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  1. #1
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    early voltage of bjt

    Hello there,

    Greetings all, I am new here so I have a few nagging questions about BJT operational characterstics.
    According to electrical theory, a current source can be constructed and understood by taking a high voltage source and placing a resistive potential divider across this source, in this way the top resistor is made very high whilst the bottom load resistor is made relatively much lower perhaps 3 decades lower. With this arrangement, it is possible to emulate a approximate perfect current source. There's no problem with that, Nortons theory privides a switch between Thevenin's voltage source and a current source. The only problem I have with that, is Thevenin's generators provides inherent boundaries whilst Norton's current generator does not.
    Assume now that I have a ce bjt set up on the bench, biased and ready to go. I purposely make the base bias source resistance very low so as to avoid reverse voltage transfer effects and to maximise the early voltage. I then take a differential slice measurement of the output, i.e., delta Vce/ delta Ic across the bjt's non saturated active region.With this, I can yield output resistance and further calculations can yield the early voltage. Now, if memory serves me correct, a 2N3904 will yield about -100V in series with about 20k Ohms. So, looking into the bjt's collector is as though I have -100V source generator, in series with 20k Ohms feeding RL. So I have a approximate current source as first outlined. Now, here's the kicker, the power being dissipated in the bjt should be many orders higher than what the bjt is actually dissipating, which would violate Ohms law, as such there is no violation of Ohms law, as the power being dissipatated in the BJT is simply Vce*Ic and not (-100V+Vce)^2/(20k).
    How does the BJT or FET etc, manage to pull this one.
    Look forward to your reply

    •   Alt25th August 2006, 23:37

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  2. #2
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    early voltage bjt

    Hey there DarrenHeywood,

    It was fun decoding your question - acronyms with an English spin.
    For the uninitiated:
    bjt = bipolar junction transistor
    ce = common emitter configuration
    early voltage = (I have no idea - maybe rising voltage)
    whilst = while

    Let's consider your first problem - you wrote:
    "The only problem I have with that, is Thevenin's generators provides inherent boundaries whilst Norton's current generator does not."
    Au contraire my friend. From a voltage and current viewpoint they are identical - perfectly identical in every way. Open circuit voltages are identical, short-circuit current is identical, internal resistance is identical - what more could one ask.
    I hope you will reconsider your stance on this important issue.


    Now let's look at your second conundrum.
    Assume a supply voltage of 12.75 V. and a load resistor, RL, of 500 Ohms.
    You can check my math, but I'll give you the results.

    (Ic = 5.5 mA)
    Power dissipation:
    P(RL) = 15.125 mW.
    P(CE) = 55.000 mW.
    ------------
    P(Total) = 70.125 mW. This is the total power dissipated.

    Now let's look at the Thevenin equivalent assuming 20K and -100V as you suggest.
    P(20K) = 605 mW. This is the power dissipated in the 20K resistor.

    This is not the same as P(CE). Could Thevenin be wrong?
    No, not this time either. Here's the kicker - power sources PROVIDE power and are negative in the power entropy budget. So the power PROVIDED by the -100 v source is

    P(-100V) = -550 mW. Note: not negative because of the negative in -100V

    And total P(CE) - this is the power dissipated in the Thevenin transistor:
    P(20K) = 605 mW.
    P(-100v)=-550 mW.
    -----------------
    P(CE) = 55 mW. the same as the first

    And we can all breath a sigh of relief.

    Your method of using a base bias current is almost never used in simple ac amplifier design since the gain is then directly proportional to Beta which varies widely from unit to unit and with temperature, voltage and frequency. Usually the base is biased at some mid-voltage and the bias current and gain are controlled with an emitter resistor. But I assume you chose it to make the question clear.

    Hope this helps.



  3. #3
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    thevenin transistor

    Hi There,

    Thanks for your detailed reply, but I think you have missed my point entirely.Why you must use starting conditions with nth decimal places and a Vce 10V is not clear as this is going to make it difficult for others to read, nevertheless, I shall use your figures.
    We have Vcc=12.75V and Ic=5.5mA, so the volt drop across RL is 2.75V and Vce therefore = 10V.
    I am not interested in the power being dissipated in RL (15.125mW), I don't have a problem with that, my problem is power being dissipated in the transistor(ignoring biasing currents and voltages).
    So, according to the early voltage of (-100V) through a resistor of 20k, the power being dissipated in the transistor is (100+10)*5.5mA=605mW which is way too high.The actual power being dissipated is simply 10*5.5mA=55mW, this latter figure is the correct one and Ohms law treats the transistor as a soggy ordinary resistor of 1.818181k down to deck!
    In otherwords, my problem still exists. I am fully aware of emitter degeneration(voltage feedback) to stabilise the transistor both for biasing and for linearisation of signal, but I do not wish to add unnecessary complications to my central issue.
    Best Regards
    Darren



    •   Alt26th August 2006, 16:11

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  4. #4
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    ohms law high frequency

    Darren,

    First, I went to some trouble to calculate the supply voltage of 12.75 V so that the current (5.5 mA) and Vce (10.0 V) would NOT be numbers to the Nth decimal point. I designed the setup for easy results. This will make it easier for others to read and follow, not harder. If I had chosen the supply to be 10 V, then the current would be 5.365853658536585365853658536585... (truly n places)

    Second, please give my explanation another chance. The power in the 20K resistor is not the only factor. You must consider the power PROVIDED by the Thevenin Voltage source and subtract if off. This is how to calculate the actual power dissipated in the device using the Thevenin/Norton parameters.

