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Discussion in 'Electronic Basics' started by [email protected], Jul 18, 2005.

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  1. Guest

    I have a bunch of LEDs that I want to wire in series of 3, and then
    wire each series to my battery in parrellel.

    I have 3 LEDs that are rated for a FV drop of 3.0-3.4V, and draw 20mA.
    If I wire them in series, and aim for 3.2V each, that would give me
    9.6V. So powering them with an 8-pack of NiMH (1.2V) AA's would be

    Would I still need to use a resistor? I'm thinking yes because of the
    current, but I don't know how... one of the formulas I found to
    calculate for a resistor would yeild a zero, because zero volts would
    need to be resisted:

    Resistance = Voltage / Current

    0 Ohms = 0 Volts / .020 Amps

    So how do I limit the current, or will it be OK to not use resistors?

    --Farrell F.
  2. John Fields

    John Fields Guest

    You have to limit the current because there's no guarantee that the
    LEDs will have a Vf of 3.2V at 20mA and if it's lower than that you
    could fry the LEDs. If it's higher than that, the LEDs might light up
    dimly. The easiest way around that problem is to wire two LEDs in
    series, assume a Vf, for the pair,of 6.0V worst case, then bite the
    bullet and use a series resistor to drop the remaining 3.6V at 20mA.

    The resistance will need to be:

    Vbat - Vled 9.6V - 6.0V
    R = ------------- = -------------- = 180 ohms
    Iled 0.02A

    And the power the resistor will dissipate will be :

    P = IE = 0.02A * 3.6V = 0.072W

    So a standard 180 ohm +/- 5%, 1/4 watt carbon film resistor will be
  3. Guest

    I ended up going with 160 Ohm resistors... 3.2V per LED is what I want
    to aim for... a little brighter, even if it will be a little more

    i just want to know if its possible to just only current, and not
    influence volatage? From what I read about Ohm's law, I'm guessing No.
    I'm just curious, since I already order'd the resistors I'm going to
    use them, but what if I needed to convserve the most amount of energy.
    Is there any other device that could be used to control the flow of
    power without using it up (ok, lets be real... using less of it up) ...
    more efficient than dissapating it as mostly heat? Something that can
    provide a fixed current and voltage?

    Sorry if I sound really new, but I sort of am, and I'm slowly
    understanding Ohm's law.

    --Farrell F.
  4. I think what you mean to ask is `Is there a device that will put a constant
    current through an LED without wasting a lot of power?'. Because the voltage
    dropped by the LED is very close to constant, and P = V*I, if you keep the
    current through an LED constant then you keep the power dissipated by it
    approximately constant as well.

    The answer is yes. A resistor will make the current through the LED somewhat
    constant, as long as the voltage drop across the resistor is large. Your LED
    might be specified as dropping 1.8 V, but it will never be exactly 1.8 V; it
    might be 1.7 V or 1.9 V, depending on temperature and process variations.
    With the simple resistor circuit, you have to make the voltage across the
    resistor large. Otherwise a very small change in the LED's voltage drop will
    result in a very large change in the current. You can do some calculations
    to convince yourself of that. As you noted, that means that a lot of energy
    will be wasted as heat, because you need a lot of voltage `headroom' to
    cover the drop across the resistor.

    It is possible to build linear constant current sources that will maintain
    an almost perfectly constant current through the LED, independent of the
    voltage dropped by the LED or the supply voltage. It is not possible to do
    this with only resistors; these circuits use transistors and integrated
    circuits. They still dissipate the power due to the extra supply voltage as
    heat, but they don't need as much headroom; you could run an LED that
    dropped something between 1.4 and 1.8 V from a 2 V supply, and the
    brightness (current) wouldn't be all over the place. Since you don't need as
    much headroom, these circuits can be more efficient.

    Still higher efficiency is possible using switchmode power supply
    techniques. Depending the supply voltage that you start with and what kind
    of LED you are running, you might be able to get efficiencies above ninety
    percent. You can buy little prepackaged modules that do this, and there's
    lots of special-purpose ICs as well. Nobody would bother with that for a
    little front panel indicator, but for the large white LED flashlights enough
    power would be wasted that people think it is worth the trouble.

  5. Yes, there is, but that is beyond your understanding at this point (see
    the following sentence for the reason why).
    Since your LED is dropping 3.2V it is likely that it is a white or blue
    LED. If so, I have some important info for you.

