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Burned out BJT?

  1. Jan 24, 2012 #1

    cep

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    Hi all,

    I was working with an emitter follower circuit similar to this one: http://upload.wikimedia.org/wikipedia/commons/b/b8/NPN_emitter_follower.svg , the only difference being that in my circuit, there is a 100kΩ resistor between the Vin site and the transistor. RE is also 100kΩ. I measured ΔV across both resistors, and now that I am looking at my results, I suspect the transistor is burned out.

    My data is attached, where VI is the voltage drop across the resistor next to Vin (not pictured), and VO is the voltage drop across the resistor past the collector. Since the voltage drop across RE is constant at all values of Vin, its current is also constant.

    Am I correct in thinking this transistor was non-functional?

    Thanks!
     

    Attached Files:

  2. jcsd
  3. Jan 24, 2012 #2
    I think so. The voltage at the base is stuck at 11.xx volt.
     
  4. Jan 24, 2012 #3

    vk6kro

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    That 100 K in the emitter is too high.

    Try 1000 ohms.
     
  5. Jan 25, 2012 #4
    You have biased it wrong. Your results show its in cutoff region. Bias it properly.
     
  6. Jan 25, 2012 #5
    I don't think it is biased wrong. There is no bias, the input is the bias and the table show all the voltages from 0 to 11V. This is only an emitter follower. I think is the transistor burnt. Assume it is NPN, the base stuck at about 11V.

    I don't know about the resistor on collector. I assume VO is voltage across the emitter resistor and is always on 11 volts of so. Everything is stuck close to Vcc. The transistor is shorted.

    Make sure there is no solder splash, make sure the transistor is solder in correctly, make sure it is NPN. Double check everything. I never seen a bad transistor out of the gate yet. You have to burn it.

    100K at the input is a little high and is not a good practice, but it is not going to give the result in the table. NPN draw current from the base, if anything if you put 5V at the input, the base voltage is going to be a little lower because the base draw current, not stuck at 11V. Actually 100K input is not the end of the world. The emitter resistor is 100K, so the max emitter current is 110uA. At this current, a lot of transistor has beta of over 100, so the base current is going to be less than 2 uA therefore the normal voltage drop across the 100K resistor is only 0.2V. Nothing really wrong about this.

    I am current working on a design I use less than 20uA emitter current and the transistor I use BC550 beta of about 450!!!
     
    Last edited: Jan 25, 2012
  7. Jan 25, 2012 #6

    vk6kro

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    Maybe the transistor is shorted, but with a 100 K emitter resistor, you would not know.

    Try reducing the 100 K emitter resistor to 1 K.
     
  8. Jan 25, 2012 #7

    jim hardy

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    what kind of transistor is it?

    some have leads in sequence EBC, others in sequence ECB

    voltages are in agreement with transistor's base & collector leads swapped .

    ie V+ is going to base and Vin to collector.
     
    Last edited: Jan 25, 2012
  9. Jan 25, 2012 #8

    AlephZero

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    Would you expect to measure the "correct" voltages with an cheap multimeter, for those low currents? I don't think so.

    For a simple demo of how an emitter follower is supposed to work, an emitter current of 20mA would be more practical to work with than 20μA.
     
  10. Jan 26, 2012 #9
    I just got my circuit running and measure every single voltage with a $39 cheap DVM. Yes, my DC bias resistor is 510K and tail current of 20uA. I am working on guitar electronics that run on 9V or even watch batteries. Also the magnetic pickup characteristic of guitar change with loading resistor. It has to be between 250K to 500K and all electronics has to accommodate these high impedance and low current design. This is nothing new.
     
    Last edited: Jan 26, 2012
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