Emitter circuit / Calculate Operating Point U(CQ)

Discussion in 'Homework Help' started by Tobias Hildebrandt, Jan 15, 2016.

  1. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
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    Hello,

    this is an old exam question. The blue writing is the solution provided to us by our Prof. In Q5 A our Prof. wants us to calculate the operation point U(CQ).

    I am puzzled!
    1) I always assumed that you get the operation point when biasing!? Apparently this assumption was wrong. I originally wanted to use the voltage divider with the resistors R1 and R2, this got me nowhere rather quickly. So, biasing != operation point?
    2) Then I looked at his solution and got even more puzzled. How does he get 0.4842mA? it seems a rather awkward number to pick randomly...
    3) It seems to me, he calculated the voltage between R4 and Collector. Is that where the 'operation point' is?

    Tobias
     
    Last edited: Jan 16, 2016
  2. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It seems that some info is missing.
     
  3. RBR1317

    Active Member

    Nov 13, 2010
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    The operating point is the quiescent collector current, Ic. The base current Ib is 200 times smaller than Ic. The base resistor network will have a Thevenin equivalent Vth & Req. The base voltage must equal Vth minus the voltage drop across Req caused by Ib. The base voltage must also equal the voltage drop across the emitter resistor R4 caused by Ib+Ic, plus Vbe=0.7V (or whatever you assume for the junction voltage drop). That is all you need to calculate the quiescent collector current.
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello there,

    First you could calculate the base voltage, then subtract the emitter voltage, then divide by the external emitter resistance of 2200.

    The base voltage is calculated by the voltage divider, but the voltage divider is not just R1 and R2, it is R1 and R2 in parallel with the input resistance of (given) 442000.
    R2 in parallel with 442000 is 42483, so the voltage divider voltage is:
    15*42483/(330000+42483)=1.7108 volts.
    Now subtracting an assumed base emitter voltage of 0.65 volts the emitter voltage is:
    1.7108-0.65=1.0608 volts
    and dividing by 2200 we get:
    iE=0.0004822
    and he may have used this current as the approximation for the collector current or else did this:
    iC=iE*200/201
    iC=0.0004822*200/201=0.4818mA
    which is also close to the text value.
    Since this does not match the text exactly, he may have included the internal emitter resistance Re for example, or maybe copied the math results incorrectly after one or more calculations.
    In any case, this is probably close enough and to find out how the instructor got that exact result you probably have to talk to him/her. Another example may be that he used the actual spice voltage drop for the base emitter voltage. I used 0.65 because that is pretty close usually if the base is not driven too hard.
    We could also calculate the value he might have used for Re too, the internal emitter resistance.
    Another example of how the translation from calculation to text could have gone wrong is he could have used the more exact input resistance of 442.2k while writing the more approximate value of 442k. Stuff like this happens all the time, but to find out for sure you would have to talk to the instructor that created this circuit problem.
    You could easily compare all this (this here and the original text) with the spice simulation using a spice simulator like the free LTSpice.

    LATER:
    I found that if i use 0.6403 for the base emitter voltage i can get the same value as the original text for the collector current, but using more precise calculations for calculating things like the input resistance. Re (internal emitter resistance) has less effect at these low currents so i left that equal to zero.
    The equation for the collector current is:
    Ic=((B-1)*(Vcc*B*R2*R4-Vbe*B*R2*R4-Vbe*B*R1*R4-Vbe*R1*R2))/(B*R4*(B*R2*R4+B*R1*R4+R1*R2))
    where B here is not the Beta, it is the Beta+1 so B=201 for this problem.
    This equation can be used to experiment with different values in the circuit.
    For example, trying Vbe=0.62v and temperature of -1.28 degrees C we get approximate Re=48.1 and that takes us very close to the original text value for the collector current.
     
    Last edited: Jan 16, 2016
  5. Jony130

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    Feb 17, 2009
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  6. dannyf

    Well-Known Member

    Sep 13, 2015
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    It is fairly simple, and your intuition isn't even wrong.

    1) Vb is determined by the bias resistors, only to the extent that the current going into the base (Ib) is sufficiently small (as a result of sufficiently small Ic or sufficiently large beta).
    2) That condition holds for most applications but not all. And this case is one where the teacher doesn't want it to hold - part of the reason is that Ib (2ua) is quite comparable to the current going through the bias resistors (=15v/(330K + 47K) = 40ua), in the eyes of the teacher.

    In cases like this, you run the calculations through iteration:
    1) calculate Ie/Ic (assuming that Ib = 0);
    2) once you have Ic, calculate Ib;
    3) once you have Ib, revise Vb and then calculate Ie/Ic;
    4) go back to 1) until the solution converges sufficiently.

    I used an excel and calculated three scenarios for you - see the attached.

    Rather than going through iterations manually, I treated the Vbe->2.2K branch as an equivalent resistance (designated as Rl-2 in my spreadsheet), and parallel it to the 47K bias resistor (Rl-1 in my spreadsheet) in order to calculate Vb. I set the spreadsheet on iteration and the results pop'd out.

