Chain of 'ideal' diodes in DC circuit

Discussion in 'Homework Help' started by taras.di, Jul 5, 2008.

  1. taras.di

    Thread Starter New Member

    Feb 27, 2008
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    0
    Hi everyone,

    I was thinking through a problem involving a chain of diodes. I assumed a model of the diode that, below 0.7V, the diode does not conduct, and above this voltage, it behaves as a resistor (with a 0.7V voltage drop).

    Now, the circuit in question is a 9V DC source with a load resistor and 3 diodes in a row. I postulated that there would be values of the load resistor which would result in a non-conducting circuit, as the value of the load resistor effects the voltage drop over the diodes.... if the load resistor is of such a large value, then *if* the circuit was conducting, the voltage drop over the resistor would be so large that the resulting voltage drop over the chain of diodes would mean that the diodes would not have enough bias. Is this correct?

    +9v ----- load resistor ------- ||> ---- ||>----||>----- ground

    I tried to check this in spice, but found that the voltage drop over the diodes only decreases with increased values of the load resistor... I assume this is because spice is using a more accurate model than in my hypothetical example.

    Thanks

    Taras
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Your model of a diode is incorrect. They do not behave as resistors or resistances. Once the 700 mv across the junction has biased them into conduction, they will conduct current with no limitation (until the junction melts).

    Your setup should measure 9 volts on the battery side of the resistor, and 2.1 volts on the other.

    Can't speak for Spice, as I don't have any experience with it.
     
  3. taras.di

    Thread Starter New Member

    Feb 27, 2008
    2
    0
    I don't think my model is incorrect.

    There are a few models of diodes that can be used when analysing circuits. From least accurate to most accurate, they are (this may not be an exhaustive list):

    * (ideal diode)

    open circuit if v_diode <= 0
    short circuit if v_diode > 0

    *

    open circuit if v_diode < v_threshold
    short circuit if v_diode >= v_threshod
    (this is the model you mentioned)

    *

    open circuit if v_diode < v_threshold
    linear relationship between v_diode and i_diode if v_diode >= v_threshold
    this linear relationship can be *modelled* as a resistor with very low resistance

    *
    exponential between the voltage over and the current through the diode

    ....

    You can see that the 3rd model resembles the fourth one more accurately, so that's the one I decided to use...
     
  4. beenthere

    Retired Moderator

    Apr 20, 2004
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    Got a meter? This would not be too hard to set up as a hands-on experiment.
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    This would require a very large resistance - on the same order of magnitude as an open circuit.

    You can calculate the effective resistance of a diode by dividing diode current into diode voltage drop. Diode current for an ideal diode can be calculated with Shockley's equation: http://en.wikipedia.org/wiki/Diode#Shockley_diode_equation
     
  6. Distort10n

    Active Member

    Dec 25, 2006
    429
    1
    Beenthere is correct. You will ideally measure 6.9V across the resistor because each diode will have 0.7V across it.
    I popped this circuit in TINA spice and I measure 7.48V since each diode (1N1183) has ~0.506V of drop when forward biased.
    The things one can do with a simple PN junction!
     
  7. Sparky

    AAC Fanatic!

    Aug 1, 2005
    75
    0
    I agree,

    I would answer this problem the same way - assume "about" 0.7v across each diode.

    I got my electronics book out and found a problem very similar to this one:
    10v supply in series with a 1k ohm and 3 diodes.

    The purpose for this problem was to calculate the resistance of the diodes.

    using rd = (nVt)/(I)

    In my book with n=2, rd = 6.3 ohms and 3*rd = 18.9 ohms.

    "I" was estimated with (10-2.1)/1k.

    Could spice be using this resistance rd = (nVt)/(I) or at least the slope of the i-v curve at a pont?
     
  8. beenthere

    Retired Moderator

    Apr 20, 2004
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    Distor10n - those 1183's are very heavy duty rectifier diodes. What do you get if you put in something more like a 1N914 or a 1N4148? How about a 1N34? What is "ideal"?
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    taras.di,

    Wrong on both counts. Don't let anyone tell you that a diode does not conduct below 0.7 for silicon or 0.4 for germanium. Its current varies exponentially from zero volts, and it becomes significant at 0.7 for silicon and 0.4 for germanium. You can see this in the graph of this link. http://people.seas.harvard.edu/~jon...de_characteristics/diode_characteristics.html . Vd is chosen because that is where the exponential rise starts to become significant. As for a resistor, have you ever come across a resistor whose voltage across it varies as the logarithm of the current?

    Three diodes or one diode, 9V or any voltage, you can add resistance to make any current in the circuit you desire. If you add enough resistance, you can "starve" the diodes from receiving at least 0.7V, so that their conduction won't become significant. If you don't add enough resistance, the current becomes high enough to burn them out. The choice is yours. Ratch
     
    Last edited: Jul 6, 2008
  10. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The datasheet for Fairchild's 1N4148 diode shows its typical forward voltage at various currents. The temperature of the ambient is 25 degrees C because the forward voltage changes with temperature. They show another graph with the current increased to 800mA and the forward voltage becomes 1.4V.
     
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