Total Newbie short circuit question

Discussion in 'General Electronics Chat' started by Docile, Aug 1, 2009.

  1. Docile

    Thread Starter New Member

    Aug 1, 2009
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    Hi, Im just out to understand electricity flow.

    If I have two LED's and two 470 Ohmn resistors in parallel, using a 5V supply, they would both light up ?

    What would the value of the second resistor have to be so that no current flowed rthrough and only one LED lights ?::(
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    Give a better explanation of how the LEDs and the resistors are connected.
     
  3. yourownfree

    Active Member

    Jul 16, 2008
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    whoa that's a little weird. only one resistor is needed for each led. For your experiment you need a potentiometer. That way you can turn it until the led goes out. Measure the resistance of the potentiometer, with power off of course. What you need is to get one of those leds that turn a different color when you reverse the current flow. I think Radio Shack might have them. By the way electron flow is from negative to positive. in the old days it was thought that current flowed from positive to negative. That is called conventional current flow. It is actually hole current. Notice on the transistors there is an arrow in the schematic of a transistor, it actually points to the direction of hole current flow. times change.
     
  4. count_volta

    Active Member

    Feb 4, 2009
    435
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    Yes please draw your circuit in paint or something and post it here.

    Some important questions for you. Do you understand that an LED is a diode and has a voltage drop? Its not just a lamp (a resistance)

    What this means is that the LED has to be connected in the proper bios and then has to have at least 2V (typically) across it for it to light up.

    Also the voltage in parallel elements is always equal. Remember that.

    Its hard to understand your question without knowing how everything is connected.
     
  5. Docile

    Thread Starter New Member

    Aug 1, 2009
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    I made my second diagram ever using tinycad, hope it explains my question ?;)

    Well it seems my picture is as clear as mud when you click on it, but if u download it and look at in Paint or any other picture viewer it is clear.

    Made a mistake in drawing, looks like both LEDS are going to be short circuited.
     
    Last edited: Aug 1, 2009
  6. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    If you build the circuit like in the drawing both LEDs will light. What is your question about the circuit?
     
  7. Docile

    Thread Starter New Member

    Aug 1, 2009
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    I think electricity flows along the path of least resistance, so I am wondering if there is a simple method of determining how much resistance would need to be applied to one of the resistors or paths so that all electricity flowed through just one path/LED ?
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    You would have to increase the value of one of the current limiting resistors quite a bit.

    Let's look at just one LED and one current limiting resistor across a 5v supply for a minute.

    LEDs have a number of specifications, but the most useful specification is the "typical Vf @ current" specification (Vf means the LED's "forward voltage").
    In other words, at a given current, a typical LED will drop a certain voltage across itself.

    A typical modern red LED might be rated for Vf=2v @ 20mA.

    To calculate a current limiting resistor, use the formula:
    Rlimit >= (Vsupply - VfLED) / Desired Current
    Let's say you're using that typical red LED with your 5v supply. So:
    Rlimit >= (5v - 2v) / 20mA
    Rlimit >= 3v / 0.02A
    Rlimit >= 150 Ohms. 150 Ohms is a standard value of resistance. If it were not, you would go to the next higher value of resistance.

    A chart of standard resistance values is available here:
    http://www.logwell.com/tech/components/resistor_values.html
    E6 thru E24 values are typically easily available to hobbyists.

    Let's look for a moment at what you are currently using. I'll assume a Vf of 2v with 20mA current.
    LEDcurrent = (Vsupply - VfLED) / Rlimit
    LEDcurrent = (5v-2v) / 470 Ohms
    LEDcurrent = 3/470
    LEDcurrent = 6.38mA (rounded off)

    You can see that the current supply to your LED is quite low. This is good for the life expectancy of the LED and low power consumption, but the LED will be much less bright than if it was receiving it's rated current.

    The brightness of a given LED is more or less proportional to the amount of current flowing through it.

    What is surprising, is that LEDs will continue to emit light even with very low levels of current flow. You may have to darken the room and allow your eyes to adjust to see it, but I've seen them continue to glow with a fraction of a mA current flow. The exact point at where they won't continue to emit visible light would have to be tested for the individual LED in question.

    However, most applications would either use a physical switch or a transistorized switch. The latter will have small amounts of what's called "leakage current", but it will be so small as to not permit the LED current sufficient to illuminate.

    A quick note on drawing schematics:
    It's customary to have the inputs and power supplies on the left side of the schematic, positive "rails" (supply voltages) towards the top, and grounds towards the middle or bottom of the schematic. If there is a negative supply, it is normally drawn below the ground "rail" The output of the circuit is generally towards the right side. This gives schematics a more consistent appearance; we're already used to reading books from left to right and top to bottom.
     
  9. sissow2

    Member

    Jul 14, 2009
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    The path of least resistance is a simplification of the behaviour of electricity, it really does flow through all paths (some more than others based on resistance). The magic resistance you are looking for is ∞, or an open switch
     
  10. count_volta

    Active Member

    Feb 4, 2009
    435
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    1 Mega ohm resistor would work. 98% of the current would flow through the other resistor. sissow2 is right, there will always be some current, unless the resistor is ∞ (open switch)

    If you want to do simple analysis of this circuit, replace the LED's with 2V (typical LED voltage drop) voltage sources with plus at anode and minus at cathode. Then you can find all the voltages and currents in the circuit using KVL, KCL, and voltage and current dividers. No formulas. No mess.
     
  11. Docile

    Thread Starter New Member

    Aug 1, 2009
    6
    0
    Thanks fellas for so much explanation. I was really confused about many things including electricity flowing through the least resistance path. Thats howshort circuits happen.

    But when i look at other circuit diagrams I picked up that current was flowing through several paths with different resistors. So I found that at odd with my first impressions of current flow.

    Thanks for clearing up many of my issues.:D
     
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