    As you can see, Ie/Ic is roughly the same for beta > 200; but at lower beta, assuming Ib = 0 yields significant errors.
     
  7. dannyf

    Well-Known Member

    Sep 13, 2015
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    The above calculation relies on iteration - which can be messy to calculate without a computer / simulation.

    As a 1st order approximation, you can "equate" the b-e junction + Re resistor as a simple resistor whose resistance = beta * Re (=200 * 2.2Kohm = 440K in this case), while ignoring the "equivalent" resistance of the b-e junction. Parallel that equivalent resistance in the determination of the bias point and you will get a result that is very close to the one using the iteration approach shown above.

    This approach is technically incorrect but I would prefer to call it "correctly incorrect", :)
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    See formula in post #4 for direct calculation.
     
  9. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
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    So, before I sit down again and study again, I want to take this opportunity to say thank you to everyone! I am overwhelmed by the support! Thanks to ALL of you! All of you have been a great help where my books failed me. So and no back to the problem.
     
  10. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
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    This is how I understood your instruction and either sth. was lost in translation or I made a mistake.

    Step 1) 200 * 2.2 kΩ = 440 kΩ
    Step 2) Did I get this right, you say this resistance is parallel to R2 (47 kΩ). This translates to a parallel resistance of 42.46 kΩ
    Step 3) Calculating bias point 15 V * 42.46/372.46 ≈ 1.71 V
    Step 4) Voltage drop 1.71 V - 0.6 V ≈ 1.11 V
    Step 5) 15V - ((15 kΩ/2.2kΩ)*1.11) ≈ 7.43 V

    Close, but no cigar. What am I? A German that can't follow orders???

    I found this in some old lecture notes from my Prof. You don't need to speak German, I highlighted in yellow the equations that I wanted to use. Unless I am very much mistaken, you can use equation I to III to solve this problem.
    1) Calc U(Th) (Equation I)
    2) Calc U(E) (Equation II)
    3) Calc U(out) (Equation III)
    If I use all the values given in the question, I get 7.023 V and that is wrong too.

    In his exam solution, he used equation IV, equation V seems helpful too.

    Couple of comments:
    -I contacted Prof. Dr. Best and I wanted to know how he got that value. If he comes back to me, I will let you know the answer.
    -I am a lot older than most of the other students, but this does not seem super easy. Do first year students in the US and UK have to deal with similar problems, is this an appropriate question for a first year student? In this exam we cover, high/low pass filter, emitter follower, emitter amplifier, FETs, Logic Gates. I am just curious.

    Thanks again for all your help.
    Tobias

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  11. dannyf

    Well-Known Member

    Sep 13, 2015
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    Not sure if I follow your professors but here is what I would do:

    The low side bias resistance is 47K//440K (not exactly but good enough) = 42.46K;
    Vb = 15 * 42.46/(42.46+330)=1.71v;
    Ve = Vb - 0.7v=1.01v;
    Ic = Ie = Ve / 2.2 = 0.4591ma;
    Vc = 15 - Ic * 15K = 8.11v.

    You can re-iterate further but that's fairly close to his answer.

    The rest of his calculation is for (ac) input impedance and the corner frequency.
     
  12. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    I think that's none of this steps what used by your professor, except step one.

    (1) Vth = 15V*47kΩ/(330kΩ + 47kΩ) = 1.87V and Rth = 47kΩ||330kΩ = 41.14kΩ

    (2) Write the KVL for the "input" equivalent circuit
    1.png

    Vth = Ib*Rth + Vbe + Ie*R4

    and

    Ie = Ib*(β + 1)

    Vth = Ib*Rth + Vbe + Ib*(β + 1)*R4 we solve for Ib

    Ib = (Vth - Vbe)/(Rth + (β + 1)*R4) = (1.87V - 0.7V)/(41.14kΩ + 201*2.2kΩ) = 0.00242065mA = 2.421μA

    And Ic = Ib*β = 2.421μA*200 = 484.2μA = 0.4842mA

    (3) And from KVL for the "output" circuit we have

    Vc = Vcc - Ic*R3 = 15V - 0.4842mA*15kΩ = 7.737V

    And this is the way you should do it.
     
    Last edited: Jan 17, 2016
    Tobias Hildebrandt likes this.
  13. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hello again,

    Yes that seems like a nice way to do it. At least we know he's not using the internal re for the calculations.

    The simplified network equation comes out to (B=DC Beta):
    Ic=(B*(Vcc*R2-Vbe*R2-Vbe*R1))/(B*R2*R4+R2*R4+B*R1*R4+R1*R4+R1*R2)

    which gives us:
    Ic=4.8414e-4 amps

    so we have:
    Vc=7.7379 volts

    Note Ic is very slightly different because there are no approximations used along the way.

    The most important point here is probably that the input resistance of the transistor itself alters the calculation because it acts in parallel with the circuit resistances.
     
    Last edited: Jan 17, 2016
  14. Tobias Hildebrandt

    Thread Starter Member

    Jan 1, 2016
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    Thanks to everyone! That solved it!!!
     
