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newbie series-parallel circuit question

Discussion in 'Electronic Basics' started by the_doug, May 24, 2004.

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  1. the_doug

    the_doug Guest

    I'm a software developer that has recently decided that electronics
    would be a *fun* hobby. i'm not too far into it and I'm already
    stumped... In this little project that I've invented for myself I'm
    trying to light an array of twelve flashlight bulbs simultaneously...
    and can't seem to manage it. :| Please excuse my idiocy in

    the bulbs say they are rated for 2.33V and 270mA. I got a 6V, 300mA
    dc wall adapter that I'm trying to use to power the deal. If i throw
    22 Ohms worth of 1/2 watt resistors in front of a single bulb - i have
    no problem lighting it... but those resistors start getting Hot.
    Eventually - I'll want to be able to selectively switch each bulb on
    and off, so i'm aware that i'll need some sort of parallel
    configuration... but I'm just not sure if i'm doing the right thing
    by generating so much damn heat.

    feel free to give me hints, suggestions, schematics - or anything else
    you have a moment to impart upon an idiot. :)
  2. bench

    bench Guest

    You can't do it with this transformer because you will not have
    enough wattage to light all bulbs. What sort of bulbs are these, 2.33V
    is a strange rating, usually one finds 1.5/3/4.5/6 V Have you considered
    working with LED's, you can do it with LEDS. If you want to do it this
    way simply connect all the leds in parallel and put a resistor of 500ohms
    after each LED. You can also have switches as required.
  3. Soeren

    Soeren Guest


    (the_doug) wrote in
    That would be 2.33 * .27 = 0.63W (or 630 mW)
    12 of these would amount to a little over 7.5W
    It is only able to deliver 6 * 0.3 = 1.8W, not nearly enough !

    Of course, it gets more than 0.8W, you would need a higher wattage
    resistor to keep ot from "melting" :)

    First of all, you would need a beefier supply to get all 12 lamps to
    light up at once.

    Resistors are not the best way to keep your lamps alive, a voltage
    regulator (delivering 2.33V @ 3.3A) would be a better solution.

    If you want to use resistors, you should choose the right values:
    A lamp of 2.33V/270mA has a resistance of 2.33 / 0.27 = 8.63 Ohms.

    If your supply is 6V, the needed resistance would be:
    (6V - 2.33V) / 0.27 = 13.6 Ohms.
    You then select the next _higher_ standard value of 15 Ohms.
    The current possible is then: 6 / (15 + 8.63) = 0.254A.

    The power dissipated in the resistor is then:
    15 * 0.254^2 = 0.97W so you need at least a 1W

    The current is slightly reduced compared to the nominal rating, but this
    will give the lamp a longer life :)
  4. bench

    bench Guest

    A voltage regulator *is* a resistor, albeit a semiconductor one.
    What is required is to match the voltage of the supply to the
    voltage of the bulbs. Therefore, it would be better to choose
    standard 3V bulbs and a 3V transformer that can supply enough
    current, any other way will be a waste of energy.
  5. the_doug

    the_doug Guest

    I greatly appreciate the help everyone. I found this interesting...
    This calculation should probably have been somewhat obvious to me -
    but I hadn't made the connection to apply Ohm's law to this info.
    It also explains why my calculations were causing me so much grief...
    I had done a manual measurement of the bulb resistance and found it to
    be ~1.6 Ohms. Is this difference between spec and measurement due to
    lamp resistance changing under heat?

    This is probably as good a time as any to post a follow-up question,
    and bear with me if i do a poor job of explaining myself... If I were
    to decide to run this circuit in a series - And Still try to
    incorporate switching the bulbs on and off individually... i'm
    imagining something like this... (don't laugh):

    |- NPN - Bulb -
    --| |-----
    |- PNP - R -

    ....with the base V coming in from a 5V pin on a pic... I'm thinking
    that a high or low signal would alternately push the current through
    the bulb -OR- the resistor... Giving me switching in a "pseudo"
    series circuit. Of course, I've been doing electronics for about two
    weeks - and made this up myself... SO - would this actually work?!

    Thanks in advance,
  6. Al Borowski

    Al Borowski Guest


    As the filiment heats up, the resistance increases.
    I have no idea what you're trying to do here, sorry :)

  7. Soeren

    Soeren Guest

    No !
    It might fill out the same function *if* the driving voltage is free of
    any fluctuations, but in any real life applications, it is a grossly
    simplified view, espscially when the mains supply deviates maybe +/-

    Well, if that means "discard what you allready have and go out and spend
    money on new stuff", your solution will be a waste of both energy (the
    energy he has to spend shopping) *and* money ;)


    * If it puzzles you dear... Reverse engineer *
    New forum: <URL:>
  8. Soeren

    Soeren Guest


    (the_doug) wrote in
    Yes, you could try to measure the resistance of a thin piece of steel
    wire (mounted in free air) and while measuring, heat (some of) it to a
    cherry red with a blow torch - quite an eye opener :)

    Do you mean something like this ?

