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Floating current source

Discussion in 'General Electronics Discussion' started by Rajinder, Dec 31, 2016.

  1. Rajinder

    Rajinder

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    Jan 30, 2016
    Hi all,
    I have the following circuit (see attached image).
    What would be the best way fro me to make the output a floating current source rather than with ground connection?


    Or use a current transformer? Note I am using currents of 100nA-300uA. Is this possible?

    Or could I use a photovoltaic method? I am not sure on this

    Lastly a opto coupler or similar?

    I look forward to hearing from you.

    best regards,
    raj
     

    Attached Files:

  2. Alec_t

    Alec_t

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    Jul 7, 2015
    A current transformer won't work for DC currents.
    An opto solution would give a floating source if powered with a non-grounded battery.
     
  3. Rajinder

    Rajinder

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    Jan 30, 2016
    Hi Alec_t,
    Please could you elaborate? Am I not sure on how to connect the opto device to this circuit. I am looking at minimising the circuitry modifications.
    Would I need to use a seperate supply for the opto device? Also would the opto dark current need to be very low as i am looking at very small currents?

    I look forward to hearing from you.
    Raj
     
  4. Gryd3

    Gryd3

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    Jun 25, 2014
    What are you powering this device with, and what do you consider 'floating'?

    The 'ground' in your diagram does not necessarily mean 'Earth Ground'. It's simply a 'common' connection point. You could re-draw the circuit by removing the 'ground' indicators and then draw a line connecting all the points 'ground' used to be.

    If you want it to float with reference to ground, then you can power it with a battery, use an AC/DC converter that is isolated, or use an isolation transformer on any other method you want to power it with.
     
    hevans1944 likes this.
  5. Alec_t

    Alec_t

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    Agree with Gryd3. If the existing supply is isolated there's no point in adding opto-isolation.
     
  6. Rajinder

    Rajinder

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    Jan 30, 2016
    The existing supply is not isolated. The unit is powered from a PSU which gives out +12/-12V and 5V. The 5V generates 3V3 from a LDO. The -12V generates a -5V from a LDO. The +5/-5V supply the opamp. These are all DC voltages from a PSU which is fed from mains. Floating for me means something that has no ground connection on one side.
    So I am still unsure how to achieve this. Are you saying use a AC/DC converter instead of my PSU?
    I am not sure.
     
  7. Gryd3

    Gryd3

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    Jun 25, 2014
    Are you confident the existing supply is not isolated?
    As a preliminary step, measure the voltage of your outputs with reference to earth ground.
    Just because something is powered by mains does not mean that it's output is in any way connected or referenced to ground.
    If it's not isolated, you can use an isolation transformer (which is a 1:1 transformer) to power your current supply.

    I should also be asking 'why' you want the output to be floating. What are you attempting to power or connect this circuit to that would require it to float?
     
  8. Alec_t

    Alec_t

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    Jul 7, 2015
    How accurate do you need the current control to be? The current transfer ratio of an opto-isolator is quite temperature dependent (e.g it could vary by ~10% over a 40C temperature range).
     
    Gryd3 likes this.
  9. Rajinder

    Rajinder

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    Jan 30, 2016
    The current control needs to be accurate. Tolerance acceptable would be around ,10%.
     
  10. hevans1944

    hevans1944 Hop - AC8NS

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    Please answer the question that @Gryd3 asked in post #7. What are you trying to DO with your "floating current source"?

    I notice that your control input is referenced to "ground" so if you want the current source to not care about that, just power up the op-amps and the mosfet with ±6 V (instead of ±5 V) from two 6 V batteries. I would suggest lantern batteries with screw-terminal binding posts if you can find them. Otherwise, a pair of 4-cell holders for 1.5 V D-size alkaline cells (8 cells total) will work fine, last a long time, cost a little, do a lot. No circuit modifications necessary, and your current source will have a quieter (less noisy) output at low current settings. Remember to disconnect the cells at the end of each day and they should last for several weeks.
     
    Gryd3 likes this.
  11. Alec_t

    Alec_t

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    Does that comma denote a decimal point?
     
  12. Rajinder

    Rajinder

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    Jan 30, 2016
    No it doesn't. Just a typo. So I canake this into a floating current source by just changing the supplies to the opamp? i.e. using separate supply or battery powered as mentioned by hevans1944?
    The control voltage for the first opamp is fed from an adjustable supply. The second opamp keeps the FET conducting whilst maintaining the same potential at its inputs. This current is fed to the unit under test.
    We have an application where the unit under test requires a floating DC current source. Hence the need to make only slight changes as the PCB has already been designed and populated.
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    You will need to ensure that your input voltage is also floating because both this circuit and your signal source must share a common ground.
     
  14. hevans1944

    hevans1944 Hop - AC8NS

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    That's not a need. It's a want because "the PCB has already been designed and populated." Some one should have considered the need for a floating current source before designing and populating the PCB. Sorry about your design oversight, but caca occurs and sometimes you have to start over...