    I have been running some white LEDs at 20mA and 30mA for a number of
    months 24hrs, 7days/wk. I have found that there isn't much "risk" in
    running LEDs at or above their maximum current. WHat happens is that
    the light output drops dramatically after a thousand or more hours.
    Some cheaper LEDs running at only 20mA drop to less than a third, more
    like 1/10, their light output after less than a thousand hours, which is
    only a month and a third, or 42 days.

    If you notice, some white LED specs give an "initial" light output which
    might be 10,000 mCd. But after a thousand hours, the output might be
    closer to 6,000 or 7,000 mCd, so they use the term 'initial' just to be
  6. John Fields

    John Fields Guest

    You misunderstand. Since the LED is a diode, you don't have direct
    control of the forward voltage; it just comes out to be whatever it is
    when you force a certain amount of current through the LED.

    For the LEDs you're using, with 20 mA going through them, the voltage
    dropped across any of them them could vary from 3.0V to 3.4V, so for
    two LEDs on the low end in series with a 160 ohm resistor, the current
    in the circuit would be:

    Vs - Vled 9.6V - 6.0V
    I = ----------- = ------------ ~ 22.5mA
    R 160R

    Which should be OK unless your LEDs are spec'ed at 20mA max, as some
    white ones are. Running them at a small overcurrent like that will
    only degrade their output over time, and shouldn't lead to a
    catastrophic failure.
    Good guess, but...

    Diodes don't exhibit a fixed resistance in the forward direction, so
    while Ohm's Law:

    E = IR

    still holds, R decreases as I Increases, so E will have to change very
    little as it causes very large changes in current through the diode.
    Forget about a fixed voltage. What you need is something which will
    provide a fixed _current_ and let the voltage across the load vary as
    it wants to.

    If you want the most efficient way of doing it, then you need to use a
    current-regulating switching supply.
  7. Guest

    Thanks everyone.

    Quoting John:


    For the LEDs you're using, with 20 mA going through them, the voltage
    dropped across any of them them could vary from 3.0V to 3.4V, so for
    two LEDs on the low end in series with a 160 ohm resistor, the current
    in the circuit would be:

    Vs - Vled 9.6V - 6.0V
    I = ----------- = ------------ ~ 22.5mA
    R 160R


    but isn't that assuming the worst... a 3.0V drop from each LED? if it
    were 3.2V that would become 20mA and if it were 3.4V it would become

    don't get me wrong, it's best to assume worst, but I just want to make
    sure I am understand correctly. since this project of mine is just to
    make some headlights, or rather stadium-type lights, for my RC car, I'm
    not terribly concerned if they slightly diminish in brightness after
    1000 or even 200 hours... as they wont be on for more than perhaps
    30-40hrs a month at absolute most, more like 5-6hrs on average.

    Thank again,
    --Farrell F.
  8. John Fields

    John Fields Guest

  9. I think the best answer for the OP has been done. But one thought
    that crossed my mind, when I allowed it to wander on handling a wide
    range of V+ and widely varying Vfwd of the LEDs (different colors,
    even) is something like the following. It's very insensitive to
    variations in the Vfwd of the LEDs and also relatively insensitive to
    the V+ supply, which I kind of guessed might vary over a span of a
    volt or more from reasonable use given that it is proposed to be a
    stack of 8 NiCds:

    / R3
    \ 270 V+ V+ V+
    / | | |
    | | | |
    | | | |
    | | | |
    | | | |
    Q2 e>| |<e Q3 |<e Q4 |<e Q5
    |---+--| ,--| ,--|
    V+ c/| | |\c | |\c | |\c
    | | | | | | | |
    | | | | | | | |
    | | | | | | | |
    | +-----+--- | ----+--- | ----' |
    Q1 c\| | | | |
    |---+ | | |
    e<| | --- --- ---
    | | \ / LED1 \ / ... \ / LED N-1
    | \ --- --- ---
    | / R1 | | |
    | \ 1200 | | |
    | / | | |
    | | --- --- ---
    | | \ / LED2 \ / ... \ / LED N
    '-----+ --- --- ---
    | | | |
    \ | | |
    / R2 | | |
    \ 10k | | |
    / gnd gnd gnd

    Basically, a cheap multiplying current mirror with Q1 designed to
    siphon off most of any current increase in R2 due to V+ changes.
    Thus, the current through R2/Ic(Q2) remains fairly fixed over V+
    changes, which tends to hold the LED current fixed. I just gave some
    gross example values for the resistors (R1 to be 0.6V/500uA, for
    example) without thinking much on them. LED strings can be simply
    added with just a BJT per string.

