Stuck on this problem on my final exam review guide!

Discussion in 'Homework Help' started by BenBa, Dec 7, 2013.

  1. BenBa

    Thread Starter New Member

    Dec 7, 2013
    Hello everyone,

    Finals are coming up and I am trying to study for my Introduction to Analog Electronics class. It is my first electronics class and we have covered simple things such as passive components, passive filters, power amps, op amps, and active filters.

    I cannot figure out this problem for the life of me.

    I was wondering if i could show you guys what I have been able to figure out for this and maybe you can tell me what I am missing to help me understand these better?

    I am currently just stuck trying to find the Q point for Transisitor 3. I know tat the DC load line has an X intercept of Vcc because that is always the x intercept of DC load lines, and the Y intercept is Vcc/(Re+Rc) but seeing as how there is no collector resistor I think its just Vcc/R7, right?
    Now i have to figure out the quiescent current and quiescent voltage from from Vc to Ve. I know that Vc is Vcc for this transistor because its collector is hooked directly to Vcc, and i know that Ve is simply Vout, this means that Vce for the DC load lines Q point is Vcc-Vout. But I have no clue how to find Vout.

    for part b of this problem I am pretty sure all i need to do is use the DC load lines Q point to calculate the AC load line, and then find the most voltage that results in a Vce that is still on the AC load line? I am not sure what "mid frequency" means exactly...

    for part C of this problem I do not know how to determine the corner frequency of this circuit, it seems way to complicated to analyze (but maybe i am over thinking it).

    Any help is greatly appreciated!
    Last edited: Dec 13, 2013
  2. WBahn


    Mar 31, 2012
    Yes, please show your work, in as much detail as you can. That will let us better identify what you are doing right and where you are getting stuck.
  3. BenBa

    Thread Starter New Member

    Dec 7, 2013
    So i have used a voltage divider to find the voltage at the base of Q1, then use the standard diode drop to find voltage at the emitter of Q1 then used the resistor value to find the current. Now using this i can find the voltage drop from R4 and find the voltage at the base. Assuming the voltage drop form R6 is negligible i found the Voltage at the emitter of Q3 and used this to find the current and Vce.

    Now i am stuck on part b because i dont know what i means when it asks to find the maximum mid frequency voltage that results in no distortion. I think they are asking me to use the AC load line but im not sure how. I also forgot how to find the AC load line.
  4. Efron


    Oct 10, 2010
    What it is requested is to identify the maximum input voltage for which the output is undistorted, considering the mid-range frenquency (for which capacitors are not having any influence in the gain).

    In a problem, when you have questions a), b), c) ... usually b) requires a) to be done first.

    First complete a), we cannot say you did it so far. Your logic is correct but where is your "ic against Vce for Q3" ?
  5. BenBa

    Thread Starter New Member

    Dec 7, 2013
    Actually i need help figuring out part a), apparently i am not allowed to assume that the current into the base of each transistor is negligible, meaning I cannot use a simple voltage divider. How would i go about finding the voltage at the base of Q1 now?
  6. WBahn


    Mar 31, 2012
    Again, show your best attempt to solve it.
  7. BenBa

    Thread Starter New Member

    Dec 7, 2013
    I just don't know how to start. I don't know how to account for the very small current that goes to the base.

    I suppose we can use kirchoff's Voltage and current laws.

    So if i start at the node after C2 we get that

    V_{Q1_b} - V_{be} - I_{Q1e}R_5 = -V_{cc}

    The we use KCL to get
     I_c = I_{R_2} + I_{Q1_b} + I_{R_3}

    But I feel like we don't know any information to solve for any of this...
    Last edited: Dec 10, 2013
  8. panic mode

    Senior Member

    Oct 10, 2011
  9. WBahn


    Mar 31, 2012
    So first solve it -- all the way -- assuming negligible base current. This will give you a good estimate to compare your more detailed analysis to and will also give you a good feel for the flow you need to take. By "all the way" I mean more than just saying that you used a voltage divider for this and use that to find this over here. I mean say that you determine the voltage at the base of Q1 to be x.xx volts and the current in R5 to be y.yy milliamps, etc.

    That will give you a reall good idea of the current each transistor is carrying. You can then go to the datasheet and get reasonable estimates for the actual Vbe and β for each transistor and then use those to go back and adjust the analysis to take those into account. If the circuit has been designed well, the difference should be marginal.
  10. BenBa

    Thread Starter New Member

    Dec 7, 2013
    But I can't use the KCL to plug into the KVL. Lets assume Ib is negligible.

    We have
    V_{Q1b} - 0.7 - I_{Q1e}R_5 = -15
    I_{capacitor} = I_{R2} + I_{R3}

    So we can't plug anything in to solve for... Am i missing something?
  11. WBahn


    Mar 31, 2012
  12. BenBa

    Thread Starter New Member

    Dec 7, 2013
    current through capacitor is zero. so R2 = -R3?

