VCCS help

Discussion in 'The Projects Forum' started by jxm1092, Aug 27, 2012.

  1. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    Looking to get some help with this circuit since it doesn’t seem to be doing what I am expecting it to do although I will admit my expectations may be wrong.

    Background:
    I’m looking at implementing a project where I can use the output of a rogowski coil, this one the JRF-2, and change it to a current output based on the voltage. For a range of 0-668mV on the rogowski I wanted to get 0-1A out of the circuit, although if I was able to get 0-5A out that would be the best. For some reason I am only getting about 14mA. The LPF is in there to attenuate anything above 60Hz.

    How I chose the components:
    For the LPF I just did RC=1/(2*pi*f) and then came up with a RC pair that worked. I did do a sweep simulation on the frequencies that performed pretty much as I expected.

    As for the op-amp portion, I always get just under 15V out and depending on the input voltage around 14mA. I looked at the data sheet for the LT1210 and it said the Av gain was 71dB. So I worked that backwards and got an Av of 3548 with Av(dB)=10^(71/20). I saw that the filter was attenuating so I had found that at 668mV my actual input would have been 455mV so I multiplied that by 1.25 to account for overshoot and whatnot. So with vi of 0.56875 I got Av=vo/vi, so vo was 2018. For the op-amp the gain is Av=1+(Rf/Ri) so found a Rf/Ri pair that you can see in the circuit. From here I run the simulation and nothing really seems to be working like I expect.
    Thank you for the help guys.
     
  2. wayneh

    Expert

    Sep 9, 2010
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    Your op-amp is pegged to the upper rail. What's the voltage on the two inputs? Check them against what you expect. You should also consider a 0.1µF bypass cap across the power pins on the op-amp.
     
  3. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    1.9V on the positive and 190mV on the negative. I don't understand why the LPF is amplifying the input voltage because before the LPF the voltage is 668mV. Based on just the LPF circuit I was seeing 3.33dB of attenuation at 60Hz so I expected something around that value. I've also updated the schematic with the recommended cap.
     
  4. wayneh

    Expert

    Sep 9, 2010
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    I said the POWER pins, not the input pin. D'Oh!

    Are the grounds actually connected, between the signal generator and the op-amp? And how sure are you that your sine wave really is a sine wave, and actually goes negative for half a cycle?
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    1. That is not a VCCS. It is just a voltage amplifier.
    2. Your power supplies are screwed up. See attachment.
    3. Your voltage gain is1 + 240k/68=3,530 (!)
    4. That op amp is a current feedback type, so the absolute resistor values (in addition to ratio) are impotrant.

    Do you want an AC output of 1A peak (2A p-p) for 668mV peak (1.336V p-p) input?
    Why do you need a current source? Does the load impedance vary? If so, what is the load impedance range?
     
    Last edited: Aug 27, 2012
  6. wayneh

    Expert

    Sep 9, 2010
    12,078
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    Talk about D'Oh! I saw the mislabeled IC pin - and didn't mention it because it's probably just a typo - but I looked right at the "extra" ground and completely missed it. Tough to beat having plenty of eyes. ;)
     
  7. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    That was my fault Wayne I misread what you had typed. And I'm positive about the sine wave because its all simulation at this point. Not going to breadboard the circuit until I have it figured out. Although looking at the rogowski data sheet it does not say that it is a sine wave just that it's 344mV AC/1000A @ 50Hz so I assume sine wave.

    Ron, I had found some VCCS circuits online that were merely expansions of this one so I wanted to be able to get this to work properly before moving on with the circuit...baby steps.

    The LT1210 has two positive inputs in LTSpice, but when you look at the actual data sheet it's a positive and negative so I assume that this was a typo in LTSpice and that it should actually be a negative.

    I got that voltage gain from the math I talked about in my first post. I assumed +/- 1% actual resistor values. Typical gain on the LT1210 is 71dB from the data sheet so 20log(3530)=70.95dB. Please let me know if my math is incorrect somewhere.

    Again I am assuming that the output voltage of the rogowski is 668mV peak so I would be looking for an AC output of 1A or 5A peak yes. From what I have read many current utility products for sampling operate on a 0-1A or 0-5A scale to operate outside this scale typically cost a significant amount of money. I am hoping that I can create this circuit so that I really just have a larger current available to sample from with more OTS products. I hope I made sense and thanks again for helping me out guys.
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Hmmm... so it does. As you said, a typo.



    You need to read up on op amps, and I know the learning curve is steep. The 71dB you saw is the typical open loop gain, with no feedback. When you apply feedback, the feedback network controls the gain as long as the open loop gain>>desired closed loop gain.
    I'm not convinced you need a VCCS.
    Whatever we come up with, it has to be able to drive the load. We need specs on whatever "utility product" you plan to drive with this.

     
  9. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    So I changed the circuit a bit. I am somewhat familiar with op-amps just not as familiar reading the data sheets I guess. So I changed the gain to 25 and moved the power supplies for the op-amp around a bit as well. I've attached the updated circuit below. I haven't decided what utility product i'm going to use yet but let's say something like this. Should I actually put a CT in the circuit or should I just put something like a 10 Ohm load resistor to simulate the CT?
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    With a gain of 25 and .668V peak input, you will drive the op amp into saturation.
    .668*25=16.7
    The op amp output will never be more than the supply rails.

