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Circuit Solving

  1. Feb 8, 2010 #1
    Basically I need to solve the following circuit:

    5V in --- 200 OHMs --- 100 OHMs---GND
    ..............................|
    ...............................-2V LED---300OHM---GND

    I need to solve for the voltage in the resistors, and I've lost my circuit solving ability. How can I solve this? From where I'm sitting I need more info.
     
  2. jcsd
  3. Feb 8, 2010 #2

    cepheid

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    Just use KCL at the node between the 200 ohm and 100 ohm resistors. I think you have to assume a certain forward voltage drop (usually 0.6 or 0.7 V) across the LED when conducting, and hence find the current through it.
     
  4. Feb 8, 2010 #3
    We assume 2V (the norm for red LEDs), but I can't use KCL because I know nothing about current and cannot find it for anything else.
     
  5. Feb 8, 2010 #4

    cepheid

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    Call the node in between the 200 ohm and the 100 ohm resistor "node 1" (i.e., this is where the anode of the LED is connected). All quantities related to node 1 will be labelled with a subscript 1. For example, we'll call the potential at this node V1.

    You know that the current flowing into this node is just the current across the 200 ohm resistor, which, in turn is given by:

    [tex] \frac{\textrm{5 V} - V_1}{200~\Omega} [/tex] ​

    You also know that the current across the branch with the LED in it depends upon the voltage across the 300 ohm resistor, which is the voltage at the cathode of the LED. Assume that this is V1 - 0.6 V (or whatever you would like the "forward voltage drop across the LED when conducting" to be). Therefore, by Ohm's law, the current in this branch is:

    [tex] \frac{V_1 - \textrm{0.6 V}}{300~\Omega} [/tex] ​

    Now, KCL says that:

    current into node 1 = current out of node 1

    The current into node 1 is the current across the 200 ohm resistor (topmost expression above), and the current out of node 1 is the current in the remaining two branches: the branch with the LED in it (lowermost expression above), and the branch with the 100 ohm resistor in it (for which I have not given you an expression). Does that help?
     
  6. Feb 9, 2010 #5

    vk6kro

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    It looks as if the LED will not turn on.

    There is 1.666 volts at the junction of the top two resistors and this is less than the 2 volts needed to turn on the LED.

    So, the circuit just becomes a 200 ohm resistor in series with a 100 ohm resistor across 5 volts. This is because the LED will stay open circuit until it gets 2 volts across it.
     
  7. Feb 9, 2010 #6
    Sorry, I'm still not getting it. I'll try it again and post back if I fail again.
     
  8. Feb 9, 2010 #7
    vk6ro,

    I think Lancelot59 intended the circuit to be drawn as shown in the top circuit in the link below:

    http://img704.imageshack.us/img704/5951/0209001155.jpg" [Broken]

    If the polarity of the diode were reversed then it would not turn on.



    Lancelot59,

    Assuming you intended the circuit to be drawn the way I've shown in the image above take cepheid's advice and solve for the voltage at node 1 using a KCL equation. This might be easier if you replace the diode with a 2v DC voltage source as I've shown in the bottom circuit of the image above. For the sake of computing the voltages across the resistors it works as a good approximation to the diode. This only applies if you are considering the diode to be ideal though. The diode model gets more complex if you assume a nonideal diode.
     
    Last edited by a moderator: May 4, 2017
  9. Feb 9, 2010 #8
    What was up before was right. The diode has a voltage drop of 2V, not a source behind it.
     
  10. Feb 9, 2010 #9
    I drew the first circuit to check the polarity of the diode in the circuit. If it was reversed the circuit would not work as vk6kro pointed out. I drew the second circuit with the 2V voltage source to model the diode with passive circuit components. You see the voltage source will act in the same way as the diode by lowing the voltage by 2 volts from node 1 to the node above the 300ohm resistor. It's a trick to make understanding the KCL equation easier. There isn't really a voltage source in the diode, but an ideal voltage source of 2v is a perfect zero order approximation to a diode with a 2v drop.
     
  11. Feb 9, 2010 #10

    vk6kro

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    LEDs are always used in the forward biased direction.

    No, there is only 1.66 volts across the 100 ohm resistor and this is not enough to turn on the 2 volt LED. So, it stays open circuit.

    You could put different resistors in to make it work, but this circuit, as it is drawn, would not turn on the LED.
     
  12. Feb 9, 2010 #11
    Ok then, that sorts it out kind of. Thanks for the help everyone.
     
  13. Feb 9, 2010 #12
    Can you show me how you are calculating the 1.66V drop across the 100ohm resistor? If we label the node at the anode of the diode as node A and apply KCL at that node we get the following equation:

    [tex] \frac{5-V_A}{100\Omega} = \frac{V_A-2}{300\Omega} + \frac{V_A}{200\Omega} [/tex]​

    Solving for VA gives a voltage of 3.09V at the anode of the of the diode. This is a 5-3.09=1.91V drop across the 100ohm resistor.

    Also I'm not too sure of your reasoning for the diode not turning on when the voltage drop across the 100ohm resistor is less then 2V. So long as the voltage from the anode to ground is greater then 2V there will be current flowing through diode branch of the circuit and the LED will be on. I would appreciate it if you could clarify your reasoning for me.

    I'm convinced that this diode will turn on based on the circuit configuration.
     
  14. Feb 9, 2010 #13

    cepheid

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    klouchis,

    v6kro is correct. I think you might have switched the 100 ohm and 200 ohm resistors' positions in your analysis. I will post a screenshot showing what at least one circuit-solving software program thinks momentarily:
     
  15. Feb 9, 2010 #14

    cepheid

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    Last edited by a moderator: May 4, 2017
  16. Feb 9, 2010 #15
    Oh your right cepheid, I did switch the 100ohm and 200ohm resistor, that's why I didn't understand why vk6kro was saying that 100ohm voltage drop was 1.66V and that it was significant to the circuit working properly. It's hard to be dyslexic and be a circuit designer I suppose :rofl:

    vk6kro is correct this circuit will not work...but if you want it to work just switch the 100ohm and 200ohm resistor :smile:
     
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