    Of base current is ignored can't we simply use a voltage divider to find the voltage at the base of Q1? It would be -15 - 2Vcc*R3/(R1+R2+R3), right? That comes out to be -6.83Volts. But i thought we weren't allowed to call the base current negligible, at least thats what i think the exam wants.
  13. WBahn


    Mar 31, 2012

    R2 = 3.9kΩ
    R3 = 8.2kΩ

    As for the current through them, you need to assign polarities to your currents. Is I_R2 going upward or downward? Is I_R3 going upward or downward? Based on the equation you gave, it is impossible to tell. The only thing that is evident is that you are defining them as going in opposite directions, which, while perfectly valid, doesn't make much sense.

    Please use units properly and look at your equation to see if your answer makes sense in comparisong to it.

    You have

    V_Qb1 = -15V - 2Vcc*R3/(R1+R2+R3)

    But Vcc = 15V, correct? So you have -15V minus something that only uses positive numbers, so you should end up with something less (more negative) than -15V, right?

    The given data isn't helping matters because it says Vcc=±15V. Well, that means that there are two values for Vcc. But the schematic clearly shows that the positive supply is "+Vcc" and the negative supply is "-Vcc". That pretty clearly requires that Vcc be the single value, namely "Vcc=15V".

    So, keeping that in mind, your equation should be:

    V_Qb1 = -15V + 2Vcc*R3/(R1+R2+R3) = -6.827V

    As you have noted. Note that unless you are keeping the intermediate results in memory somewhere, it is best to record them with one or two additional sig figs to prevent excessive round off error.

    You are doing fine. So keep going.
  14. LvW

    Active Member

    Jun 13, 2013
    Hello Benba,

    at first: It seems to be a rather challenging task. This comment seems to be appropriate because it is even not a simple task to calculate the bias point and all the other properties (gain, input impedance) of a simple gain stage consisting of only one single transistor. The other way round - to find the parts values for a given operating point is much easier.

    Secondly: In any case, your calculation must be based on two assumptions, which I didn`t see up to now in your steps to approach a solution:
    You need a realistic assumption for the
    * DC current gain B (in case you are not allowed to neglect base currents), as well as for the corresponding
    * base-emitter voltage Vbe (can be ssumed to be 0.65...0.7 volts).

    Otherwise, you have to use the characteristic curves Ic=f(Vbe) from te data sheet. But this wouldn`t give you more information.
    Without B and Vbe you cannot solve the problem.

    That`s what I can say at the moment.
  15. BenBa

    Thread Starter New Member

    Dec 7, 2013
    I used the value -6.827V at the base to find the emitter voltage (-6.827V - 0.7V) = -7.53V. Then using this i found the current through R5. (-7.53V- -15V)/2.7kOhm = 2.77mA.

    Then I assumed this was the same as the current flowing through the collector of Q2. Knowing this i found the voltage at the collector of Q2 by finding the voltage drop across R4 (2.77mA)(2.7kOhm) = 7.479, then i subtracted this from 15 to get the voltage at the collector 15V- 7.479V = 7.521V.

    Now I assumed that there was negligble voltage drop across R3, so the voltage at the base of Q3 is also 7.521V, making the voltage at the emitter of Q3 7.521V - 07V = 6.821V. Now using this i found the current through R7 using (6.821V - 0V)/3.6Kohm = 1.89mA. This gives me the current for my Q point. Knowing the Voltage at the emitter of Q3 is 6.821V i used 15v-6.821V = Vce = 8.179V.

    So my q-point is (1.89mA, 8.179V)

    But I had to assume negligible base current to get that, which is what i believe i am not supposed to do.

    @LvW, I believe we are to use beta = 100 for our transistors, and Vbe = 0.7
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    I would check that the R5 current of 2.77mA makes some sense.

    Assuming Beta values are 100 and VBE values are 0.7 Volt......

    This would make VE1=-7.521 Volts and VB1=-6.821 Volts.

    Continuing on, the current in the R3 [8.2k] would be 997.44uA. Adding this to IB1 [2.77mA/101=27.43uA] would make the current in R2 = 1.0249mA, giving a VB2 of VB1+3.997 Volts = -2.824 Volts. This leaves the current in R1 as 990.22 uA. And this doesn't seem to quite make sense, as the current in R1 should be greater than the current in R2 by an amount equal to Q2's base current.

    I believe this indicates a modest error in your analysis. A current of 2.77mA in R5 probably isn't far off - perhaps around 110uA too high. Presumably the error arises due to assumptions about the base currents being small enough to neglect.

    Of course this is rather academic since the small error encountered in the analysis would probably matter little given variations in beta and VBE in a practical situation.
    Last edited: Dec 12, 2013
  17. WBahn


    Mar 31, 2012

    Good, but you need to do a couple of sanity checks at this point. First, find the voltage at the base of Q2 and, from that, determine the voltage at the collector of Q1 and be sure that Q1 is, indeed, in the active region of operation. Second, use the collector and base voltages of Q2 to verify that it, too, is in the active region.

    I think you meant to say R5 and not R3.

    That all looks reasonable.