    OTOH... If your "constant current" source must not have a reference to "ground" then the control input must also "float" without a reference to "ground". There are several ways to make this happen, none of them simple. The easiest approach would be to "float" the control input, perhaps derive it from a battery-sourced voltage, as I suggested for the op-amps and mosfet. If this is impossible, because of other considerations, then the control input must be isolated or floated.

    This could be done by converting the control voltage to a digitized value (for 10% accuracy, referenced to full-scale, an 8-bit ADC would be adequate). You then, for galvanic isolation, optically couple the digitized value to a digital-to-analog converter, powered in isolation by the same batteries that power the op-amps and mosfet. Voila! Everything on the output side of the optical isolator(s) is divorced from "ground," including your current source connections. This can all be done with a PIC microprocessor and a few lines of code: analog control signal in, referenced to "ground," and a floating analog control signal out. The floating part of the circuitry needs to be battery operated with the batteries well-insulated from that pesky "ground". Or you could power from an isolated power supply, but that's a can of worms in itself if very good isolation from "ground" is a requirement.

    If you don't care much about temperature drift and problematic accuracy, you can use an optical isolator as an analog device. The transistor conductivity is fairly proportional to its LED illumination current, so you could use that as your control input to your "ground isolated" circuitry. I did this once to convert a 4 to 20 mA control signal to a variable frequency using a 4N24 and the ubiquitous 555. The result of frequency output versus current input was surprisingly linear. Converting from a variable frequency output to a variable voltage (or current) output is the next step. There are both analog and digital solutions available for that.
     
  15. Rajinder

    Rajinder

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    Jan 30, 2016
    Hi all,
    Thanks for your help. The PCB was designed with the circuit as shown. Then 6 months later the requirement for a floating current source was required. Hence the problem of keeping the mods to a minimum.
    At the moment the input opamp is controlled by an adjustable voltage at its input, which is from a DAC controlled from a PIC micro.

    I have a few questions,
    1. Do I have to float this input to the first opamp? If so what will happen to my DAC values output from the PIC?
    2. if I simply power the whole opamp stages and MOSFET from separate supplies, how can I control the DAC output as currently there is a common 0V between PIC and DAC and opamp stages? I simply can't understand this. Sorry.

    I appreciate any help.
    Raj
     
  16. Gryd3

    Gryd3

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    Jun 25, 2014
    Like you, I am having a bit of a hard time with this.
    I know you want it floating, but you haven't told us why other than it being a 'requirement'.
    If you power the whole thing from batteries then it will be isolated from ground...
    And if you power the whole thing from a transformer, rectified and filtered, it will be isolated from ground.

    What would happen if your device was not isolated/floating?
    You want a solution that requires minimal modifications to you board, you need to be very specific on why you need isolation so a possible alternative can be supplied
     
  17. Rajinder

    Rajinder

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    Jan 30, 2016
    If you can imagine a 3V battery with a 1M resistor connected in series. This give 3uA current. If I connect the battery terminal ends to the UUT, this then gives me an alarm a CO alarm.
    The requirement is to have a constant current source not referenced to 0V but floating as the battery and resistor mentioned above.
     
  18. hevans1944

    hevans1944 Hop - AC8NS

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    Jun 21, 2012
    Imagine, if you will, that your original circuit with +Ve, -Ve, GND, Vin and it's associated GND, and Iout, taken between the mosfet drain and -Ve, are all enclosed in a "black box" completely insulated from the outside world. +Ve and -Ve are supplied from batteries or from well-isolated line-operated power supplies. GND is just a name for a node shared by Vin and current-sensing resistor R3.

    If you could create Vin inside the "black box" without a galvanic reference to anything outside the "black box" then your current Iout would be totally floating without reference to any "ground" outside the "black box." You could connect it as you would with the 3 V battery and series 1 Meg-Ω resistor with the same effect. So, how do you create a Vin control signal inside the "black box" without any reference to anything outside the "black box?" There are many possible solutions, and I have described two (using optical isolators) in post #14. You could also use transformer isolation, but this is a little more difficult.

    Since you are already comfortable using a PIC and a DAC, simply place both of these components inside and power them from the "black box." Communicate with this PIC using an optical isolator (or a pulse transformer) to transfer a digital representation of Vin to the inside of the "black box." The DAC output then becomes your Vin control signal. If the signal Vin changes slowly, you can implement a simple serial data protocol requiring only one optical isolator. Some PICs have serial UARTs "built in" that make this approach particularly easy. You don't need the UART duplex capability since data flow occurs only from the external real world into the "black box." The transistor in the optical coupler should be able to directly control the serial data input to the PIC.

    If you decide to use pulse-transformer isolation, more signal conditioning is required between the transformer secondary winding and the PIC input, and the software is usually more complicated. I would choose the optical isolator approach for the sake of simplicity.

    Please let us know what you decide to do.
     
  19. elebish

    elebish

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    Aug 16, 2013
    Does your requirement have any thing to do with the mains ground or common?
     
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