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Current bypass

Discussion in 'General Electronics Discussion' started by Sravanthi, Feb 4, 2015.

  1. Sravanthi

    Sravanthi

    72
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    Feb 4, 2015
    HI...


    I want a method to bypass the current when it reaches more than 1A.

    In my design I have one source,two components(Say K1 and K2),and load in a loop.When the current reaches >1A ,I dont want to pass through K2 but I want to pass through K1.

    So can you suggest a method to bypass the current or to shunt the current when >1A.

    Thankyou....
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Start by showing us a schematic of what you want to do.

    Include information such as the voltages, whether it's AC or DC (and if AC, the frequency)

    How quickly do you want the bypass to activate and deactivate?

    What is this project for? Tell us about the REAL components and the reason for doing this. Don't just tell us "I have two components". You know what these "components" are, don't you? Tell us! Playing guessing games just wastes time.
     
    davenn likes this.
  3. Gryd3

    Gryd3

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    This is common for overvoltage conditions... which may be good enough depending on what K1 and K2 are...
    As Kris suggested, we need details. A suggestion meant for one situation could fail horribly and spectacularly if used in a different situation. No one wants to assume, it could lead to property damage or personal harm.
     
  4. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015

    Hi...

    File I am not able to upload. So I will explain clearly...

    In a loop DC voltage source(variable), Relay,hall sensor and resistive load are connected.

    Rating of the hall sensor is 1A.
    For some testing purpose of relay I want to pass current more than 1A and upto 10A.So in that case I want to bypass hall sensor.
    So for this can you suggest a method for bypassing the current or is there any methods to shunt the current if >1A.

    And this bypassing has to be done in less than 10 micro seconds or still less if possible.

    Thank you....
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Your description isn't clear. You need to upload a diagram.
     
  6. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015
    upload_2015-2-5_9-32-1.png

    When loop current >1A then hall sensor has to be bypassed or current can be split but max current can be passed through hall sensor is 1A.
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    That doesn't help much! It's just a bunch of rectangles!

    What's the voltage?
    What's the relay type? Is it the coil or contacts that you have in the circuit?
    What's the part number of the Hall sensor? Which pins are connected in the circuit? Or are you using it to measure the current through a piece of wire?
    What's the resistance of the load?

    Which part of the circuit do you want to bypass when the current exceeds 1A?
    How is the relay controlled?

    Come on! You can do a lot better than that! Remember, we have no idea what you're trying to do. You need to tell us all the details!
     
  8. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015
    HI...I am new to this forum so I am not able to give the required details...Thanks for your patience and immediate replying...

    Coming to the u r questionnaires

    Voltage value is adjusted as per current what we want to pass.
    Coming to the relay part contacts are closed and i don't think so controlling part is required to mention.

    Hall sensor I am using here to measure the amount of current. But when the current is >1A I don't want to measure.

    Resistance load can be also varied (upto ohms only).This I kept freely because of constraints in current shunting method. So as per convenience in current shunting method resistive load can be chosen.

    Hall sensor limit on measurement is 1A ,so to protect this when current is >1A this I want to bypass. or a method can be proposed such that current can be split in to hall sensor path as <1A and remaining current in shunt path.

    I am attaching the schematic where part numbers are specified.

    Thank you
     

    Attached Files:

  9. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK, that's better!

    The LA25-NP current sensor has a very low insertion resistance and can withstand pretty large overloads. Unfortunately the data sheet is pretty poor quality, and it doesn't give an absolute limit, but it does say that even with the five primaries connected in series for maximum sensitivity, the insertion resistance is only 6.3 mΩ. If the total power dissipation in the primaries is to be limited to 100 mW (a pretty low value), the peak current can be as high as 4A.

    This assumes that the primaries are rated to withstand that much current. If they aren't, or if a higher current is needed, primaries can be connected differently to reduce the insertion resistance and increase the current capability. See the table on the third page of the data sheet.

    So I don't see that there's any need to bypass it, even if that was feasible. No electrical or mechanical bypass circuit can achieve resistances so much less than these numbers in the milliohm range that they could effectively bypass the Hall sensor.