    +V O----+
    +--+--+ First stage
    | |
    b |/e c\| b
    --| PNP NPN |--
    |\c e/|
    | |
    | |
    [Resistor] [Bulb]
    | |
    +--+--+ Last stage
    | |
    b |/e c\| b
    --| PNP NPN |--
    |\c e/|
    | |
    | |
    [Resistor] [Bulb]
    | |
    -V O----+

    If so, I'm sorry, but it won't work.

    What is the actual purpose, perhaps there are better ways to do it.

    If you are using a PIC, you could use PWM to reduce the current through
    the lamps.


    * If it puzzles you dear... Reverse engineer *
    New forum: <URL:>
  9. the_doug

    the_doug Guest

    No problem - somehow - the lines got messed up upon saving. what the
    diagram attempted to show was two branches of a "pseudo" parallel
    circuit attached to a larger series circuit. I say pseudo because my
    intention would be to use switching to have the current run through
    only one "branch" of this parallel hookup at a time - thus, either
    lighting the bulb or running through the parallel resistor.

    My question is simply: is this sort of configuration simply the
    imaginings of a deranged amateur? =)

    Thanks again,
  10. the_doug

    the_doug Guest

    Do you mean something like this ?
    Yes - this seems like a pretty accurate description. Won't work, eh?
    I was afraid of that...

    A concise description of the project would be this:

    3x4 array of lights which will end up mounted pixel-like behind 1-inch
    square pieces of frosted glass - it will need to generate enough light
    to function as an ambient end table lamp.

    I've got a PIC driving 12 leds directly now - and doing all the
    patterns i want - etc (the programming aspect is the only part i've
    got experience with - so it was no problem). The problem is - I need
    it to be brighter. So I started playing with the idea of a separate
    power loop being controlled by the PIC via transistors... little did
    i realize how much Amperage it was going to take to drive 12 bulbs
    independently in parallel. The reason i was trying to make a series
    solution work is because higher voltage seems to come cheaper and in
    smaller packaging than higher amperage (and this could be BAD newbie

    I don't really want to swallow a lot of your time on this... but if
    you're enjoying the discourse - I'll take all the help I can get.
  11. Rich Grise

    Rich Grise Guest

    Probably. :) What you've described will probably work - apparently
    you want to put resistor/lamp pairs in series, and swap out the
    bulb with a resistor to get the same series resistance, ergo the
    same current, right? Well, it would waste power and be a clooge
    to drive.

    Since you have a pic, just use 12 NPN transistors with a suitable
    current rating, connect the bulbs directly to the supply, and
    PWM them to cut down the _average_ voltage/current. Voila! Total
    new parts count: 12 transistors. And that's only if you need new
    ones. :)

    And with a suitable input, you could ramp up the pulse width
    slowly, and watch the bulbs get brighter until they're at about
    95%, then define that as the upper limit.

    And you _might_ be able to get by with your existing supply, as
    the total load will be a percentage (the duty cycle) of the bulbs'

  12. If you are using LEDs now, you should know that you can get really
    bright ones (that will, sadly, take more current to light.) Take a
    look at for some ideas. Also, these are sold on

    Using incandescent bulbs should work, but its going to take 'more
    power, Scotty!'.

    You can get cheap wall warts that put out 12V with higher current
    ratings at various places, including here:

    I see a nice 12V 800mA adapter for $3 USD. (plus shipping, sigh)

    Bob Monsen
  13. Soeren

    Soeren Guest

    Hi Rich,

    12 lamps of 2.33V/270mA = ~7.5W
    1 power adapter of 6V/300mA = 1.8W

    Now I am very anxious to get the specs of your power-from-thin-air
    component ;)


    * If it puzzles you dear... Reverse engineer *
    New forum: <URL:>
  14. Rich Grise

    Rich Grise Guest

    Well, I said "might," since I didn't remember the spec from the
    1st post. Just sloppiness on my part, sorry. :)

  15. Soeren

    Soeren Guest

    Hi Rich,

    Just wondered :)


    * If it puzzles you dear... Reverse engineer *
    New forum: <URL:>
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