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Led driver of constant current - is it really constant ?

Discussion in 'Power Electronics' started by kellogs, Jul 25, 2020.

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  1. kellogs

    kellogs

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    Jan 7, 2014
    I have never used led drivers and after some good hours of searching I am still at loss. In the specific case of the ZXSC380FHTA led driver, will the current be constant at the output and if it will be, to what value ?
    It says that

    [​IMG]
     
  2. Martaine2005

    Martaine2005

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    May 12, 2015
    The datasheet says for 18mA to use a 100μH inductor.
    By changing inductor value, the current increases or decreases.

    Martin
     
  3. kpatz

    kpatz

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    Feb 24, 2014
    To me that isn't really a "constant" current source, since the current varies with supply voltage. It's not a regulated current source in other words. It's sort of like a fancy, efficient resistor.

    Its purpose in life is to boost the supply voltage and allow LEDs to be driven from a single 1.5V cell, and it also controls (limits) the current, but you'll still find the LED will get dimmer as the battery voltage drops. Depending on your application this may not matter. I think this IC is commonly used in LED flashlights.

    To set the amount of current, you'd have to select an inductor value corresponding to the supply voltage. For example, for V = 1.5V, and L = 100uH, you'd get 18mA to the LED. But as the battery drains to 1.2V, the current would drop to 13mA, so you would get some dimming over the life of the battery.
     
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  4. kellogs

    kellogs

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    Jan 7, 2014
    Alright. Then it looks like i'd better go with a switching regulator for constant voltage. Or are the led drivers still a better choice for driving a led at constant current / voltage ?
     
  5. kpatz

    kpatz

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    Feb 24, 2014
    You want constant current, not voltage for driving LEDs. Voltage just has to exceed the forward voltage of the LED, which the chip you listed will provide even from a single cell.

    What's your application? That chip appears to be geared toward smaller LEDs at low mA usage, so something like solar garden lights or a small flashlight.
     
  6. kellogs

    kellogs

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    Jan 7, 2014
    my application is 3.2 V alkaline batteries into a 20 mA @ 2V led
     
  7. kpatz

    kpatz

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    Feb 24, 2014
    You could simply use a series resistor with a 3+V supply though a switching IC will be more efficient, thus better battery life.
     
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  8. kellogs

    kellogs

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    Jan 7, 2014
    I surely could, less the series resistor which i think not required for a voltage regulator sized at precisely 2V output. Just that it has more parts to solder.
     
  9. kpatz

    kpatz

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    Feb 24, 2014
    A LED, like all diodes, don't act like a resistive load. You can't reliably use voltage to control current through a diode. The Vf of a LED isn't always consistent across samples, or at different currents or temperatures. Without a resistor or some other form of current limiting, the LED will go poof.

    You'll need a resistor if you use a voltage regulator. A current controlled source is better, like the IC you referenced in your original post. With that chip you'll get some dimming as the battery voltage drops, since it's not holding current to a constant value when voltage changes. I don't know if this is a problem for your application or not.
     
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  10. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    You may want to read up on how LEDs work. We have a good resource covering this topic: "Got a question about driving LEDs?"
    Look for "LED driver ic" using the search engine of your choice. You'll be overwhelmed by the results.
     
  11. kellogs

    kellogs

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    Jan 7, 2014
    Yes, got it. Unfortunately not many led drivers out there that will drive just one led and do it properly. Back to SMPS.
     
  12. kellogs

    kellogs

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    Jan 7, 2014
    Or...

    Is there a minimum output voltage for the MIC2287 ? Datasheet only specs a max of 34V.

    [​IMG]

    Also, is an ~84% efficiency decent or kind of low for these chips ?
    Looking at the 3 series LEDs @ Vin = 3.0V, 20 mA graph. Judging by the graphs below I am speculating maybe 85% efficiency for just one LED.

    [​IMG]
     
  13. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Output voltage cannot be lower than input voltage with this boost topology.
    Minimum input voltage is 2.5 V so that sets the limit.

    In my opinion 84 % is acceptable given the comparatively simple circuit design. More complex designs claim to achieve up to 98 %. Is it worth the effort? It's up to you to decide.
     
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  14. kellogs

    kellogs

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    Jan 7, 2014
    How about this one ?
    LM334_current_source.png
    I am confused by this diagram. Would the voltage drop on R_set be 67.7 mV (@ 25 deg C) ? And the voltage across D1 1V perhaps, and then what, how does this work as a zero temperature coefficient current source ?
    And why are they redefining the V+ current as 2 * I_set when in the datasheet the V+ current is defined as I_set ?
     
