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Collector current of an NPN common emitter circuit

Discussion in 'General Electronics Chat' started by MrAl, May 29, 2013.

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  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello there,


    We have a 2N22222A transistor and 1k collector resistor (Rc) and a base emitter voltage (Vbe) of 0.65 volts. This is powered by a 10v supply source Vs. What is the collector current Ic ?

    Feel free to use whatever method you care to. In another thread we were talking about using a voltage control method so that's why i stated a base emitter voltage (Vbe) of 0.65 volts. We would like to stay away from Spice for now though, and use that later to check our results.
     

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    Last edited: May 29, 2013
  2. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    Hi Al,

    I cannot calculate the collector current based on those parameters.:rolleyes:

    Eric
     
  3. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi there Eric,


    Can you tell me why that is? What would you like to see changed?
     
  4. dave

    Dave New Member

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  5. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi,
    For starters the transistor hfe and Base current due to the 0.65v voltage on the base pin.
    E.
     
  6. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again Eric,

    Couldnt the HFE be found from the data sheet?

    So how would you change this circuit to make it more 'solvable' ?
     
  7. audioguru

    audioguru Well-Known Member Most Helpful Member

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    If you apply 0.65V to the base-emitter of a transistor then some transistors will have a high base and collector current and maybe will burn out but other transistors will have a low base and collector current because each transistor is different even if they have the same part number.
    Also, a transistor conducts more when it is warm and conducts less when it is cool.

    Therefore, we feed a current (not a voltage) to the base of a transistor and add an emitter resistor to swamp the difference of Vbe of different transistors.
     
  8. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    Hi
    I could get the hfe from the datasheet, I could also get an estimate of the Ibase from the d/s plots if they were shown.

    To make it more solvable I would calculate and add biassing resistors and perhaps an emitter resistor if I wanted to have a known stage gain.

    E

    What would you do to make it more 'solvable' ?..:rolleyes:
     
  9. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,

    audioguru:
    This is more of a theoretical question where we hold certain things constant like the temperature, and ask about the behavior when we look at the base emitter voltage.

    Eric and audioguru:
    I have attached another circuit, with an added base resistor. Now we are MEASURING Vbe rather than applying it directly. The measured value with Rb=518k is 0.65 volts again. So the base emitter voltage is again 0.65 volts. See if this makes it more simple now. But keep in mind i am not looking for super super accurate numbers here, my main goal is to (possibly) show that using the Beta vs using the "voltage control" is not that much different because the unknowns are still unknowns or rather the knowns are not as well known as they are claimed to be. In fact, i think there is a spread for the voltage control parameters as well as the well known spread of Beta.
    So anyway, i would think that this circuit should not throw us 'professionals' in 'electronics' too far, that we should be able to come up with some reasonable result because after all it is just a one transistor circuit right?
    Feel free to agree/disagree on any item, we can discuss it in more detail.
     

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  10. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi,

    Suffice to say, the circuit as drawn is unrealistic and its performance unpredictable.

    The Base current will be ~18uA.

    Just using the d/s, no equations, for Vbe versus Ic the 2N2222, gives approx 5mA for Vce=10 at 25C.

    Also from the d/s using the hFE versus Ic gives an hFE of approx 180, which gives a IC figure of approx 3.2mA

    E
     
  11. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again Eric,


    Thanks Eric, you are the only one on this board who can answer this question :)

    Yes this is more or less a theoretical question where we hold certain things constant like the temperature, so nothing can change once we have established a working point.

    I also point to this link for the total voltage control formula, where we simplify kT/e (which is more usually expressed as kT/q) as 0.0257 at about 25 degrees C. And that temperature is assumed to be held constant. If you care to take a look at this too...

    http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html

    Scroll down for the last circuit on the page.
     
  12. audioguru

    audioguru Well-Known Member Most Helpful Member

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    How can you hold the temperature of the transistor chip constant when the current though the transistor causes it to warm up?
    Also, you do not know the hFE of a transistor unless you measure it. But the hFE also changes with temperature changes.

    I was taught to NEVER bias a linear transistor like that.
     
  13. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello there audioguru,


    First, thanks for taking part in this discussion. I think the more it is discussed the more details come out.

    As i was saying, this is a theoretical problem i am looking at. That means we can keep certain things constant regardless what happens in the real world. But if you feel more comfortable, then here's what we will do:

    Connect a Peltier device with heatsink to the transistor package. Connect the Peltier device to a control circuit that controls the current through the Peltier device. Using a microcontroller with a 16 bit A to D converter, measure the base emitter voltage Vbe. The uC programming is such that if the Vbe goes down then increase the current through the Peltier device. If the Vbe does up then decrease the current through the Peltier device. In this manner keep the Vbe constant at 0.65 volts. Keeping the 0.65 volts constant in this manner keeps the die at the constant temperature.

    What Eric did was he found spec's on the data sheet that specify the collector current with Vbe. Note that we dont have to be perfect here, but we should be able to at least match up with some data sheet specs at least in theory.

    What i would like to do later is take a look at the same thing except using a well known formula, or at least trying to use a well known formula. We should be able to get results either way, then compare techniques using the formula vs using the Beta method. Because some day we may want to use this in a real circuit, it's ok if there is a Beta spread too. We just have to make sure one day that the Beta spread does not hurt the operation too much.
     
