Current Limiter

Discussion in 'General Electronics Chat' started by edliu, Mar 16, 2011.

  1. edliu

    Thread Starter New Member

    Feb 22, 2011
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    Hi All,

    I need to design a circuit to limit a current to around 13A constantly to drive a hacker BLDC motor of model B40-20L. The input is about 11.1V. I have attached the circuit diagram with the parts I think I need.

    I know that I may not be correct and thus will appreciate if someone can advise me on what I may have missed out?

    Help is appreciated!

    Thanks! :)
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What is the value of Vcc? A linear voltage regulator such as this is going to dissipate a lot of power.
     
  3. Jaguarjoe

    Active Member

    Apr 7, 2010
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    The 2.2 ohm zener resistor is waaaaay too low. With an 11v diode driven by a 12v source(way too low, I know) you'll have 1v across the resistor and will have ~1/2 amp flowing. An 11v zener @ 1/2 amp is 5 watts dissipation. Totally out of the ballpark for a reference voltage. Higher source voltages only make it worse. FWIW, that is not the sense resistor. The unlabeled resistor across the B-E junction of the little transistor is.

    You show a 1k resistor in series with the load. To get 13 amps flowing, you will need 13,000 volts across it.
     
  4. edliu

    Thread Starter New Member

    Feb 22, 2011
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    I am sorry, but can you advise what reference voltage is needed? My objective is to get 11V and 13Amps out constantly.

    The unlabeled resistor, can you please advise what is the sense resistor in the B-E Junction? Can you advise what to change in the circuit?
     
    Last edited: Mar 16, 2011
  5. edliu

    Thread Starter New Member

    Feb 22, 2011
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    The Value of Vcc is 11 - 12V. Thanks for replying.
     
  6. Jaguarjoe

    Active Member

    Apr 7, 2010
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    You will need more than 11 volts into the regulator to get 11 volts out. Try 15 or so because the pass transistor will have a voltage drop across it. If you go too high the transistor will eat the difference in excess heat.
    You can not have constant voltage AND constant current at the same time unless you have a constant load, like a fixed resistor. The circuit you have here is a constant voltage regulator.
    Pick a zener diode that has a Vz of about 1/2 of the output voltage. 6.2 volts will be fine.
    Pick R1 to provide 10 or 20ma Iz to the zener diode: R = E/I = (Vin- 6.2Vz)/Iz.
    If your input supply has ripple you should put a filter capacitor across the zener.
    If there is a 6.2 volt reference on the (+) input to the op amp then the (-) input needs 6.2 volts of feedback from the regulator's output to balance out the op amp's inputs. Put a voltage divider across the output of the regulator. Set it up such that 11 volts out of the reg creates 6.2 volts out of the divider. 10ma or so is OK for the divider current. If you stick a trimpot in the middle of that voltage divider you will be able to tweak the output voltage to get exactly what you want.
    Pick R2 to just turn on the current limit transistor at the desired Cl (13a). The transistor turns on when its Vbe is ~0.6v, so R = E/I = 0.6/13 = ~0.05 ohms.
     
    Last edited: Mar 17, 2011
  7. edliu

    Thread Starter New Member

    Feb 22, 2011
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    What is a pass transistor?


    Why do i need to put a 6.2V zener as a reference voltage? Can anyone explain?

    "Set it up such that 11 volts out of the reg creates 6.2 volts out of the divider. 10ma or so is OK for the divider current. "

    Where do I set it up in my circuit?

    Thanks to everyone who looks as well as response to this thread! =)
     
  8. Jaguarjoe

    Active Member

    Apr 7, 2010
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    You can use any zener diode you want but 6.2 volt zeners have better characteristics than other voltage zeners because the drift coefficients cancel out at 6.2v. Even better yet, look for a 3 terminal adjustable voltage reference like a TL431.

    The pass transistor is the big regulating transistor that the 13 amps "passes" through. Depending upon the Vin, it may need a heatsink.

    The output of the regulator is exactly that, the regulator's output. The voltage divider goes across the output so it can provide a scaled down version of the output voltage that can be compared against the zener diode voltage. It goes right across the output, from (+) to (-). A fixed resistor on top from output (+) in series with a low value trimpot in series with another fixed resistor on the bottom to output (-). The wiper of the trimpot goes to the op amp. Pick a standard value for the trimpot, like 500 ohms. With the trimpot wiper centered at 6.2v, there is 250 ohms on either side of it. For a +/-1 volt trimpot range, I = E/R = 1v/250 ohms = 4ma. This will be the current through the whole divider. The voltage at the top of the trimpot will be 6.2v + 1v = 7.2v. The top resistor will be R = E/I = (11-7.2)/4ma = 3.8/0.004 = 950 ohms. The voltage at the bottom of the trimpot will be 6.2 - 1 = 5.2v so the bottom resistor will be 5.2/0.004 = 1.3k.
     
    Last edited: Mar 17, 2011
  9. edliu

    Thread Starter New Member

    Feb 22, 2011
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    With a 6.2V ref voltage, is my output to my device constant at 6.2V or 11V assuming that I have a Vcc of 15V?
     
  10. Ron H

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    Apr 14, 2005
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    I didn't see a driver transistor.:confused:
     
  11. Jaguarjoe

    Active Member

    Apr 7, 2010
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    That is the object of the voltage divider. It DIVIDES 11 volts down to 6.2 in direct proportion. Whatever the regulator output does at 11 volts, will be mimicked by the voltage divider at 6.2 volts so if the 11 volt output rises, the 6.2 volt divider output (the wiper on the trimpot) will rise also. That upsets the balance at the op amp input because the (-) input is now higher than the (+) input. The op amp output decreases causing the pass transistor to throttle back which lowers the regulator output voltage until the op amp inputs are balanced again. If the output of the regulator falls, the opposite occurs causing the pass transistor to open up just enough to get the output back to 11 volts.
    The zener diode will have a voltage tolerance, either +/-10% or +/-5%. This will make your reference voltage anywhere between ~5.6 to ~6.8 volts if its a 10% diode. The trimpot in the divider gives you a range of ~5.2 to ~7.2 volts to allow you to set the feedback voltage equal to the zener voltage as long as it is within a 5.6 to 6.8 range.
     
  12. Jaguarjoe

    Active Member

    Apr 7, 2010
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    Maybe after all these years of being told, "you must be halucinating" they were right :)
     
  13. Ron H

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    Apr 14, 2005
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    We should have advised you in the beginning that a switching regulator would be much better suited to your needs.
     
  14. edliu

    Thread Starter New Member

    Feb 22, 2011
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    Please advise the transistors to be used in order to support such current pls.

    Thanks for those reply...
     
  15. Ron H

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    Apr 14, 2005
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    Did you order heat sinks?
    My point was that a linear regulator is a huge waste of power, compared to a switching regulator.
     
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