  15. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Utterly inefficient: This is a linear design that will dissipate the most power. Would you really need zero tempco and accept the power dissipation instead?

    No. It is 64 mV as stated in the datasheet:
    Look at the internal diagram of the LM334 (datasheet figure 1). Rset connects to the "+" input of an op-amp. The "-" input is at Vref = 64 mV. The op-amp tries to get the voltage difference between "+" and "-" input to 0 V.

    The voltage across D1 is more like 0.6 V to 0.7 V. The absolute value is not important. The relevance of D1 is for temperature compensation. The LM334 has a temperature coefficient of +0.336 %/°C. A typical silicon diode has a temperature coefficient of -2 mV/°C. At a pass voltage of 600 mV this translates to -0.333 %/°C See how these values almost cancel?

    From the datasheet:
    So we have 1×Iset out of V- plus 1×Iset though Rset which adds up to 2×Iset.
     
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  16. kellogs

    kellogs

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    Jan 7, 2014
    Aha!
    Still, I am not on the same page as the datasheet. it clearly says

    Furthermore:

    Are they not defining I_set as the total current flowing into V+ pin ?

    Aside from that, if the diode does all the temp compensation, what would the R1/R_set divider purpose be then ? From the picture above:

    I might make it more efficient with a similar design to this one, yet maybe also incorporate the zero drift circuitry.
     
  17. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Hmmm As you phrase it it sounds different. To be honest, I have no clear idea.

    You cannot make a linear current source considerably more efficient (assuming the first design was not completely underengineered). Check this diagram:
    upload_2020-8-4_13-34-0.png
    With a linear circuit Iin = Iled, so total power dissipation is Ptot = Iin × Vin = Iled × Vin. Regardless of the construction of the current source. It can only become worse if the current source has its own path to ground and adds power dissipation on top of that.
    To raise efficiency you have to move on to a switched topology where Iin < Iled (for a buck regulator with Vin > Vout).

    If Vin < Vout, a linear current source will not be possible. You'll have to use a switch mode boost regulator in that case anyway.

    What for? If you want to go for temperature independent brightness of the LED(s), then you will also have to take into account the brightness vs. temperature characteristics of the LED. Keeping the current as constant as possible will not help.

    It might help to know what your underlying issue is at all. LED lighting? Constant current? Whatever else.
     
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  18. kellogs

    kellogs

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    Jan 7, 2014
    Of course... you are right, back to SMPS. Again.

    The thing with those SMPS (I am talking about voltage regulators as I was unable to find a suitable led driver for just one 2V led) is the utput voltage problem. Vf varies between 1.6 and 2.4V with the leds I want to use. I guess I could check each batch of leds bought for its Vf and use proper resistors to set the output voltage for minimal losses.

    Surely I would like the luminosity constant, and it does vary with temperature to about 10-15%, which is sort of ok on the limit. More variance than that is not ok, hence the need to regulate that current too. Also, getting the most juice out of a couple of Alkaline AAs is among my goals.

    Thanks a lot!
     
  19. kpatz

    kpatz

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    Feb 24, 2014
    I don't know what the application for your circuit is (what is it being used for) but maybe look at circuits used in solar garden lights, like these (there are 2 different circuits described on this site): http://www.talkingelectronics.com/projects/SolarLight/SolarLight.html

    These circuits include a solar charging circuit for a 1.2V cell, an oscillator that boosts the voltage to drive a white LED, and a circuit to cut off the LED during daylight. If you're looking for continuous operation (day or night) the solar cell and photocell circuits could be taken out.

    These solar light circuits don't have a current limiting resistor on the LED since it's driven with pulses, making it more efficient, and it's a low power circuit running from a single AA cell.
     
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  20. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    That should be no issue. This is why current drivers are used. Current is controlled, voltage follows, dictated by the characteristic of the LED.

    About I being 2 × Iset. I think this is how the calculation goes:
    The sum of the current out of V- plus the current through Rset equals Iset.
    The diode drops ~ 600 mV which is approx. 10 × Vref (not exactly). R1 = 10 × Rset, so the current through R1 is Ir1 ~ (10 × Vref)/(10 × Rset) = Vref/Rset ~ Iset
    The total current out ov -Vin is Itot = Id1 + Ir1 = Iset + Iset = 2 × Iset.
    Note that R1 requires some tweaking for best temperature stability.

    I think this circuit is not one of the best application circuits at all.
     
    kellogs likes this.
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