    Finally, even if it was feasible to bypass the Hall sensor, as soon as you did that, you would lose the current measurement from the Hall sensor so you could never tell when to remove the bypass. You would have to add a second Hall sensor, for example with a much higher rating, so you could tell when the current had dropped far enough to remove the bypass from the first Hall sensor. But that's beside the point because bypassing the Hall sensor is not feasible anyway.
     
  10. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015
    Hi...

    Your seeing the datasheet of LA-25NP but I am using LA 25-NP/SP11.

    Circuit can be modified in any way means adding any resistance and any other component to bypass the current.

    Thank you
    For your reference I am attaching the hall sensor data sheet of LA25-NP/SP11.
     

    Attached Files:

  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    OK, the /SP11 version has a primary resistance of "less than 51 mΩ" so it could be damaged by currents higher than the specified 1.5A, although again, the data sheet doesn't state a maximum allowable input current.

    Also, with the higher input resistance, it may be feasible to connect something across the device's input to shunt most of the current away from it. If you want a response in the microsecond range, you certainly can't use a relay, so a MOSFET with very low ON-resistance (RDS(on)) would be the obvious choice.

    But that would require quite a lot of complexity and would affect the circuit in many ways, so I would like to explore other options first. Please answer some more questions.
    • Do you have any thoughts on how you would decide when to turn the shunt OFF?
    • What is the maximum actual current that could flow in the circuit?
    • What factors limit this current? The power supply? The load resistance? Anything else?
    • Is the load inductive? Perhaps you could stop playing guessing games and tell us what the load actually is.
    • Does the current always flow in the same direction?
    • Is the current smooth DC? If there's any signal superimposed on it, what is the nature of that signal?
    • What is the maximum voltage that could ever appear across the load?
    • Do you have to use that particular Hall sensor?
    • Would you be able to connect a resistor of around 1Ω in series with the Hall sensor? (This would increase the required output compliance voltage of the power supply significantly.)
     
    Last edited: Feb 5, 2015
  12. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015
    Thank you for replying...

    If the relay contacts are closed and current only changing then, I think its possible to continue the relay. If not please tell me the suggestion.

    Coming to your questions
    1. If current is <1A then I want to make the shunt OFF. In this case entire current has to be passed through hall sensor.
    When >1A shunt should be ON.(Either current bypass or current shunting method can be suggested).
    2. Maximum actual current can be passed through this loop is 1A but for some testing purpose I want to send upto 10A with provision of hall sensor protection.
    3 .Load is resistive in the range of ohms only.
    4. Yes,current is always flowing in the same direction.
    5. Current is smooth DC only.
    6. Because better sensitivity in the measurement of current in range of 1A I am using this particular hall sensor. Is there any better way to measure the current in this range.
    7.Yes I can include 1Ω resistor in series with hall sensor.
     
  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    If you can include a resistor in series with the Hall sensor input, I think your best option would be a voltage clamp in parallel with the combined resistor and Hall sensor input. The simplest type of voltage clamp is a diode. The circuit looks like this:

    272558.001.GIF

    The Hall sensor input has a certain resistance, RHALL (the data sheet says it is "less than 51 mΩ"). A fixed resistor, R1, is connected in series with the Hall sensor input.

    These resistances add together to make a total shunt resistance that I will call RS. The voltage across RS is VS (marked on the diagram).

    A high-current diode, D1, is connected across RS. As the current through RS (R1 plus RHALL) increases, VS increases in proportion, due to Ohm's Law. At a certain point, VS will reach a voltage where D1 will start to conduct, and it will start to shunt current away from the Hall sensor circuit.

    For example, if D1 begins to conduct when VS reaches 0.5V, then you choose a value for R1 so that RS is 0.5Ω. Once the current through that path reaches 1A, VS will reach 0.5V and the diode will start to conduct.

    I've had a brief look for a suitable diode and found a possible option - the Vishay FES16BT. It's available from Digi-Key for USD 2.17. See http://www.digikey.com/product-detail/en/FES16BT-E3/45/FES16BT-E3/45GI-ND/2152949. Here is the VF vs. IF typical graph for that diode.

    epoint 272558 FES16BT I-V typical graph.png
    This graph shows the typical relationship between VF, the voltage across the diode in the forward direction (which is equal to VS in the circuit above) and IF, the current through the diode in the forward direction.

