Simple current regulator

Discussion in 'The Projects Forum' started by kyle7119, Feb 11, 2011.

  1. kyle7119

    Thread Starter Member

    Feb 11, 2011
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    I need to make a simple current regulator.

    Input voltage: 12V
    Max current: 550mA

    The device which I am powering normally draws about 450mA under normal operation. Since the voltage source is a battery, it will range from 12.5V - 10V during the course of several hours. I need to prevent inrush current spikes at power-up and would prefer to avoid using thermistors for limiting inrush current.

    I am still in high school and have very little experience in circuit design. Any help would be greatly appreciated. Learning is just as important as finding a solution to my problem. :)

    This is the simplest design that I have found so far, but I have not idea what values I need for the resistors.
    http://caladan.nanosoft.ca/c4/ccorner/4.php
     
    Last edited: Feb 11, 2011
  2. Wendy

    Moderator

    Mar 24, 2008
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    Welcome to AAC!

    You have given a power supply voltage, 12VDC.

    What is the max voltage out for the current regulator? It will be less than 12V, whatever it is, if you want 12V as a max then you will likely have to have more input voltage.

    I think you have given a maximum current required, 550ma. What is the minimum current required? Your wording suggests variable.

    There are some very good ways of doing this, but we need to pin the numbers down a bit.

    Note: I can't casually click on links, because I'm pretty paranoid about viruses. It is better you post the schematic.

    ***************

    Just reread your post, if someone doesn't beat me to it I'll see what I can come up with.
     
  3. kyle7119

    Thread Starter Member

    Feb 11, 2011
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    I don't need to regulate the voltage at all. When the battery is at 12.5V, the output voltage can be at 12.5V and when the battery is at 10V, the output can be at 10V.

    I'm not sure of the minimum current. We'll say 100mA for these purposes.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    It would help to know what kind of load you are powering.

    Perhaps the most simple current limiter would be a power resistor in series with your load. 12v limited to 550mA current would require roughly 12v/0.55A = 21.8 Ohms resistance, useful if your load acted as a dead short on start-up; for example, a large capacitor.

    Radio Shack sells a 2-pack of 10 Ohm 10 Watt power resistors; wired in series they would give you 20 Ohms resistance. Link:
    http://www.radioshack.com/product/i...1&filterName=Type&filterValue=Power+resistors
    You could then use a single switch across both resistors to remove the current limit (shorting across the resistors), or perhaps a couple of switches to gradually decrease the limiting from 20 Ohms to 10 Ohms, then just the resistance of the switches.

    The current regulators that you found will have a significant voltage drop across them, and will dissipate power as heat.
     
  5. kyle7119

    Thread Starter Member

    Feb 11, 2011
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    Well, the device is powered off of 12V and the normal current is about 450mA so that would mean that the load is about 26.7 ohms. On power-up, though, the load is much, much smaller than this.

    I am not sure how you figure that the current regulator will have a voltage drop. As I understand it, the only resistance that you affect the output voltage would be the top transistor (the bottom right resistor is of negligible resistance) and this would normally be open unless the current starts to spike. What I am trying to build is a circuit that will only provide resistance during current spikes.
     
  6. Audioguru

    New Member

    Dec 20, 2007
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    The transistor is an emitter-follower. When the load is 500mA then the 220 ohm resistor will have a voltage drop of 11V max plus the saturation voltage of the base-emitter which is 2.6V max on the datasheet. So the total voltage drop is 13.6V max. Therefore some transistors will do nothing with a 12V supply.

    The voltage drop will be about 1.9V max when the current is only 50mA.
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    I've taken the schematic you found and plugged it into LTSpice using some commonly available transistors; a TIP41 can be substituted for the TIP31.

    C1 is in parallel to Rload to simulate basically a "dead short" on startup. Note that the peak current in R2 (shown as a dark pink trace, I(r2), on the right) is approximately 570mA, eventually decreasing to ~460mA as C1 becomes charged, and all current flows through Rload.

    Note the peak power dissipation for Q1, shown as V(12v,Q1e)*Ic(Q1)+V(Q1b,Q1e)*Ib(Q1) on the right, will exceed 6 Watts during startup, eventually dropping to ~600mW.

    The lower set of plots shows the voltages at various points in the circuit.

    Note that V(out), the yellow trace, eventually settles to about 10.1v, or 1.9v below the source voltage.
     
  8. Audioguru

    New Member

    Dec 20, 2007
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    The simulation shows "typical" transistors that you cannot buy. Some have much worse spec's.
     