    As I suppose you know, this is still a simple voltage amplifier, not a VCCS. And I don't understand your question about the CT. Where in the circuit would you put it?
     
  11. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    The max going in would be 668mV, however the filter has shown some attenuation to about 455mV so going into the op-amp is only 455mV which should show up on the output as 11.375V, although I am having issues with that as well. Sorry if I was not clear on that earlier.

    I guess it would not be a true CT. The idea would be to take the output current from the op-amp and that would be the input to the GE monitor via pin 9 and then pin 11 on the GE monitor would go to GND.
     
  12. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    So I ran the circuit and looked at the current through the load resistor which I set to 1 Ohm and it seems reasonable. I've attached the current output I got when I ran the simulation. On the high end I'm getting ~1.45A when I have an input of 700mV and as low as 85mA when the input voltage is 50mV. I guess the only question I have is if that is an accurate load model? I know that if I increase the value of the resistor then the current is going to go down, but is that an accurate representation of the meter I would hook it up to? I would think you could put the input resistance of the meter, but it's not listed anywhere so I assume its negligible so that's why I used the 1 Ohm resistor. Is this correct?

    I attached the final test circuit I had as well as the current output through the 1 Ohm resistor.
     
  13. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I don't have any idea how you would scale the input to the meter.
    Are you aware that a Rogowski coil requires an integrator to convert current to voltage? Of course, if the current is a pure sine wave at a single frequency, I suppose you could get by without the integrator, but power line currents typically have harmonics, which makes the integrator a necessity.
    I think a different op amp, like OPA548, would be more suitable. The one you chose is optimized for wide bandwidth. Also, it's a current feedback amplifier, and as such, the resistor values affect the open loop gain, and consequently may cause gain error if you choose them incorrectly. All in all, it can be a can of worms.
    Why don't you just use a CT and simplify your life?
     
  14. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    If only it was that easy, the idea is what if you can't disconnect the utility line into building X because they can't shutdown production how can you get around that. Hence the rogowski, which has an internal integrator as well so that is not a concern for me.
     
    Last edited: Aug 29, 2012
  15. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I'm sure you know you can buy clamp-on CTs. Are they not flexible enough for your application?
    If I were in your shoes, I think I would call the equipment mfr (e.g., GE) and ask them how to interface a Rogowski to their unit. If they can't tell you that, they should at least be able to tell you the value of the burden resistor.
     
  16. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    Well this just points out that the voltage amplifier is not what I want which I should have realized yesterday. So if I switch gears to something like the circuit below how do I adjust the resistors to get the output I want? I know the transistor isn't enough so I found a D44H11 from Onsemi which can do 10A. I know the op-amps won't be enough so I'm working on that part.[​IMG]
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You need an AC current source. That circuit only has the capability to source current.
    I think you need a Howland current pump, or, more specifically, an improved Howland current pump.

    Attached is my first cut at this. Note that the sine wave is actually five overlaid waveforms, one each for loads of 1 through 5 ohms.

    I think the zip file contains all the files you need to run the simulation.
     
  18. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    So is there any advantage over your circuit with the one I've attached? It looks to be exactly the same for the most part except for the transistor. Assuming ein is grounded and in e1 is coming from the LPF portion of the circuit.

    Also if I'm looking to have my fc at 60 Hz would I be better off putting a passive filter just before the load instead of an active one before the Howland? Then I could just use an op amp for gain only before the Howland.

    The other issue I'm starting to see as well is with R1 having to handle the entire current. At a full amp it has to be a 5W resistor and if I wanted to configure this circuit to work up to 5A I would probably need double that as well right?
     
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    Last edited: Aug 30, 2012
  19. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    They're basically the same. The transistor can only source current, so it can't handle AC. You could probably use a standard op amp with a push-pull power transistor emitter follower in place of the OPA548. I could show you how to do that (I think).
    If you really want infinite output impedance, you will need another pot, to make up for resistor tolerances.

    Are you sure you want a filter? Power is affected by harmonics. You will be filtering out some of the harmonics.
    A filter in series with the load would require some humongous parts, due to the low impedance of the load.

    I'm attaching a modified schematic which changes the current sense resistor to 0.1Ω. I noticed that there was significant load current when the input was zero, so I added an offset adjustment. Most of the offset is actually occurring in the OPA548, but the easiest place to add an offset adjustment without affecting gain is on the summing node of the "preamp".
    You can create a single-pole lowpass filter by adding a cap across R7.
    Fc=1/(2πRC).
    Don't forget to add 0.1uF bypass caps from each op amp power pin to GND.
     
  20. jxm1092

    Thread Starter Member

    Jul 16, 2012
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    So I adjusted the circuit a bit to use the OPA549 so that I can got up to 5A if I wanted to. I changed the value of the load to match the burden of the power monitor. I also adjusted the offset resistor because the high and low values of the waveform were off by a bit.

    I also added the cap to adjust the filter, I'm unable to get the proper output though while still having the filter operate properly. When I set the cap to 0.27uF the filter dB is good but the output current is to low. And exactly the opposite for setting the cap to 0.027uF.

    I would want the filter in there to protect the load monitor in the case of some failure in the rogowski that was producing a signal frequency that was greater than 60Hz.
     
    Last edited: Sep 5, 2012
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