    EDIT ***** Previously (prior to 2015-02-10) I had estimated VF for IF=10A as 0.84V but I misread the graph; it should be about 0.88V, and the calculated value for RS is cleaner if I assume it's 0.885V. I have updated the following calculations to fix this error. *****

    If you look at the right hand curve (for TJ=25°C) you can see that an IF range of 1~10A corresponds to a VF range of about 0.7V to 0.88V (0.885V makes the calculations come out cleaner).

    If we want to limit IS, the current through R1 and the Hall sensor, to a maximum of 1.5A when D1 is shunting 9A or 10A, we need to choose R1 so that when VS is 0.885V, IS will be 1.5A. That will work with a typical diode in the D1 position.

    That works out to RS = 0.59Ω. If RHALL is 50 mΩ, that would mean R1 would be 0.54Ω. This would ensure that no more than 1.5A flows through the Hall sensor even when the total load current is 10A.

    Edit: The maximum power dissipation in R1 will be quite low, because VS is limited to about 0.9V and IS is limited to about 1.5A, so a 2W resistor will be fine for R1. /edit

    The problem is that we need the Hall sensor to measure up to 1A without interference from the diode. With RS = 0.59Ω, 1A current corresponds to 0.59V across the diode, and at VF = 0.59V, the diode will still conduct some current. The graph doesn't go below VF=0.6V, but it also doesn't show what happens below IF=0.1A.

    So with this solution, D1 will still pass some current and affect the Hall sensor's accuracy at currents of 1A and less. What is needed is a clamp with a VF vs. IF graph that's a lot closer to vertical.

    I think it would be possible to make a clamp circuit using a MOSFET with very low VGS voltage and a few other components to provide the steep V-I curve. The MOSFET I'm looking at is the Vishay Siliconix SIA436DJ which is available from Digi-Key - product page at www.digikey.com/product-detail/en/SIA436DJ-T1-GE3/SIA436DJ-T1-GE3CT-ND/3471119. This device is in a compact no-lead SMT package that would be hard to work with. Also, the clamp circuit would need about 2.0~2.5V across it before it could clamp; this would mean R1 would need to be around 2.2Ω.

    Please read this post thoroughly until you understand it, then tell me what you think.
     

    Attached Files:

    Last edited: Feb 10, 2015
    Sravanthi and BobK like this.
  14. Sravanthi

    Sravanthi

    72
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    Feb 4, 2015
    Thank you very much....

    Diode protection will work fine for my design.
     
    KrisBlueNZ likes this.
  15. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015
    HI...

    Can you suggest me a precision resistor of 0.51 ohms with wattage as 50W and tolerance as 1% or 0.5%.
     
  16. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I don't think you'll find a single resistor that meets those requirements. You should connect several resistors in parallel - say, nineteen 100Ω 3W 1% resistors in parallel and an 82Ω resistor and 50Ω trimpot in series, in parallel with the resistors. Then you can adjust the resistance to get within 1%. Also consider whether inductance will be a problem or not; standard wirewound resistors are inductive but you can get wirewound resistors that are made so they're non-inductive, and thick film and metal film resistors are also non-inductive.
     
  17. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015
    The configuration what you have given is giving 1 ohm resistor. This you have given as example to explain or actual arrangement.
    Can you clear a doubt that when we are connecting parallel and series configuration how to consider the wattage.
     
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    The resistor in series with the Hall sensor doesn't need to be much more than 1W. 51ohm resistors are common enough that you can get a few 1/2 w devices and connect them in series parallel.
     
  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Sorry, I meant nineteen 10Ω resistors in parallel, not 100Ω resistors.

    Did you read and understand this section of post #13?
    If you're using a simple diode clamp, your resistor doesn't need to be rated for 50W. At 10A diode current, that diode's typical VF is about 0.88V (not 0.84V as I originally said in post #13 - I have corrected it.) With VS = 0.88V and RS = 0.59Ω, IS will be about 1.6A and total power dissipation in RS will only be about 1.4W. Not 50W!
     
  20. Sravanthi

    Sravanthi

    72
    1
    Feb 4, 2015

    Sorry I have done some wrong calculation so mentioned 50 W.
    Yes I need 1.5W and 0.51 ohms resistance.So for this can you give suggestion in configuration...
     
    Last edited: Feb 10, 2015
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