  10. I meant R1/Ic(Q2), as it is the current through R1 that must flow
    through Ic(Q2), except for a little to Q2 and Q3's base.

    Here's my simplified design steps:

    (1) Note the desired LED current: 20mA.
    (2) Estimate the base drive current for the BJT sourcing that
    LED current. I estimate the beta at 200, so I get 100uA.
    (3) If more than one string may be added and all of them must
    work about the same as if there was only one string, then
    estimate Ic of Q2 at say 50X as much -- 5mA.
    (4) Guess at the Vbe of Q1 at 0.66V, for now.
    (5) Set R1 = 0.66V / 5mA or 132 Ohms. Round up to 150 Ohms.
    (6) Compute new estimated I(R1) at .66V/150 = 4.4mA.
    (7) Figure R2 at the minimum voltage to drop at this current.
    With an estimated 8.2V minimum (let's say) and a gross
    estimate of .7V+.66V for the two Vbe's of Q1 and Q3, this
    means (8.2-.7-.66)/4.4mA = about 1550 Ohms. Call it 1500.
    (8) Estimate Ic(Q2) as being about R1's current minus two base
    drive current loads for modest measure, or 4.4mA - 2*100uA
    or 4.2mA.
    (9) Since Vb(Q2)=Vb(Q3), Ic(Q2)*R3+Vbe(Q2)+Ve=Vbe(Q3)+Ve. I'm
    calling the I*Re=Ve, which is a constant 26mV, or so. They
    cancel so, Ic(Q2)*R3=Vbe(Q3)-Vbe(Q2). But this Vbe differ-
    ence is 60mV*log10(Ic(Q3)/Ic(Q2)), so we compute R3 as
    or about 9.7 ohms. Call it 10 ohms.

    All that will probably not give 20mA exactly. But what it does
    provide will probably be pretty insensitive to the Vfwd of the LEDs,
    up to a limit of course.

    / R3
    \ 10 V+ V+ V+
    / | | |
    | | | |
    | | | |
    | | | |
    | | | |
    Q2 e>| |<e Q3 |<e Q4 |<e Q5
    |---+--| ,--| ,--|
    V+ c/| | |\c | |\c | |\c
    | | | | | | | |
    | | | | | | | |
    | | | | | | | |
    | +-----+--- | ----+--- | ----' |
    Q1 c\| | | | |
    |---+ | | |
    e<| | --- --- ---
    | | \ / LED1 \ / ... \ / LED N-1
    | \ --- --- ---
    | / R1 | | |
    | \ 150 | | |
    | / | | |
    | | --- --- ---
    | | \ / LED2 \ / ... \ / LED N
    '-----+ --- --- ---
    | | | |
    \ | | |
    / R2 | | |
    \ 1500 | | |
    / gnd gnd gnd

    So, how better might one BJT per LED string plus two extra BJTs be
    used to improve the insensitivity to Vfwd and V+?? No op-amps please.
    That's cheating, big-time! And no JFETs. Just BJTs. I'm interested
    in learning.

  11. ehsjr

    ehsjr Guest

    For a 1 off, if the resistor/2 LED series combo is not
    to your liking, I'd do it like this:

    ----- 250
    +Vcc -----in|78L05|out---[R]---+---LED--LED---Gnd
    ----- |
    | |

    and repeat as often as needed to run all the LEDs.

    I like it better than the mirror, as each LED string
    is a "2 wire module". The 78L05's are 4 for a dollar
    so the cost isn't great, and it's only 2 parts per
    string, so its simple. (The mirror saves on
    parts count once the number of strings exceeds 5, if
    that's a factor.)

  12. Actually, I was more interested in playing around with using BJTs, as
    I'm just a hobbyist and still at that level of things. I like to
    learn to be more facile with the basic tools and I don't learn much
    when I use opamps with many dozens of transistors, some of them in
    configurations I cannot possibly buy and play with, and yielding
    abstract behaviors that are much easier to design with; or with
    voltage regulators, themselves of similar complexity and thus both
    easier to use but also not nearly as educational to use.