    If you can assume fixed values for β and Vbe, then you can do the analysis with that information directly and analyze the circuit to get the answers. The most basic way to do that would be to, having determined that each transistor is, indeed, in the active region, to replace them with the large signal linear model consisting of a DC voltage source equal to 0.7V between base and emitter and a CCCS between collector and emitter with a current gain of β.

    An alternative is to walk into the Q point by making incremental adjustments to the no-base-current data. This doesn't take as long as it might sound, provided the circuit is reasonably designed in the first place, since the adjustments are very minor.

    In this case we have three branches with currents (from the no-base-current analysis) of 997μA, 2.77mA, and 1.89mA from left to right.

    The base currents, with β=100, would be 27.7μA for Q1 and Q2 and 18.9μA for Q3.

    Notice that the base current for Q1 has to flow through R1 and R2 while the base current for Q2 only has to pass through R1. Let's call the base current amounts for Q1 and Q2 Ib and the amount that through downward though R3 I3. Yes, the base currents for Q1 and Q2 are not identical, but they difference is way down in the noise (as we will hopefully see).

    That means that we have

    (Vcc - -Vcc) = (I3+2Ib)R1 + (I3+Ib)R2 + (I3)R3

    2Vcc = I3(R1+R2+R3) + Ib(2R1+R2)

    I3(R1+R2+R3) = 2Vcc - Ib(2R1+R2)

    I3 = 2Vcc/(R1+R2+R3) - Ib(2R1+R2)/(R1+R2+R3)

    You can see that the first term is the no-base-current result and the second is the adjustment to it.

    From this we can get the current flowing downward in R5, which we'll call I5) by first finding the base voltage of Q1.

    I5 = (2Vcc*R3/(R1+R2+R3) - Ib(2R1+R2)*R3/(R1+R2+R3)-Vbe)/R5

    I5 = [2Vcc*R3/(R1+R2+R3)-Vbe]/R5 - Ib(R3/R5)(2R1+R2)/(R1+R2+R3)

    Again, the first term is the no-base-current case and the second is the adjustment.

    Also, note that this is of the form

    I5 = I5o - η*Ib

    where I5o is the no-base-current value of I5 and η is a constant that depends solely on some resistor values:

    I5o = 2.777 mA

    η = (R3/R5)(2R1+R2)/(R1+R2+R3) = 4.026

    Now, Ib is related to I5 (which is just Q1's Ie) by

    Ic = βIb = Ie - Ib => Ib = Ie/(β+1)


    Ib = I5/101

    giving us

    I5 = I5o - η*(I5/101)

    I5(1 + η/101) = I5o

    I5 = I5o / (1 + η/101)

    I5 = 2.777 mA / ( 1 + 4.026/101) = 2.7777mA / 1.03986 = 2.671 mA

    You can see that the effect is about a 4% reduction in I5, which is likely overshadowed by the tolerance of the resistors involved.

    The base current of Q1 is therefore:

    Ib1 = I5/101 = 26.44μA

    which is about 1 μA less than our initial estimate.

    The collector current of Q1 is thuse

    Ic1 = 2.644 mA

    which is also the emitter current of Q2, meaning that it's base current is

    Ib2 = 26.18 μA

    Notice that the assumption that we used that Ib = Ib1 = Ib2 was off by only 260 nA, which is completely lost in the noise.

    The collector current of Q2 is therefore

    Ic2 = 2.618 mA

    I'll leave it to you to follow this same approach to find the quiescent current in Q3.

    So let's recap what we have so far and see how we can iterate in on the solution.

    I5 = [2Vcc*R3/(R1+R2+R3)-Vbe]/R5 - Ib(R3/R5)(2R1+R2)/(R1+R2+R3)

    Next we have I4, the current downward in R4. This is the current in R5 less the base currents in Q1 and Q2 plus the base current in Q3, which we'll call Ib3.

    I4 = I5 - 2Ib + Ib3
  18. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    This makes a reasonable challenge worth attempting. With some effort, I managed to reduce the solution to two simultaneous equations with the two unknowns being the Q1 & Q2 base voltages. Once those voltages are known it is a reasonably simple task by way of algebraic manipulations to find all other unknowns of the circuit quiescent conditions.

    It seems improbable that a teacher would set such a problem for exam conditions, even making some acceptable assumptions. The likelihood of making errors in the substantial but necessary algebraic manipulations is potentially rather high. Notwithstanding this, the practice problem at least ensures the diligent student candidate is well prepared in bias analysis methods.
  19. WBahn


    Mar 31, 2012
    I agree. A question like this really isn't fair game on an exam -- plus it would be a royal pain to grade (unless you don't assign partial credit and track the work through to determine which parts are done wrong and which are done correctly but just with wrong partial answers from earlier parts that have already been deducted for).

    But, as you note, it does provide a preparation problem that gets the students to work through stuff at a deeper level. Then on the exam you can ask just for one part of it or the setup for part of it or give them the solutions to part of it and have them do the next step. All of those, and more, are reasonable and fair game.