Quick question about a Constant Current Source

Discussion in 'The Projects Forum' started by ke5nnt, Oct 28, 2011.

  1. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    I've never toyed really with a constant current device. I'm understanding that the MOSFET Q1 will dissipate power based on (Vin - Vdrop at load) times current. Question is, if there's a larger than nominal voltage left over that Q1 has to absorb, can I put a resistor in series with my load to drop a couple more volts in an effort to reduce the power dissipation of the MOSFET? I'm assuming I can, but assumptions have gotten me in trouble before.

    Thanks.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    You can a a resistor to reduce the transistor power dissipation as long as the voltage drop across the resistor does not reduce the voltage drop across the transistor to less than zero at maximum load and minimum supply voltage.

    Note: Don't know where you got that circuit but it is incorrect and will not work. If you want to use a positive supply voltage then the MOSFET must be an N-channel, not the P-channel you show. Also you would need to reverse the collector and emitter connections of the NPN transistor (forward biased current flow in a bipolar transistor is always in the direction of the base emitter arrow).
     
  3. Wendy

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    Mar 24, 2008
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    That will work, as stated, and I am puzzled by your constant current source also.

    I prefer to use diodes, since they tend to be more fixed voltage, while resistors aren't.
     
  4. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    That must have been an error on my part when drawing the schematic last night. Did it just before bed and was in a rush. THIS circuit is what I am using, just with a different load.

    Bill, can you provide either an example or an explanation of your diodes solution?
     
  5. crutschow

    Expert

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    Don't understand your comment.:confused:

    The OP's circuit uses the base-emitter diode as the reference voltage which has the same stability as diodes. When the voltage across R3, due to current through Q2, equals the base-emitter junction voltage of Q1, Q1 starts to turn on and reduce Q1's bias voltage, which limits the current. Thus the current limit point is ≈ 0.65V/R2. R1 is just a bias resistor and its value has only a small effect on the current limit point.
     
  6. Wendy

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    OK, lets start off with this circuit...

    [​IMG]

    You have a 12V power supply, and are feeding a blue LED 700ma at 3.6VDC Vf.

    The LED is dropping 3.6VDC X 0.7A, or 2.5W.

    Mean while the LM317 is dropping 8.4V at 0.7A, which works out to almost 6W.

    If you put a diode (such as a 1N4007) inline with the constant current it will absorb ½W, this heat will be removed from the LM317.

    It so happens I have a working example of something like this, although the reason I'm doing it is different.

    [​IMG]
     
  7. crutschow

    Expert

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    Ok, now it's quite clear what you meant by using diodes instead of resistors. I was just a little slow on the uptake earlier.:rolleyes:
     
  8. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    In your example Bill, does the diode absorb 1/2W because it produces a "set" voltage drop. In your example of 700mA, the diode would be producing about a 0.7V drop across it at 1/2W.

    Assuming my theory is correct, I have a 14V source, a 6V load leaving 8V to be dissipated by Q1, at 85mA, that's 680mW. If the Diode Vdrop is set at 0.7V, placing it in series with my load, the diode would dissipate 59.5mW. Not overall that advantageous in my circuit without placing many diodes in series. That assumes that 680mW being dissipated in Q1 which has a 2W max power rating is really even an issue. I just worry about the .68W making Q1 really hot, considering a 2W max is most likely the max when the device is heavily heat-sinked.

    If I place 2 one-watt resistors in parallel, and the parallel bank in series with the load to drop about 6 more volts:

    110Ω parallel to 200Ω = Req of 71Ω. 71Ω at 85mA total current = 6V drop across each resistor. The 110 would dissipate about 327mW. 200 about 180mW and Q1 about 170mW. With 1W resistors, these values shouldn't be a problem, even factoring fluctuations within' the resistors due to heat or other factors. Everything should have a decent "safety cushion".

    Of course, all of that depends on the 2 things stated earlier.
    1. That the diode drop is fixed at 0.7V
    2. That 680mW dissipation in Q1 is even a problem

    Thanks for the help! :)
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    That was just an example, Zener diodes are also an option, the idea is to remove some of the loading from the power component, but to do so using fixed voltage increments. A resistor will vary a bit, but will work OK.

    In your example the power is quite low, so it probably isn't a problem.

    Just curious, I tend to use single BJT (or high gain variates) as constant current sources, because a BJT is basically a constant current source from the git go, and can drop a minimum of voltage doing so. Why add a MOSFET?

    Can you show the source for this circuit? I would like to investigate the advantages/disadvantages.

    Here are a couple of other designs. The first is eminently practical. The second works, but tends to have problems with temperature stability. Inside an IC this is not a problem, and you can get transistor arrays. With two discrete matched transistors the partial solution is to glue the cases together to keep them at the same temperature.