    Let's try another example:
    Suppose I have an ideal 1 V supply with an ideal 1 Ohm resistor. What is the open circuit power? Zero, right? Now convert this to the Norton equivalent: A 1 Amp current source with a 1 Ohm resistor in parallel. Same v & i characteristics. But the power is 1 Watt. How do you explain this? Well I explain it by understanding that I must subtract the power provided by the current source as 1 Watt to get the real power dissipated (0). But you will declare that there is 1 Watt being dissipated in this Norton device open circuit, but the Thevenin device has no power being dissipated and so something is amiss with the theorem.

    How is this explained?
    Easily - Thevenin and Norton translations are not directly concerned with power. There is no guarantee of equal power dissipation in the equivalent or actual resistors.

    A side note:
    From your tone it sounds like you are somewhat miffed at my reply. My apologies if my attempt at humor have offended you. Most of us are here to learn and help and maybe have a little fun. So sorry if I crossed the line. We should be able to have a friendly debate don't you think?

    Dennis



  5. #5
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    early voltage 2n3904

    Hello Dennis,

    First of all, I would like to apoligise for sounding abit stern maybe?, secondly I don't consider that you crossed my line except that you may have come across as overly confident about a problem that has puzzled me for a very long time.Thirdly, I think this is a great site where interested parties can debate central issues, both hot and cold.Forthly, I too hope that I did not offend you in anyway whatsoever.
    Lets get back to my problem in question, to make things even less ambigious, let me propose that we will put the Thevenin, Norton generators on the back burner, as I do not require either to force home my problem.
    Lets assume I have 112.75V supply with negligible internal resistance, i.e., it's very low. I now connect a 20k resistor in series with 500 R resistor down to deck, i.e., so I have a ordinary potential divider across the 112.75V supply. If I increase or decrease the 500R resistor in small decrements/increments I find the current, which is common to both resistor remains more or less the same, in other words, I have an approximate current source. If I short out the 500R resistor completely, the current will rise only slightly, but thats to be expected since my supply voltage is only 112.75V. Now, how much power is being dissipated in the 20k resistor?
    As before, this will be 605mW, this is real power, I know this because my 1/2w rated 20k resistor gets hot.
    Getting back to my transistor, this 20k exists inside the transistor, in otherwords it should be dissipating 605mW power and we both know it isn't, otherwise I could announce to the world that I've created free energy. Secondly, Ohms law comes to rescue and furnishes the correct power dissipation of 55mW but only at the expence of saying, there is no early voltage of 100V nor is there a 20k resistor present but in fact just a soggy ordinary 1.81k resistor to deck.
    Just to add icing on the cake, it's possible to wrap an op amp around any transistor, under 100% feedback, whereby the Alice in wonderland early voltage then becomes tens of thousands of volts at d.c. e.g., 4 to 20mA signal generation.
    Also, both rf and high Vce(max.) type transistors yield early voltages between 500 to 1000V.
    I look forward to any comments anyone may have.
    Best regards
    Darren



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    calculate in circuit power dissipation bjt

    Hello again Darren,

    Sorry that I came across as overbearing. In my defense, I have a Masters Degree in EE and over 40 years experience as an electronics engineering consultant, blah blah, blah. So sometimes I tend to just blurt out things the way I see them.

    Now back to our little discussion. I think I have determined the root of this thing.
    You say "Getting back to my transistor, this 20k exists inside the transistor..." and this is the crux of the matter. There is no 20K inside the transistor. If you look up the equivalent circuit for a transistor, you will see that it is a current controlled current source for the most part. That's what's really inside - a junction. If you setup all the parameters of the transistor parametric circuit and calculate the power, you will get the actual power dissipated. The Thevenin equivalent is simply a mathematical representation for calculating voltage and current as a function of load impedance.

    So does that bring us closer to the same thinking? If not maybe someone else will chime in with a better explanation.

    BTW what does early voltage mean?
    And soggy?
    I assume deck mean ground or common?

    Good Luck,
    Dennis



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    current source bjt has resistances

    Hi dennis,

    I am not sure what you mean by BTW, what I mean by 'soggy' is just an ordinary passive positive gradient resistor that you buy at the shops, 'Deck' is indeed ground or 0V potential. When you refer to Ohmic models of the transistor and there are many to suit a variety of situations, be careful you don't fall into the trap of ac signals as opposed to dc signals, where resistors become parallel as opposed series. Just to give you some more food for thought, according to electrical theory, there is no power difference between the 'useful' power in a 20Ghz signal as opposed to a 20Hz signal, assuming both have the same resistive load and potential be that r.m.s. or pk to pk values.As such, electrical power is proportional to voltage^2 or current^2, in every other area of physics it's proportional to either frequency^2 or amplitude^2. When an audiophile thinks he has the best sounding amplifier in solar system, is he aware that electrical signals, in terms of frequency, are not related power in any way, but when his music undergoes the transfer function of his speaker set, power 'in air' is proportional to the frequency^2. Just an observation.

    My original problem remains unanswered.

    Best Regards
    Darren



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