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Help Getting Started - Simple DC Circuit

Discussion in 'Electronic Basics' started by Mark Jerde, Nov 11, 2005.

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  1. Mark Jerde

    Mark Jerde Guest

    (I apologize for posting this basic circuit question. 25 years ago in
    college I had to know E=IR etc. Just having problems getting started...)

    I'm trying to "invent" a mechanical device. I've been having some alignment
    problems with it and this afternoon it occured to me some simple electronics
    would help a lot. But it has been many years since I've done anything more
    complicated with electronics than change batteries in the household smoke
    detectors. ;-) I'm looking at the Jameco web site, seeing if I can find
    enough info to play with E=IR but there are just too many options &

    I want to go to my neighborhood Radio Shack and buy
    - 30 ea SPST NO switches
    - 30 ea green LEDs
    - 30 (?) ea resistors to limit current to the LEDs
    - A circuit board to solder the LEDs & resistors to
    - A battery case (e.g. 4 "D" cells) or 9v clip
    I have an electronics soldering iron and plenty of wire.

    The goal: When everything is lined up right on my mechanical device, all
    the switches will be closed and all 30 LEDs will be glowing. Then I'll
    unplug the battery, as the machine is ok once all the switches are closed.
    (E.g., no long-term lighting requirement.)

    How can I get started with this? For the battery which value to use, 1.5,
    3, 6 or 9 volts? Does each LED need its own resistor or is one resistor
    enough? (I don't care how bright the LEDs are so long as they are visible.)


    -- Mark
  2. Randy Day

    Randy Day Guest

    2 'C' cells in series should give you enough power
    for short periods, and 3 volts will be sufficient
    to light the LEDs. You could try 2 'AA' cells, but
    I'm guessing they won't handle 30x0.02A = .6A for
    very long (if at all).

    Each LED&switch will need a separate resistor.
    Don't try to parallel them thru 1 resistor; you'll
    let the magic smoke out of the LED's. :(

    To calculate the resistor values:

    R = (Vcc - Vled) / Iled

    where Vcc is the supply voltage, Vled is the
    forward voltage drop of the LED, and Iled is the
    operating current of the LED.

    You can find a minimum R value for your LED's,
    and try larger values that still give acceptable
    brightness with less power consumption.

  3. Guest

    You will want to make sure the source voltage is a greater than the
    led's forward voltage and the current limit resistor voltage drop. I
    think white Leds are about 3V and Red Leds are 1.7V, but be sure to

  4. Mark Jerde

    Mark Jerde Guest

    Randy & David -- Thanks for the replies.
    I'm just back from Radio Shack with a package of these to get started
    20 assorted LEDs, 2 - 3V, 10-20 mA

    (They were the only thing economical. Other LEDs were one or two per
    package, $1.98 to $4.98. I'll get another package when I go to get the
    battery holder & resistors.)

    Am I reading the specs right to use Vcc = 3 V, Vled = 2.5 V and Iled = 15
    R = (3 - 2.5) / 0.015
    = 33.33 Ohms

    When I did electronics many years ago most of the resistors were in the
    kiloohm and megaohm range. 33 ohms seems wrong. Am I missing something?

    Thanks again!

    -- Mark
  5. tempus fugit

    tempus fugit Guest

    If you need that much quantity, don't go to RadioShack. Check out - they are WAY cheaper and have no minimum.
  6. Red, yellow and green LEDs are around 2 volts, and I find they are
    quite bright enough with 10 mA, so, with 3 volts, you could use 100

    Blue and white LEDs want about 3.6 volts, I believe, so you'd need
    another cell or two for them.

    Resistors under 1K are certainly not uncommon - I think I've got a few
    values under 10 ohms in my stock at work (although I'll admit that I
    don't use such low values very often.)

