Am I Drawing Current from 5V or 9V Rail?

Discussion in 'The Projects Forum' started by the kidson, Dec 31, 2014.

  1. the kidson

    Thread Starter Member

    May 15, 2011
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    Thanks for reading. I made a simple flash pattern using an ATTiny85. The circuit is on my breadboard. My problem is this. I'm not sure where I am drawing my current from for the leds. Not sure if it's +5V or +9V. I know...I know, but I'm a total noob. So please forgive me. I'm assuming it's +9V, since it's coming straight down the rail.

    The Vcc of the ATTiny85 is connected to the output of the 7805 which is straight in from the +9V rail. The output (pin 2) of the ATTiny85 is connected to a bc337 to its base pin with a 560 ohm resistor. The emitter is grounded, and, the collector is connected to the leds cathodes. The leds are connected straight to the power rail with 47 ohm resistors on the anode side in two groups of three in parallel. I know I figured the resistor for +5V because of its value. But, I'm confused as it's directly connected to the +9V rail running down, not through the output of the 7805....

    The problem is as so. If the circuit is using the +5V for the leds, is there a way to keep the same circuit and power it with the +9V so I can wire in series? The leds are all white and SMD ones that need t fit in a tight space.

    Here is a schematic of basically how it is on the board. Please note this is my first attempt at this. I used design spark to do it. The leds are meant to say white not red. I was having trouble trying to figure out how to change names on stuff.

    Again, please forgive the ignorance and thanks for your time and insights.
     
    Last edited: Dec 31, 2014
  2. blocco a spirale

    AAC Fanatic!

    Jun 18, 2008
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    The LEDs are sourcing current from the 9V rail........ I mean, the 9V rail is sourcing current to the LEDs:)
     
    Last edited: Dec 31, 2014
  3. the kidson

    Thread Starter Member

    May 15, 2011
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    Thanks for the timely response. It's much appreciated. However, I fell on stupid again. If the leds are being sourced by the +9V with the 47 ohm resistor, wouldn't that be too hot for the smd leds? I figured around 100 ohms for +9V at 20mA. Do all white leds burn hotter than 20mA? Or am I just that dense and missing something? Again, thanks for the input.
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    I cant read the .sch file. Have no idea what produced it...

    To blocco: I would have said that the 9V rail is sourcing current to the LEDs :D...

    (I can say this. English is not my native language).
     
    Last edited: Dec 31, 2014
  5. MikeML

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    Oct 2, 2009
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    Most White LEDs have a forward drop of ~3.6V, so (9-3.6)/0.02 = 270Ω
     
  6. blocco a spirale

    AAC Fanatic!

    Jun 18, 2008
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    20mA seems to be typical for most LEDs. I can't open your schematic so I don't know whether it's just one LED or two in series. But yes 47R appears to be too low; I would say 270R for one white LED and 100R for two in series.
     
  7. the kidson

    Thread Starter Member

    May 15, 2011
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    Thanks. So it would be more efficient to have them in groups of 2? I also updated the the schematic to a picture. I should be able to be referenced.
     
  8. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    You really should have a resistor in series with each led; not one resistor and then three leds in parallel. My calc. assumed one resistor per led.
    If you want to save some resistors, then put two leds in series with each resistor: (9-3.6-3.6)/0.02 = 90Ω, so use a 91Ω resistor.
     
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