    [​IMG] [​IMG]
     
  10. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Thanks again Bill.

    The original constant current idea comes from This Page

    Attached is my application in an abbreviated circuit. The intent is to limit current in the LEDs to around 85mA. As discussed earlier, and subject to change, is the 1W resistor bank which drops 6V and divides power dissipation between Q1 and the 1W resistors.

    Voltage enters this portion of the circuit above R5, which is a 100K resistor. Q2 and Q1 control the current. Q3 is a small signal NPN MOSFET for high-speed switching of the LEDs. Switching is controlled by the micro, and rate of switching can be changed by depressing S1.

    In the original circuit for the current source, too much current would start to turn on Q2, which would turn off Q1 keeping current constant.

    I added a TVS diode at the source in reverse bias to shunt transients from the battery. I worry about the effects of placing the electrolytic capacitor C1 there in the event of a transient voltage, and whether or not the over voltage would just destroy the capacitor or not.

    Anyways, suggestions about improving this by reducing parts and or increasing efficiency/effectiveness would be great if you have any. Seems threads I start always end up 2 or more pages long :rolleyes:
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Are the LEDs red? They will drop 2.2V to 2.5V I'm thinking. 85ma?

    I have a design I've never used I would like to run pass you, it is for a floating constant current source.

    I'll post the drawing when I make it.

    BTW, using a CMOS output from the pic, R3 is not needed. Might even be a bad thing.
     
  12. Wendy

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    BTW, why not use a LM317? Two parts, and the switching MOSFET. Much simpler.
     
  13. Wendy

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    OK, here is how I would do your LED drivers. You could eliminate R4, it is to tweak the value. Since the LM317 is 1.25V ±0.05 V and the resistors are 5% there are minor tolerance issues.

    [​IMG]

    If the LEDs are 2.2V Vf (worst case), and the power supply is 13.7V, then the LM317 would dissipate 0.6W, a very manageable number. You can use TO220 heat sinks such as these to raise that value.

    http://www.bgmicro.com/to-220sliponheatsink.aspx

    I just picked up around 10 or so for myself.

    The floating current regulator looks something like this.

    [​IMG]
     
  14. crutschow

    Expert

    Mar 14, 2008
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    MOSFETs also have a high drain impedance in the active region so act as a constant current source, similar to BJT. The only advantage of using a MOSFET, that I can see, is a slight increase in efficiency since the wasted bias current is less.
     
  15. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    The LM317 looks promising the more I look into it. I have 2 strings of LEDs. Either string can be Vf of about 2.0 to 2.2 or 3.0 to 3.2V. I've read in one location that using the LM317 as a constant current device, the LM317 will have a forward voltage drop itself of 3V. I'll be running into problems with not enough voltage on the 3.2V string if that's the case. 3(3.2V) + 3V = 12.6V total drop. With a supply voltage that can range from 10.5V worst case... that could be a problem.

    That assumes the 317 has that kind of voltage drop.
     
  16. Wendy

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    That sounds about right. What is your minimum voltage requirement?
     
  17. ke5nnt

    Thread Starter Active Member

    Mar 1, 2009
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    Minimum voltage of about 10.5V. Entire circuit operating temperature range ideally -40 to +85 C.
    0 degree components would be unacceptable for this application.
     
  18. Wendy

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    0 degree components?
     
  19. steveb

    Senior Member

    Jul 3, 2008
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    He's probably talking about commercial grade operating temperature range of 0 to 70 degrees C.

    Industrial grade is -40 to +85, which is a good choice when you have the option for choosing components.

    I'm sure he would have told you that soon, but by me adding a comment it makes it easier for me to track this thread, which i'm reading out of interest.
     
  20. mixed_signal

    New Member

    Dec 5, 2009
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    This circuit in post #10 is right out of Gray & Meyer, Chapter 4 appendix in the 3rd edition, with the output device replaced with a MOSFET. It is a low cost current source, with the issues noted that the current will have some dependence on the supply voltage and the device characteristics over temperature.

    However, it is a current source, rather than a voltage source, which is good for LED applications, since the light intensity varies with current (not voltage). (And it's cheap, too.)

    By the way, one advantage of a MOSFET here is that the lack of gate current means the circuit will be more 'stable' (constant) over conditions compared to the BJT equivalent where the high current output device will have a relatively high base current compared to the low current reference side's quiescent current.

    Also, if the MOSFET you're using can handle the power dissipation (Vds * Id) then you do NOT need series resistors with the LEDs, for further cost savings. They can help lower the power dissipation in the MOSFET if needed, but since you're regulating the current through the LED string they serve no other purpose. That is unlike the voltage-mode circuits mentioned by Bill and Carl which rely on the resistor drop and the regulated voltages along with the LED voltage drops to define the current... you can see how direct current regulation should be a little better.

    -Scott
     
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