    Also, I get 2N3904 and 2N3906 BJTs at one cent per, on sale. This is
    $10 for a thousand of the things. Dirt cheap. So rather than 25
    cents per string, it's one cent per string. May not make any real
    difference as the labor time is everything, if you are building this
    as a one-off for yourself. But still, they are cheaper.

    Mostly, though, it's just about learning to think in various ways with
    simplified (and somewhat wrong) models. Then learning to build on
    that to think about 2nd and 3rd order effects and to deal with those,
    etc. I'm curious, right now, about what effects hamper the design I
    mentioned and how those effects can be better managed with a different
    topology that I haven't considered, using the same number of BJTs to
    get there. That is the kind of nuanced thinking I'd like to listen
    more to, just now.

  13. Bob Monsen

    Bob Monsen Guest

    It wastes a bit of energy. An LM317 would be a better choice, due to its
    lower voltage differential.

    Also, if you want to vary the brightness on N strings, you need to swap
    out N resistors.

    If you combine your idea of using the 7805 (or 317) with Jon's circuit,
    you get a stable current sink which has relatively good tempco with the
    ability to adjust the current on all the LED strings at the same time.
    The only thing I'd change in Jon's circuit would be to add some 10 ohm
    emitter resistors to even out any differences between the transistors.
    Also, the more strings he adds, the worse the matching between source
    and string gets, due to base currents. This can be fixed using a
    transistor as a follower, in the traditional fashion.

    He would have to recalculate the appropriate multiplier resistor (R3) in
    this case. He wants 20mA through each string, and 500uA through the
    control transistor, so

    Ia/Ib = 40 =~ e^((va-vb)/vt)


    vt*ln(40) = va-vb

    but using 10 ohm emitter resistors,

    va + 20mA*10 = vb + 500uA * R3


    vt*ln(40) = 500u*R3-20m*10 ==> R3 = 587 ohms

    Or something like that. Then, the resistor between Vout and ground for
    the 7805 would be 10k.

    Bob Monsen

    If a little knowledge is dangerous, where is the man who has
    so much as to be out of danger?
    Thomas Henry Huxley, 1877
  14. I am just jumping into this thread, and suspect that others may have
    already advised against matching voltage. Reason: Current actually drawn
    is probably excessively inpredictable and varying excessively with
    temperature. Combine with this: LEDs get more conductive as their
    temperature increases.

    Better off to:

    Put these LEDs in series pairs, with each series pair in series with a
    resistor, and then put all of these "series strings" in parallel with each

    Sounds to me like 3.2V is typically across each resistor, and for 20 mA
    the resistance would be 3.2V/.02A which is 160 ohms, and the wattage would
    be 3.2V*.02A which is close to 1/10 watt.
    Closest popular resistor alue is 150 ohms and for conservative operation
    the next one up is 180 ohms. Resistor reliability is best when wattage
    rating is well above actual dissipated wattage, so I advise to use
    resistors of wattage rating more than 1/8 watt - as in the next popular
    wattage rating up from that, which is 1/4 watt. Although I do expect (but
    not guarantee) you to get away with resistors rated 1/8 or 1/10 watt in
    prototypes or a small quantity production that will not have to survive
    adverse conditions (in temperature and/or supply voltage and/or thermal
    design of the product).
    If the product needs to survive being operated in sunlight or a
    sun-baked car when the air temperature is 40 degrees C and you need LED
    life in the 10,000's of hours and component packing density is high, then
    I recommend using 470 ohm 1/4 watt resistors each in series with two
    LEDs [assuming 9.6-10V supply] and hope the LEDs are bright enough at 8-10
    mA (which they probably will be).
    I advise to redesign to use resistors.

    Some guidelines I have in:

    - Don Klipstein ()
  15. This requires looking into "LED driver" circuits. Some of these
    circuits use "LED driver" and "white LED" ICs. This is an extra step of
    Please consider that with diodes including LEDs, it is better to
    consider current to be the independent variable and voltage to be the
    dependent variable as opposed to the other way around. Voltage as a
    dependent variable changes at least sometimes annoying little with current
    and at least sometimes annoying excessively with temperature and with
    production tolerances of the diodes!
    Circuits that regulate (even if only somewhat) current or wattage
    generally do well for powering LEDs. Feeding LEDs regulated voltage or
    voltage as an independent or regulated variable tend to be less

    - Don Klipstein ()
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