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Vbe temperature coefficient of transistors

  1. Jun 28, 2009 #1
    Something I'm having trouble grasping - if the Vbe temperature coefficient is -2 mv/degree C, why does the collector current INCREASE instead of decrease with increasing temperature? If the temperature rises and the voltage base to emitter drops, wouldn't the smaller Vbe cause less emitter current to flow?
     
  2. jcsd
  3. Jun 28, 2009 #2

    ranger

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    A decease in VBE means an increase in Ib. This increase causes the collector current to increase. Also beta increases with temperature. The increase in base current is like saying the input resistance has decreased with rising temperature giving a net effect of a decreased VBE. These are all factors that cause the collector current to increase from its Q-point value.
     
  4. Jun 28, 2009 #3
    When we say that Vbe goes down 2 mV per degree C, that is based on a specific current value. Any p-n junction exhibits this characteristic. At 25 deg C ambient, with 1.0 mA of forward current, a p-n junction measures 0.65 V forward drop. If the temperature is elevated to 50 deg C, then the forward voltage drop of the junction decreases (2 mV/deg C) * 25 deg C = 50 mV, as long as the current is the same. Thus at 50 deg C with 1.0 mA of forward current, the forward voltage drop is 50 mV less than the 0.65 V at 25 C, which is 0.60 V.

    At higher temperature, a p-n junction has more carriers available due to increased thermal energy. Hence at 50 C, the junction is easier to drive than at 25 C. A given current incurs a smaller voltage drop at higher temp.

    Does this help?

    Claude
     
    Last edited: Jun 28, 2009
  5. Jun 28, 2009 #4
    I think I understand now. Say I have a transistor in CE configuration with an emitter resistor, biased for some current. I have the base biased to say 1.6 volts at ambient temperature, so the transistor is in its active region, so the voltage across that emitter resistor 1.6-0.6 = 1 volt. Now, if the temperature increases, the voltage drop across the base to emitter junction might decrease to 0.5 volts, in which case I have 1.6-0.5 = 1.1 volts across the emitter resistor, which tends to increase the emitter and hence the collector current. This could lead to thermal runaway if there were no resistor to apply negative feedback where the increased current generates heat, which causes the Vbe drop to decrease more, generating more heat, etc. Does that sound about right? I think I was confused because I was mistaking the intrinsic base to emitter drop, which changes with temperature, to the applied voltage base to emitter, which determines the emitter current through the Ebers-Moll model.
     
  6. Jun 28, 2009 #5
    You've got it right. To determine the actual Ie via E-M model involves a transcendental equation solution, not straightforward. The 2 equations are

    1) Vbb - Vbe = Vre = IeRe, or Vbe = Vbb - IeRe.

    2) Vbe = Vt * ln((Ie/Ies(T)) + 1)), where Ies(T) is a strong function of temperature, the saturation current (aka "scaling current"), Vt = thermal voltage = kT/q, as k = the Boltzmann constant, T = absolute temp, & q = charge of 1 electron.

    The difficulty is that Ies is temperature dependent. The best method is what you did. Use the spec sheet graphs to find Vbe at various temps. But your thinking is sound. You seem to have a good grip on the basics.

    Claude
     
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