Theoretical question about diodes

Discussion in 'General Electronics Chat' started by gerases, Feb 24, 2013.

  1. gerases

    Thread Starter Member

    Oct 29, 2012
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    This question has been bugging me: in a circuit consisting of a power supply (Vs), a resistor R and a diode D, what is the maximum resistance of the R that will allow a 0.7V drop on the diode (let's assume it's silicon)?

    In other words, if we start with an R too big, there will not be enough of voltage to go around for the diode. As we decrease the value of R, eventually we'll reach a resistance that will allow enough voltage drop (0.7V) over the diode. That's what my question is: how do we calculate that R?

    It seems like there should be something like a "diode equivalent resistance", but I couldn't google anything to that effect.

    Thanks!
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    You got this wrong, as long as Vs is greater than Vf of the diode, there will allways be some current no matter what the resistance is. The only thing that changes with current is the Vf, because at low current it is the typical 0.7V, but as the current increases this voltage rises, so say for a 1n4007 it could be as high as 1.1V at 1A.
     
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  3. MrChips

    Moderator

    Oct 2, 2009
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    This is not an uncommon question for someone learning about semiconductor characteristics.
    To solve this problem you need to specify the value of Vs.
    You also need to know the I-V curve for the specific diode. To determine this, you need to know the diode equation which includes the reverse bias current and the temperature.

    As you can see in the graph below, the I-V curve (green) is highly non-linear.
    The red line is called the load line and is determined by the end points on the x and y axes.
    The intercept on the x-axis is determined by I = 0, V = Vs
    The intercept on the y- axis is determined by I = Vs/R, V = 0

    Your question is, what value of R will give Vd = 0.7V?
    Since the diode characteristic curve is non-linear and without knowing this curve, you have to find the answer by trial and error, i.e. change R until you get the desired Vd.

    [​IMG]
     
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  4. gerases

    Thread Starter Member

    Oct 29, 2012
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    Thanks for a detailed reply. The diode in question is 1N4001. In the case of that diode, what number in the datasheet am I looking for or how do I use the similar graph for my model to answer my question? I know about the I-V graph of course, but it wasn't helping me. What is the significance of those two intercepts and what's is the difference between Vs and V? I'm a bit confused.
     
  5. MrChips

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    Oct 2, 2009
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    What is the value of your Vs?
     
  6. crutschow

    Expert

    Mar 14, 2008
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    The diode current is a logarithmic function of voltage so the voltage starts increasing across the diode at a very low current. Here's a typicial graph of that. At some current the voltage will reach 0.7V which varies slightly depending upon the current rating of the diode. According to the data sheet for the 1N4001 it reaches 0.7V at about 10mA.
     
  7. gerases

    Thread Starter Member

    Oct 29, 2012
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    Sorry, 9V.
     
  8. MrChips

    Moderator

    Oct 2, 2009
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    Connect a 1kΩ resistor in series with the diode and 9V supply and you should get close to 0.7V across the diode.
     
  9. #12

    Expert

    Nov 30, 2010
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    Trying to visualize a diode or base-emitter junction as an equivalent resistance is a beginners attempt to organize the world of electronics. The problem is that the apparent resistance is different for every circuit, and often, different from moment to moment. Better not to try to pin it down because it is a dynamic property. You will know when you actually need the answer, and you will calculate it for that circuit at that moment.
     
  10. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Here's the data sheet for that diode. You will note there is a "Maximum instantaneous forward voltage" spec of 1.1V for 1A of current. For your 9V source, you would figure R = (9V-1.1V)/1A=7.9 ohms. However, since the voltage spec is a MAX, you would note a voltage smaller and a current larger then your computations.

    There is also a relevant "Fig. 4 - Typical Instantaneous Forward Characteristics" where the diode current gets so large (20A) that secondary effects are predominating and the V-I curve no longer looks like a nice exponential.

    That's OK... when designing with a diode you can plan for the worst case (the max voltage), and if that makes enough voltage for the circuit that follows you can stop there.

    And of course there are exceptions to that (like the power is a linear regulator following the diode).

    Generally an engineer literally takes a guess at the diode voltage and goes from there. I've used .6, .7, .75, and 1V as a diode drop depending on the usage, from some low low current to a medium current power supply. Measurements on a prototype (or spice model) may indicate a different estimate needs be used.
     
  11. JMac3108

    Active Member

    Aug 16, 2010
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    Diode and R in series ... here is a simple way to figure it out ...

    1) Assume the diode will drop 0.7V.

    2) Subtract .7V from the power supply voltage. This is the voltage across the resistor, Vr.

    3) Calculate the current in the circuit using I = VR / R

    4) Look up the I-V graph in the diode datasheet and see what the actual forward drop of the diode will be at this current.

    In most simple cases, this will get you the answer you need.
     
  12. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    That is assuming the resistor value is high enough and/or the supply voltage is high enough for the actual drop to not matter that much. But anyway drawing the load line like MrChips suggested could get you there even a bit faster and more accurate, unless the graph is logarithmic.
     
  13. JMac3108

    Active Member

    Aug 16, 2010
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    Yep. But in most practical circuits, this is the case. :) But this brings up a good point ... One of the most valuable skills in circuit design is to understand enough about what is going on to know when to perform a detailed calculation and when to estimate. In other words, what's important and what isn't.
     
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