    Peter Bennett, VE7CEI
    peterbb4 (at)
    new newsgroup users info :
    GPS and NMEA info:
    Vancouver Power Squadron:
  7. Mark Jerde

    Mark Jerde Guest

    Thanks for the link. The last time I did any serious soldering the 80186
    and 80286 were brand new. ;-)

    -- Mark
  8. Mark Jerde

    Mark Jerde Guest

    Peter --
    So I didn't make some kind of math error. Thanks.

    -- Mark
  9. Randy Day

    Randy Day Guest

    [piggybacking - for some reason Mark's reply didn't
    show up here...]

    Since they are assorted LED's, I think you can
    count on using 'assorted' values for the resistors.
    Try finding a set of resistors from 47 ohm to maybe
    150 ohm. Test each LED with the higher value
    resistors and step down until a reasonable
    brightness is achieved.

    Nope, because you're dropping such a low voltage
    (0.5 volts) at such a low current (.015 A).

    A Vled of 2.5v just seems a bit high to me,
    though; I'd suggest using 2v in your calcs.
    You can always step down to a lower ohms
    value if an LED is visibly dim.
  10. The_Truth

    The_Truth Guest

    alright... for radioshack green led's...

    Typical MCD: 620 30 mA current (max.) Typical wavelength: 570mm Size:
    T-1-3/4 (5mm) Viewing angle: 12° Sold in package of 1 Typical voltage:
    2.1V, with a maximum of 2.8V

    chances of running 30 led's from 1 nine volt are slim to none... at
    least not for very long... you could put 2-3 nine volt batteries in
    parallel... and that would divide the current draw between them...

    ya know... i'm not big on radioshack prices... so i found a nice little
    can't get them all at once.. but free is a good thing...
  11. Ken Moffett

    Ken Moffett Guest

    wrote in
    Check this URL for the Vf of different colored LED's:
  12. ehsjr

    ehsjr Guest

    Radio Shack is far too expensive for toggle switches, as you
    discovered. You can get them for 40 cents each in lost of 10
    from Allelectronics -
    catalog # MTS-75PC

    You can add the 31st LED and a simple circuit to tell you
    when the other 30 LEDs are lit. That way, you need glance at
    only 1 LED to see if all the others are on instead of needing
    to look at all 30 of them. You'll need 30 diodes - catalog #
    1N914TR (100 for $2.00) an NPN transistor (any NPN would work -
    catalog # PN2222A is a suggestion) and a couple of resistors.
    The single LED can serve as a "run" light. When it is lit, the
    battery must not be disconnected. When it goes out, the battery
    can be disconnected. (You could use it as the basis of an
    automatic battery disconnecting circuit if you want.)
    Here's the circuit:

    +3 ---+------------------} }----+--------------------+
    | | |
    [LED1] [LEDn] [LED31]
    | | |
    [R1] [Rn] [R31]
    | ->|- | ->|- |
    +---[Diode1]---+ +---[DiodeN]---+ |
    | | | | |
    [Switch1] | [SwitchN] | |
    | | | | |
    Gnd | Gnd | |
    | | |
    +---} }-------------------+ |
    | |
    [4.7K] |
    | /c
    +----| NPN
    | \e
    [100K] |
    | |

    Use 100 ohm resistors for R1 through R31 and a 3 volt supply.
    (If you don't want to use batteries, you could use catalog #
    DCTX-330, which is a 3 volt 300 mA wall wart power supply.)
    That will limit current through each LED to about 10 mA,
    which is plenty bright enough for red LEDs. Catalog # LED-1
    gets you 10 standard red LEDs for $1.00. The catalog numbers
    for the resistors are 100-1/4 (100 ohm, 1/4 watt), 4.7K-1/4
    and 100K-1/4 They cost 50 cents for 10 and must be ordered
    in lots of 10.