  9. kyle7119

    Thread Starter Member

    Feb 11, 2011
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    Hey, I actually just installed LTSpice the other night and was trying to hook up the same circuit. I didn't have as much luck, though, because I was very unfamiliar with how to use the software. Thanks for doing that for me.


    Is is possible to do the same thing while giving R2 a value close to 0.5 ohms? That should get rid of most of the voltage drop.

    I was showing this to my EE buddy today at work, and he seemed to think that it would work.
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    Another way to do it might be like this...

    [​IMG]

    Q1 needs to be a power transistor, and heatsink. It will drop around ½V at .5A. You need to have CR1 thermally connected to Q1, glue the diode to the transistor case.

    Took longer than usual for me, its been a long day.

    .
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    No problem, my pleasure.
    Be sure to join the Yahoo! group, LTSpice Users Group.
    Lots of models and symbols to download and add to your LTSpice, and lots of reading in the messages. Unfortunately, Yahoo! has started sending daily summaries of Yahoo! group activities, but I've routed those messages to my spam folder.

    That'll cause the maximum current to increase considerably. Iout ~= 0.75v/R2; so as it is now, Iout ~=0.75v/1.4 = .535mA. If you go 0.75/.4, then max current increases to 1.875A.

    If you want to have a better route, it'll involve using an opamp and a MOSFET to sink current from your load, and a resistor to sense the current. However, I don't know if you want to get involved in opamps at the moment.

    Like I said, decreasing the value of R2 will increase the maximum allowed current.

    If you want to try a MOSFET current sink version, I'll post that up. You'll need an N-ch power MOSFET, a low-value resistor, a voltage reference and an opamp that has a common mode input that includes the negative rail; an LM358 is a cheap dual opamp that satisfies the criteria; an LM324 quad opamp, also cheap, could be used as well.
     
  12. kyle7119

    Thread Starter Member

    Feb 11, 2011
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    Where do you find the TIP31C transistor on LTSpice? I have the lastest version (v 4.11), and I can only find the 2N2222.
     
  13. kyle7119

    Thread Starter Member

    Feb 11, 2011
    71
    17

    Hahaha, my EE buddy also told me the same thing. He encourged me to look into the MOSFET route because he thought that it would be a better design. He even started to whip up a schematic for me, but then got halway though and was like, "you get the idea." Yeah, I wish... :)

    Yes, I would be interested in see that schematic.
     
  14. kyle7119

    Thread Starter Member

    Feb 11, 2011
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    SgtWookie,

    Would you give me a quick explanation on how to run the simulation to get the output windows that you got. I'm not clear on what all of the different settings on the "run simulation" box do.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Bill,
    Your NPN transistor is backwards (emitter-collector swapped), so the emitter-base junction will avalanche once it exceeds ~6v.
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    I just did a simple transient analysis.
    Click "Simulate" on the menu bar, then "Edit simulation command"
    A pop-up dialog appears.
    For the box "Stop Time", type in 10m (that's for 10mS)
    Click the box to the right of "Start external DC voltages at 0v" so there is a checkmark.
    Leave the rest blank.
    In the box at the bottom, you'll see ".tran 10m startup" appear; leave that alone. Then click the OK button.

    Then on the menu bar, you can click the little black running man to start the simulation. It should either finish within moments, or error out if you've done something wrong or forgotten to select values for things like caps and resistors.

    You must have at least one ground symbol in the schematic somewhere or it won't run. The ground symbol should be connected to a wire; otherwise it'll run forever and you'll get strange error messages. Everything in the circuit must have a path to ground somehow.
     
  17. kyle7119

    Thread Starter Member

    Feb 11, 2011
    71
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    New design.

    This one is very similar to the first but doesn't have the voltage drop problem. However, the current limiting characteristics lead to some interesting patterns...
     
  18. kyle7119

    Thread Starter Member

    Feb 11, 2011
    71
    17
    I would still be interested in seeing that MOSFET/opamp circuit if someone has a minute to draw it up.
     
  19. wayneh

    Expert

    Sep 9, 2010
    12,122
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    I'm using the attached current regulator circuit. It works fine, at least for my use. But I may have missed the point of all this.
     
  20. kyle7119

    Thread Starter Member

    Feb 11, 2011
    71
    17
    Thanks, I'll take a close look at that!

    The point of all of this is to make a current regulator which prevents current spikes, especially at power on. I have a device for which I swapped the stock power supply (12V) for my own 3 cell Lithium Polymer battery (12.6V - 9V). The problem is that this battery is not regulated like the original power supply and will supply much more current (44amps max).
     
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