  14. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You are chasing a question that has little practical value. The question should not be about the difference between pure "current control" and pure "voltage control" for biasing since both vary and neither gives sufficiently stable results to use in practice. The question should be about the stability of properly designed bias circuit due to both Beta and Vbe variations. A good bias circuit is relative insensitive to both these factors. That is why local feedback using an emitter resistor is typically used to bias a transistor since it minimizes the effect of both.
     
  15. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello Carl,

    The main reason i havent included a emitter resistor is to make this problem simpler from a formulaic point of view. We could include an emitter resistor i guess but then we'll have to calculate Vbe too.
     
  16. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi Al.
    Checked your link and the last circuit on the page.

    Its all pretty standard information.

    I noted in your other Thread on this topic, some members were fussing about your approximations.

    All our known theoretical and practical physics are based on approximations, all the decisions we take in our daily lives are based on approximations.

    There seems to be some confusion regarding the required degree of accuracy and what is actually needed in order to give a workable solution.

    E
     
  17. Winterstone

    Winterstone Banned

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    Yes - I think, this contribution has a real practical background and is worth to be discussed further.
    The necessity of an emitter resistor has been mentioned already by Eric (see post#7).

    In the following I present an example, which clearly shows how dc feedback drastically reduces the uncertainties connected with the spread of beta and Vbe (see Is in Shockley`s famous formula).

    For the following example the only information regarding the BJT is that the base-emitter voltage in the active region will be between 0.65 Vand 0.7 Vs.

    Example:

    *Transistor stage in common emitter configuration
    * DC feedback with Re=1kOhm
    * Supply voltage: Vcc=12 V.
    * Given collector current: Ic=1mA.
    * Resistive divider for dc biasing assuming a divider current Id=0,1*Ic
    (this assumption results in a resistor niveau that seems to be acceptable with regard to the input resistance of the circuit)
    * Note that no information about the DC gain B=Ic/Ib is used

    (a) Assumption Vbe=0.65 V

    It is an easy task to compute the values for the divider:
    R1=103.5 kOhm and R2=16.6 kOhm.

    Simulation result for 2N2222: Ic=0.92mA.

    (b) Assumption Vbe=0.7 V

    R1=103 kOhm and R2=17 kOhm.

    Simulation result for 2N2222: Ic=0.96mA.
    _______________________________________

    Summary: Based on a rough estimation for Vbe (0.65...0.7V) and without any knowledge/estimate for B=Ic/Ib it is possible to design a common emitter stage with dc feedback.
    A circuit simulation for a real transistor shows that the calculated resistor values for BJT biasing lead to a collector current that deviates from the target value by only (4...8)%.
    This deviation has the same magnitude as other uncertainties within the whole circuit (resistor tolerances, transistor parameters).
    Of course, this deviation (error) can further reduced for increased DC feedback (Re increase).

    W.
     
  18. crutschow

    crutschow Well-Known Member Most Helpful Member

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    You can ignore β in your calculations but that doesn't mean it has no effect on the bias point. The value of β affects the voltage output of the divider bias circuit as determined by the divider equivalent resistance. Variations in β will cause the divider voltage to vary, which will affect the bias point. It may have a small (or negligible) effect, but it does have an effect.
     
  19. Winterstone

    Winterstone Banned

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    Hi crutschow,

    it seems I have expressed myself not clear enough.
    Of course, you are right that B (resp beta) has an effect on the bias point. Did I state there would be no influence?
    My only intention was to show that the result is a good approximation of the target value for Ic - even in case
    * B=infinite (and that is the background for ignoring Ib), and
    * Vbe (rough estimation) between 0.65 and 0.7 volts.

    This example should demonstrate the immense effect of dc feedback on Ic stability - and is one answer to the question raised in post#5 ("So how would you change this circuit to make it more 'solvable" ?)

    If - in addition - a very rough assumption for B=Ic/Ib (50 or 100 or even 200) is included in the calculation the resistor ratio will be somewhat larger and bring the actual Ic value closer to the target value.
    Of course, from the system point of view this result is not surprising.
    We all know the background of the operational amplifier: Feedback drastically reduces the influence of the active device and its properties.

    W.
     
  20. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello again,


    Yes the span of B has quite an effect. From the data sheet (or at least one of the data sheets on the 2N2222A) i find that B can go from about 50 to about 300 when we are dealing with collector currents in the range of this circuit. That's because the span of Beta is not specified very well, as well as the variation with current, but even just looking at the manufacturing span we'd have to use that range if we wanted to be sure. And see this goes with any circuit. But lets look at this for a minute.

    The input voltage is specified as 0.65 and since we have a 10v power supply that means we have a fixed voltage across the base resistor which means we have a fixed input current of around 18ua.
    Now take that 18ua and multiply it by 50 and we get 0.9ma, which is less than 1ma. But take that 18ua and multiply it by 300 and we get 5.4ma, which is a considerably different Ic.

    But here's the catch, and whole point of this thread...

    The voltage control 'people' would like to say that by using the voltage control method of calculating the collector current is somehow more accurate. But there's no way to ignore Beta regardless how we do it. We dont suddenly loose Beta. It's still part of the transistor operation, or at least a way of looking at it, and it doesnt go away just because we use a formula without Beta in it.

    Am i making this point clear now?

    Anyway, for those who would like to see an emitter resistor in the circuit i've included that in this next circuit. The values are:
    Rc=1k as before,
    Re=100 ohms
    Rb=497k
    Vbe=0.65 as before.

    If you would like to throw a few calculations for this circuit in, we can take a look.
     

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    Last edited: May 30, 2013
  21. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    Al,
    Its reporting 'Invalid attachment'

    E

    OK now!
     
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