porportionately.

    If the inputs of the INA114 are outside of the supply rails, you will be well outside of the linear region of the instrumentation amplifier. What kind of output you will get at that point could be anything, even slamming back and fourth from nearly one rail to the other, which seems quite likely due to the ringing you're observing when Q1 stops conducting.

    The output capacitance of Q1 is somewhere in the region of 5nf-6nF, which would make up an RC network with your 0.05 Ohm resistor, on which the 30+v ringing of the injector is surging back and forth over.

    Seems like you need to detect the start and stop of Q1's conduction, and ignore any output from the IA when Q1 isn't conducting. This would be the first positive edge after Q1 was turned on, to the first negative edge after Q1 was turned off.

    Better yet, move Rsense out of the loop - below Q1's source. That will also remove a lot of stress from the INA114's inputs.
     
    Last edited: May 9, 2008
  14. Audioguru

    New Member

    Dec 20, 2007
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    But the voltages are higher and lower than the allowed common mode voltage range so the inputs of the operational anplifier don't work.

    The inputs of the instrumentation amplifier don't work below about +1.5V.

    [/quote]Ten Q1 is off the voltage on both sides of the resistor is 14.4v, no diff so no diff to amplify (but shouldnt the oa work if i had 14v on +in and 13v on -in ??).[/quote]
    With a 5V supply, the inputs don't work when their voltage is higher than +3.5V.

    Yes.
    With a -5V supply then its inputs would work when their voltage is between -3.5V and +3.5V. But the inputs go up to 14.4V then they won't work.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    I mentioned this before, but you might've missed it.

    Take a look at Linear Technology's LT1054 switched-capacitor voltage converter with regulator:
    http://www.linear.com/pc/productDetail.jsp?navId=H0,C1,C1003,C1039,C1014,P1258
    IMHO, if your total current demands for a negative rail are < 70mA, this would be a viable option. It would be much less noisy than an inductive solution, and would require very few external components. Yes, it's rated for 100mA, but let's be conservative.
     
  16. DC_Kid

    Thread Starter Distinguished Member

    Feb 25, 2008
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    thanks for the replies.

    in an attempt to save the 30 prototype boards i will 1st try to get +/-9V into the amp, then i'll take a look at it's output. after these 30 boards are gone i'll do a redesign and put the sense resistor along with amp on the source side of the fet. as it is now the voltage on the amp input will spike up to the zener clamp, ring off (which i'll damp with a snubber), and then ride at 14V until the next pulse, etc. placing Rsense and amp after the fet means i have to add more components to the board, but probably well worth it.

    thanks again...
     
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