  15. WBahn

    Moderator

    Mar 31, 2012
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    NOTE: This is being posted well after it was intended do to Internet outage at the house.

    The term "operating point", in general, refers to the entire set of voltages and currents throughout the circuit in the absence of any signal. If you absolutely have to pick one thing to be THE operating point of a transistor circuit like this, the best choice is (usually) the collector current. But really any quantity from which you can calculate all of the other operating point parameters works and sometimes other choices have more meaning in the context of the discussion. In this case the one parameter that the prof has provided is Ucq - the quiescent collector potential.

    The 0.4842 mA is NOT a number picked randomly, it is the quiescent collector current (he probably would have labeled it Icq). It's just that the computation of Icq isn't shown in what you provided. Let's see if we can get a good approximation of it quickly.

    Ignoring the effect of the base current, the base voltage is 15 V * (47 kΩ / (47 Ω + 330 kΩ)) which is 1.870 V. If we assume a Vbe of 0.7 V then we have an emitter voltage of 1.170 V appearing across a 2.2 kΩ resistor giving an emitter current of 0.532 mA. That's within about 10% of his result, which is almost certainly good enough given the uncertainty in several of the parameters, such as the supply voltage and the resistor values. One quick check is to see if the base current is negligible compared to the voltage divider current. The base current would be about 532 uA / 200 = 2.66 uA. The voltage divider current is about 15 V / 377 kΩ = 39.8 uA, or about 15x the base current. So it's in the border realm of what is negligible and what you need to take into account.

    We could take it into account by explicitly taking the transistor beta into account (notice that we are still assuming a particular value for Vbe and 0.7 V is probably not the best choice for this transistor at this operating point, but we work with what is available). But since we already have a good idea what the numbers are going to work out to, we can use an iterative approach quite effectively -- one iteration will almost certainly put us in the realm of where the other uncertainties dominate.

    We know that any current siphoned off into the base is going to reduce the current going into the bottom resistor which is going to lower the base voltage which is going to increase the voltage drop across the top resistor. But we also know that the effect will be to increase the current through the top resistor by only a fraction of the base current. So let's take an easy guess and assume that the top bias resistor has 40 uA of current in it and that 2.5 uA of current is diverted to the base. This leaves 37.5 uA through the bottom resistor giving a base voltage of 1.763 V and an emitter voltage of 1.063 V which, in turn, gives an emitter current of 483 uA. Taking the beta into account, this yields a collector current of 481 uA, which is right in the ballpark of what the prof used. We can determine if we are high or low by looking at what this result says about the base current. A collector current of 481 uA is a base current of 2.4 uA, or just a bit lower than what we guessed. That means that the actual base current is between 2.4 uA and 2.5 uA. But doing another round of iterations won't really do us much since we are almost certainly already in the realm where a bigger factor is what the actual Vbe is compared to the 0.7 V we have been assuming.

    For a beta of 200, that would be a collector current of
     
  16. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Internet go out again? Geeze what a pain...wonder what service provider you have.

    I had forgotten about that question, about the so called "operating point". I have to agree that there is no one node that can be called the operating point and that it is really about the circuit as a whole because every node is affected by the "operating point". In particular we can narrow it down by splitting it into two parts: the input bias and the output set point. The input bias point is partly set due to the requirement of the output set point voltage. For example, many times we want to bias at 1/2 Vcc. So the input bias would be set so that we get Vcc/2 as the output and those two would collectively make up the "DC operating point" of the circuit.
    To contrast, the operating point of the input capacitor would be the voltage across the capacitor when the input is zero volts. So if you were to solve for the operating point of the circuit the voltage at the base of the transistor would be the operating point of that node and of the capacitor.
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    We only have two choices for internet -- satellite or microwave. We had satellite for several years but in addition to the loss of service anytime it rained, snowed, or was excessively cloudy we also seldom saw speeds in excess of 100 kb/s. There is only one microwave service that we can tap and that is only because the provider set up a repeater on the roof of someone else to get a signal down into our valley for about four or five homes. We are shooting through trees so the signal isn't great at the best of times, but at least we usually have 1.5 Mb/s when we have service at all. Unfortunately any snow in the trees or even much more than a mild breeze shaking the limbs causes the signal to drop out.
     
  18. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Oh, sorry to hear that. I am a little surprised to hear that they dont have some kind of cable installed by now. Fios gets to a lot of places, but i think they reduced their expansion recently so not sure how long that will take to reach you, if they ever do go that way.
    Good luck in the future with it in any case...
     
  19. WBahn

    Moderator

    Mar 31, 2012
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    I doubt either cable or DSL will ever get up this way. To much wire run to serve too few people. They tried to see about running cable along the power lines but the lease was so high that there was no way to justify it. The best we can realistically hope for is that either cell coverage gets good enough to tether (and right now most places on the mountain have no cell coverage and you have to be in just the right place in or around our house to get any, although it has improved somewhat) or that they set up another few repeaters for the microwave link. Either of those are going to be slow in coming because there's just too few people here that would be served by any infrastructure expansion.
     
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