Confused about diodes

Discussion in 'General Electronics Chat' started by shaqywacky, Feb 8, 2011.

  1. shaqywacky

    Thread Starter Active Member

    Apr 1, 2009
    To start off I'd like to clear something up for me. A diode will always drop .7V(Silicon) no matter the voltage or the current, correct? If that is true than a diodes resistance must always be changing.

    Next I'm very confused about the question in the attached picture. In the picture there is a resistor in series with another resistor and diode in parallel. The first problem I have is in step one where it says Vd = R2 * I2. I don't understand how the diode's voltage drop is used. I would expect it to be the resistors voltage drop not the diodes. Why is this?

    Next, It used KVL to find the voltage for the second resistor. I don't understand how the total voltage was used because shouldn't the first resistor have dropped the voltage?

    Is there a name for this type of problem so that I could look up some other ones that explain the problem in more detail?

  2. Jaguarjoe

    Active Member

    Apr 7, 2010
    The voltage across the diode (Vd) is fixed at 0.7 volts. The resistor in paralell with the diode will also have that same 0.7 volts across it because parallel circuits have the same voltage, so if R2 is 70 ohms, than I2 = E/R = 0.7/70 = 10ma. The power dissipated by R1 will be P = EI = 0.7 * 0.01 = 7mw.
    The total voltage of the circuit must equal the supply voltage so the voltage across R1 and the voltage across R2 || D must equal 5 volts. Shuffle this around and the voltage across R1 will be 5 - 0.7 = 4.3 volts. The power disipated by R1 will P = E^2/R or P = (4.3)^2/43 = 0.43 watts.
    The current through the entire circuit will be the same as the current through R1 because series circuits have the same current. Using I = E/R = 4.3/43 = 100ma.
    If the total current is 100ma and 10ma of it flows through R2, then the current through the diode will be 100 - 10 = 90ma.
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    No the the forward voltage isn't always 0.7V. As with any non-linear device it depends on the current flow in the diode.

    You may encounter the Shockley diode equation sometime. This general equation attempts to model the relationship between voltage & current in the diode.


    The assumed constant 0.7V is a reasonable value for most simple diode modeling requirements.

    In this situation the expectation seems to be that you treat the diode as if it were a 0.7V DC source with zero internal resistance.
    Last edited: Feb 8, 2011
  4. Wendy


    Mar 24, 2008
    Diode start at 0.6V, and can quickly go over 1V. It is very current and temperature dependent. Temperature is important enough that a simple diode can be used as a temperature sensor for a household thermometer.
  5. Audioguru


    Dec 20, 2007
    The datasheet for a 1N4148 silicon diode shows graphs of its forward voltage at various currents.
    Its typical forward votage at 1uA is only 0.275V and is 1.45V at 800mA.