Trying to understand a 2N2222 transistor

Discussion in 'General Electronics Chat' started by impaJah, Aug 7, 2012.

  1. impaJah

    Thread Starter New Member

    Aug 2, 2012
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    Okay from what I've gathered it's the base to emitter voltage that determines the amount of current that's let through from the collector to the emitter.

    If this is true then why isn't the base to emitter resistance given? How can you set up a voltage divider circuit to get the voltage (and consequently the current through C to E) where you want it if you don't have the resistance across BE?

    And on top of this question, where is the ratio between BE voltage to CE current? I can't find it anywhere on this datasheet:

    http://pdf1.alldatasheet.com/datasheet-pdf/view/15067/PHILIPS/2N2222.html

    One last thing, does a "-" instead of a number in a table indicate that it's the same as the number near it?

    Thanks in advance, I know these questions are really nooby but I haven't been able to find the answers anywhere.
     
  2. Audioguru

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    Dec 20, 2007
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    The required base-emitter voltage changes when the transistor is changed and changes when the temperature is changed.
    The base current should be used as the input signal instead of the base-emitter voltage.

    If the input signal modulates the base current then the collector signal is inverted and has a negative gain number.
     
  3. MrChips

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    The base to emitter current-voltage characteristic is highly non-linear, i.e., the resistance is not constant.
     
  4. #12

    Expert

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    Exactly true. We speak of base to emitter voltage because we are familiar with the non-linear characteristics, and some voltage is required, but it is the current through the base to emitter junction that allows a proportional amount of collector to emitter current.

    Same statement, different words, trying to see which description makes sense to this particular person.
     
  5. Markd77

    Senior Member

    Sep 7, 2009
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    The "-" means something like not specified or not applicable or zero, you have to interpret which one from which column (max/min/typical) it's in.
     
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  6. impaJah

    Thread Starter New Member

    Aug 2, 2012
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    I just measured the resistance from B to E on a 2N2222 I have and it was over 20MΩ. So how is any current even getting through to perform its "valve" function?
     
  7. #12

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    How are ohmmeters designed to intentionally not activate diode junctions?
     
  8. MrChips

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    You cannot measure the resistance of a PN junction with an ohmmeter.

    If you would like to conduct an Electronics 101 Lab Experiment, use a variable power supply, a series resistor to limit the current and an ammeter, all in series with the base-emitter junction.

    Vary the voltage from the power supply. Measure the current in the circuit and the voltage across the junction.

    Plot the current versus voltage.

    You will observe that it is an exponential relationship.
     
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  9. Audioguru

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    My Fluke meter uses 200mV to measure resistance so that it does not turn on a diode.
    Its "diode test" uses much more, maybe 5V.

    A silicon diode (or silicon transistor base-emitter diode) turns on and conducts when the voltage is about 0.6V or 0.7V.

    Here is a graph of the forward voltage of a 1N4148 small silicon diode.
    Its voltage increases 1.46 times when its current increases 100 times. the base-emitter diode of a small transistor is almost the same.
    The voltage changes when the temperature changes and two diodes or transistors might not be the same.
     
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  10. MrChips

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    [​IMG]

    Note that the Forward Current is plotted on a log scale.

    While the voltage increases from 0.5V to 0.75V, the forward current changes by two orders of magnitude.

    The fact that the relationship is a straight line indicates that the forward current is increasing exponentially as the voltage increases.
     
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  11. impaJah

    Thread Starter New Member

    Aug 2, 2012
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    Hey Mr Chips, I just tried to do the experiment you recommended and I encountered a problem I couldn't account for. I put a 100Ω in series after ramping down from 5kΩ and reading no current draw on my meter. At 7.5V I read .01mA on my meter and was about to measure the voltage across EB when I smelled something burning and realized it was coming from the resistor and so I quickly turned off the power.

    This resistor is rated for 1/4W and at .01mA/7.5V I wasn't even close to that. Why was my resistor getting fried here? Am I missing something?
     
  12. #12

    Expert

    Nov 30, 2010
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    Obviously.

    You will have to re-check the operation and accuracy of all your meters.
    I do that every time I do a research project like this.
     
  13. impaJah

    Thread Starter New Member

    Aug 2, 2012
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    Well I'm an idiot. I accidentally put the positive and negative terminal of the meter in parallel instead of in series - DOH! :p
     
  14. Audioguru

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    Without your schematic then we are just guessing.

    I assume that you connected a 100 ohm resistor in series with the base and connected the resistor and emitter of the transistor to 7.5V.
    Then the resistor had (7.5V - 0.7V)= 6.8V across it and it dissipated (6.8V squared/100 ohms)= 0.46W so of course the 1/4W resistor got burnt.
    Its current was (6.8V/100 ohms)= 68mA, not only 10mA.

    Like this?
     
  15. #12

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    I actually did this experiment and arrived at an equasion to describe the results. In other words, it works.
     
  16. impaJah

    Thread Starter New Member

    Aug 2, 2012
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    Yah I got it to work too! I understand transistors much better now!!! :D :) :D
     
  17. MrChips

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    For anyone else who wants to try this experiment and you only have one DVM, here is another way of doing this.

    Set the DVM to a voltage range and take two voltage measurements for each setup ( i.e. you can vary supply voltage or series resistor R).

    Measure voltage across R and calculate current I = V/R.
    Measure VBE (or diode voltage if you are using a discrete diode).

    Plot I vs VBE.

    Do show us your data and graph.

    (Note that VBE will not exceed 1V.
    7V supply is way too high).
     
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  18. impaJah

    Thread Starter New Member

    Aug 2, 2012
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    I did some further testing because I wanted to figure out what factor the current gain is ruled by and I would just look at a datasheet for this but I can't figure the thing out.

    Anyway I tried two different BE voltages and I found that the current gain is a factor of ≈200. So .53mA at BE can trigger a max current gain of about 106mA.

    Now here's where I'm confused. I then measured the voltage across CE and it was 2.72V and falling. So the total dissipation of the transistor in this circuit was ≈299mW. Now I started to smell some burning and I felt the transistor and it was getting hot. So I checked the datasheet and this 2N2222 transistor has a max total dissipation of 1.2W.

    See for yourself: http://pdf1.alldatasheet.com/datasheet-pdf/view/15067/PHILIPS/2N2222.html

    I'm not even near that dissipation so why was this load too much for it? :confused:
     
  19. Audioguru

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    Read the datasheet for the 2N2222 in an old metal case:
    1) Its maximum power dissipation is 500mW when the air is 25 degrees C.
    2) Its maximum power dissipation is 1.2W when its case is cooled to 25 degrees C somehow.

    Dissipating 299mW then of course it will be very hot.
     
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  20. Markd77

    Senior Member

    Sep 7, 2009
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    Also note that the maximum junction temperature is 200 Celcius (the case will be slightly cooler than that) - at 500mW dissipation the case will cause pretty painful burns, not just feel quite hot.
    The maximums in the datasheet are usually to be avoided, they are not a guarantee that the part will last long under those conditions.
     
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