Calculating a diode's Thermal Voltage based on current and voltage

Discussion in 'Homework Help' started by TimNJ, Mar 8, 2015.

  1. TimNJ

    Thread Starter New Member

    Dec 31, 2013
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    Hi everyone,

    I'm working on a lab that uses a 1N4733 Zener diode.

    Based on Shockley's diode equation which is Id = Is* e^(Vd/Vth), I'm supposed to figure out the thermal voltage of the diode. Id is the diode current, Vd is diode voltage, Vth is thermal voltage. Is is the reverse leakage current.

    I have a whole table of measurements I made using a multimeter for the diode current and the voltage across the diode. This includes both reverse and forward bias orientations.

    The thermal voltage is supposed to be a constant at a given temperature, but I can't seem to get anything constant for both forward and reverse. They are also orders of magnitude off from the expected value of 26mV.

    I solved for Vth = Vd/ ln((Id/Is).

    Should I be able to use any measurement to calculate the thermal voltage? What might I be doing wrong?

    Thank you!
     
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  2. crutschow

    Expert

    Mar 14, 2008
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    I think that equation only applies to the forward bias condition.
    Is is the reverse leakage current at a low voltage, not at the zener breakdown voltage.
     
  3. WBahn

    Moderator

    Mar 31, 2012
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    Also, Is is STRONGLY temperature-dependent. So much so that it is what gives a diode a negative temperature coefficient for the voltage vs current characteristic (the exponential part of the diode equation would yield a positive temperature coefficient).
     
  4. TimNJ

    Thread Starter New Member

    Dec 31, 2013
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    Thanks! So should I only select values from the forward bias region?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    Yes. Also, you need to keep in mind that the diode junction temperature will increase due to self-heating as you go to higher currents. So you need to make your measurements in a pulsed mode as much as possible. Establish the circuit, make the measurement quickly, then take the current to zero and let the diode cool before making another measurement. If possible put the diode in a thermal bath of some kind to keep it at the same temperature as long as possible. A simple way to get fairly good results is to get a metal tube (or a stack of washers) that the diode just barely fits within and then pack the hole with heat-sink compound and put the diode in there. Even better is if you can find split lock washers that you can then pinch down against the diode (but not so much as to crack it, of course).
     
  6. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    If we could get some data on the temperature dependence for Is that would be good.

    Also, heat sinks on the leads themselves often proves valuable because the body is more of a heat insulator than a conductor, while the leads are decent heat conductors. The heat sinks would look like 'fins' around the wire, soldered onto the lead close to the diode body.
    Nothing works perfectly though because there is always distance between the die and the cooler air, which means thermal resistance and thus a thermal gradient.
    Alternately, put the diode into a bath of well distilled water. Water makes an excellent heat sink for short time periods because the specific heat capacity is pretty high.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    I agree -- heat sinking the leads is probably the better way to go. Another alternative is to clip on alligator clips onto the diode leads or even the cheap heat sink clips that Radio Shack used to sell intended for protecting the device from the heat while soldering. I'm sure those are still available from someone.

    I had thought about the water bath, but didn't want to suggest it because of the need for very low conductivity water. An option there would be to use alcohol -- I don't know what the conductivity of typical 70% rubbing alcohol is, but it is probably pretty low.

    A pretty good rule of thumb is that for silicon and germanium that Is doubles for every 7°C to 10°C of temperature rise.
     
  8. TimNJ

    Thread Starter New Member

    Dec 31, 2013
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    Thank you everyone! Seems strange to ask for the thermal voltage because getting a proper measurement relies heavily on your testing methodology. I don't have to time to make new measurements, but I'm planning on using a small positive Vd value and subsequently small Id. Using the smaller values is working out to a voltage atleast in the same order of magnitude. (Around 44mV)
     
  9. #12

    Expert

    Nov 30, 2010
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    Hasn't anyone mentioned an oil bath? That and stirring...which brings up using a fan with air.
    None of these are perfect, so a pulse generator with a one-shot scope trace would be even better.
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Well if you are worried about the small conductivity of well distilled water then perhaps get a little more elaborate and build or buy an aquarium such as used to breed the betta splendens (variant spellings), which have a divider in the center of the tank. With a sealed plexiglass divider, drill a small hole in the center, push the body of the diode through using a small amount of silicone rubber, let cure, then fill both sides with distilled water. This way each lead of the diode has it's own bath, separated by the plexiglass.
    Of course small fins soldered to the leads still helps to dissipate heat into the water, as well as gentle water circulation.
    It all depends how elaborate we want to get here :)

    It is interesting that in some of the writings on diodes and transistors they neglect to mention the temperature dependance of Is, yet they include temperature in the basic formulas. If Is varies and it is not shown how it varies with temperature, then the equation only works at the same temperature as that for which Is was determined.
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    That's why you seldom see Is listed in datasheets. You'll see leakage currents under full reverse voltage, but I don't believe that is truly the saturation current (frequently called the scaling current).

    For temperature-sensitive applications you usually use circuits that make the dependence on the scaling current go away. The type of circuit you use depends on whether you can have two junctions that are at the same temperature or two junctions that are at different temperatures. If you just want to determine the temperature of something, then you can put different current densities through identical devices that are that the same, albeit unknown, temperature. Two ways to do this is to put the same current through two series-connected devices that have different areas (such as one diode in series with a parallel combination of ten diodes) or to use a current mirror to pass one current through one diode and a scaled version of that current through another identical diode. The voltage difference between the two diodes can then be used to determine the temperature.
     
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