  13. Just a thought - it is much easier to see one lit LED among many unlit
    ones than it is to see one unlit LED among a bunch of lit ones, so I'd
    wire the thing so that all lights are out when things are all properly
    aligned. Also, if you do it this way, there is no need to remove the
    battery while running the machine, and you will get a warning if
    anything does slip out of alignment when it shouldn't. You would,
    however, want to remove or switch off the battery while the thing is

    Peter Bennett, VE7CEI
    peterbb4 (at)
    new newsgroup users info :
    GPS and NMEA info:
    Vancouver Power Squadron:
  14. Jasen Betts

    Jasen Betts Guest

    you're pushing the envelope there...

    with a 3.0V supply and a 1.7V led drop (typical for green LEDs) and a 0.6V
    drop in the diode and in the be junction of the transiistor there's only 0.1V
    left through 4.7K that's about 20uA at 0.6v the 100K will pass 6uA leaving 14
    to flow into the base of the transistor.

    unless that pn2222 has an hfE around 500 that last LED could be pretty dim.
    and that's assuming 0.6V Vbe is sufficient

    sticking a 1K resistor in parallel with each LEDn-Rn pair would be one
    way to fix that.

  15. Mark Jerde

    Mark Jerde Guest

    Ed --
    Wow, great idea. Thanks!

    -- Mark
  16. Mark Jerde

    Mark Jerde Guest

    Peter --
    Excellent point. Thanks.

    -- Mark
  17. ehsjr

    ehsjr Guest

    Yes. I had a "window" of only roughly 3.25 to 2.25 volts. Your
    excellent suggestion (more on that later) fixes that nicely.

    The trick was to get the single LED to light without pulling enough
    current in the base path to make a "main" LED (one of the first 30)
    glow even a tiny bit.
    Not quite. The diode drops about .45 volts at that very low current.
    I guess "pretty dim" is relative. On the breadboard, the LED
    glows merrily, and is easy to see. I used 2 10K in parallel
    in place of the 4.7 K, and 150 ohms in series with the "main"

    500 is way too high for hfe. As I recall, a 2N2222 is good for
    about 300, max. On the breadboard, without the 1K in parallel
    with the main LED, base current measures ~23 ua and LED current
    measures ~3.69 mA so hfe ~ 160. With the 1K in parallel with
    the main LED, base current is ~280 ua and LED current is
    ~ 7.52 mA, so gain is ~27

    It fixes the "window" problem very nicely. The window problem
    was that the single LED would glow with the battery voltage no
    lower than 2.75. Below that it was too dim. And above roughly
    3.25, the "main" LED would start to glow dimly. With the 1K
    resistor across the main LED, the circuit works down close
    to 2 volts, and will work without causing the main LED to glow
    above 6 volts. 3 volts is still a viable option, but he could
    use 6 volts if he changes the value of R1 through R31 to 220.
  18. ehsjr

    ehsjr Guest

    Here's an updated schematic to incorporate Jason's idea
    of adding a 1K resistor in parallel with each of the first
    30 LEDs. It will allow the circuit to work as the batteries
    drain below 2.75 volts. It will also allow you to use a higher
    supply voltage if you want, like 3.6, 4.5 or 6 volts, but you
    would need to change the value of R1 through R31 using the
    values below:
    3.6 volts use 150 ohms; 4.5 (or 4.8) volts, use 270 ohms;
    6 volts, use 330 ohms.


    +3 ---+-------+----------} }----+-------+------------+
    | | | | |
    [LED1] [1K] [LEDn] [1K] [LED31]
    | | | | |
    +-------+ +-------+ |
    | | |
    [R1] [Rn] [R31]
    | ->|- | ->|- |
    +---[Diode1]---+ +---[DiodeN]---+ |
    | | | | |
    [Switch1] | [SwitchN] | |
    | | | | |
    Gnd | Gnd | |
    | | |
    +---} }-------------------+ |
    | |
    [4.7K] |
    | /c
    +----| NPN
    | \e
    [100K] |
    | |
  19. Mark Jerde

    Mark Jerde Guest

    Ed --
    Much appreciated!

    -- Mark
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