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Simple Diode Circuit

  1. Feb 14, 2007 #1
    I have an interesting question about this circuit... if I were to calculate the voltage from point X to ground... essentially all it is, is voltage across the 4 kilo-ohm resistor because if it went the route of the diode, it would lose an additional 2.7 volts.. so its going to take the better path.. right? Thats how it works doesn't it?

    so... we have... [tex]\frac{4k}{2k+4k} * 8 Volts = 5.7 V[/tex]

    Now... lets say we reversed the diode... wouldn't then.. we push 1.7V towards point X right? Now... does that 1.7 volts add up with the 5.7V, does the 5.7 volts push against the 1.7V. The voltage I know is still 5.7 volts even if the diode was reversed... but is that because the voltage won't go that way due to there being more across the 4 kilo-ohm resistor?

    Picture of circuit:
    Last edited: Feb 14, 2007
  2. jcsd
  3. Feb 14, 2007 #2
    I got 5.3333333333 Volts using your method, and if the diode was forward biased, why wouldn't point X be 2.7 Volts? I think this will cause some current to oppose the 2V source
  4. Feb 15, 2007 #3


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    For the original circuit the Diode/Battery is the functional equivilent of a zener diode.
    Try reworking your answer in terms of this.

    For the case where the diode is reversed.
    What is the basic function of a diode?
    What happens when a diode is reverse biased?
  5. Feb 15, 2007 #4
    oops I ment to write 5.3 not 5.7 for everything

    When a diode is reversed current doesn't go through it.

    But what I'm wondering is why you don't add the resultant voltage with the 5.3V
    Last edited: Feb 15, 2007
  6. Feb 17, 2007 #5
    Yes, the voltage at X is 5.3V. Subtract 0.7V from 5.3V and it's still more than 2V. So the 8V battery is charging the 2V battery through the diode, which if forward-biased.

    I'm not sure I understand your last question. Please clarify.
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