painful CE amplifier

Discussion in 'Homework Help' started by stupid, Mar 18, 2010.

  1. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    [​IMG]hi, i desperately need help.

    Q. assume that hre=hoe≈0, hie= 1.1kΩ, hfe=50, C1 & C2 tends to infinity,
    Rb=100kΩ, Rs=5kΩ & Rc=RL=10kΩ. Using CE h-parameters, find & evaluate expression for,
    a. Ai= IL/Is
    b. Ai= IL/Ib
    c. Av= VL/Vs

    the cct is reduce for DC analysis in 2nd diagram.

    thanks
    stupid
     
  2. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Since your transistor parameters are given (hie, hfe, hre, hoe), you don't need to consider the DC analysis, which would allow you to determine re, the intrinsic emitter resistance; the effect of re is included in the h-parameters.

    This problem is substantially more complicated than usual for this forum.

    Do you already have the answers? If so, would you post them so I can see if I'm on the right track myself!

    Do you have a plan of attack for the problem, so far? You're supposed to show some work of your own before asking for help.

    Has your textbook addressed the use of circuit parameters other than h-parameters? In particular, has it is discussed how to convert from one set of parameters to another?

    You might have a look at this thread:

    http://forum.allaboutcircuits.com/showthread.php?t=26710&page=5

    starting with post #45.
     
    Last edited: Mar 18, 2010
  3. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    i m at a lost as to address this problem.
    that is an assignment & has no answer yet to that.

    i will upload a part of my relevant course material for reference.

    thanks, the electrician.

    stupid.

     
  4. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Don't forget to do this so I can see just what you might be expected to know.
     
  5. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    a CE equivalent cct is attached.
    i dun know where Re & Rc fit. Pls advise.

    the formula for iL/ib=-hfe/(RL hoe+1)
    since Rc=RL, IL/2

    iL/ib= -hfe/2(RL hoe+1)
    = -50/2=-25
    is that correct?

    for voltage gain, Av=Vce/Vs
    since Vs=Vbe

    voltage gain=Vce/Vs
    =Vce/Vbe

    the vlaue of Vs is not given.
    how should i proceed?

    thanks
    stupid

     
  6. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    OK. I can see what model they're using. By the way, on the third page of the images you posted there is an error. Equation 3.18a is given as:

    Vs = hfe*ib+hre*Vce

    This should be:

    Vs = hie*ib+hre*Vce

    This error propagated to Eq. 3.18b and 3.18d

    Vs doesn't equal Vbe because there is a resistor, Rs, in the path.

    These simple formulas aren't going to work on this problem. It's significantly more complicated than problems you've posted here recently. This circuit involves feedback through two paths--Rb and R4, which complicates things.

    The value of Vs doesn't matter, because it appears in the expression you're asked for in the form of a ratio. You're asked for VL/Vs. Since this is a linear circuit, whatever Vs is, VL will be larger by a constant ratio, and you don't need to know Vs, only its ratio with VL. So you can just take Vs to be 1 if you want a number.

    I have a couple of questions for you.

    Have you studied any of the methods for solving complicated networks, in particular have you studied the nodal method of circuit analysis?

    What kind of mathematical tools do you have for solving simultaneous linear equations? Do you have a calculator that can do that, or do you have access to a PC program such as MATLAB?

    I hope the answer to my first question is that you know how to do nodal analysis, because you're going to have do that. I notice in the third page you posted, they mention using KVL, so apparently you've studied KVL and KCL. Nodal analysis is essentially application of KCL.

    I've attached an image showing what you're going to have to do. They've made things a little easier for you by setting hre and hoe to zero. You can see in the second item in the image how that simplifies the transistor model.

    The third item in the image is the network you're going to have to solve. I've numbered 4 nodes on that circuit, and if you write 4 equations using KCL at each of the 4 nodes you will be able to determine the 3 quantities you've been asked for.

    The capacitors have been replaced by shorts. Rc and RL are in parallel because Vcc is at AC ground; it's assumed to be heavily bypassed to ground by a big filter capacitor.

    It will make things less confusing for me to help you if you will follow the node numbering I've shown in the image. The voltages at each node are assumed to be designated by the letter V in front of the node number. In other words, the voltage at node 1 is V1, at node 2, V2, etc. Please use those in equations you write.

    You are in for some serious algebra here! To get you going, I'm going to give you the equation for node 3.

    You write expressions for the current through each resistor and current source connected to a node, and add them all up and set the sum equal to zero. Currents leaving a node are taken to be positive.

    The current through the resistor labeled hie is given by (V3-V2)/hie. The expression is just the voltage at the node we're working on minus the voltage at the other end of the resistor, divided by the resistance.

    The current through R4 is given by (V3-0)/R4. Since R4 is grounded at one end (the voltage at the grounded end is zero), the numerator of that expression subtracts zero from V3.

    The dependent current source (due to the transistor action) injects a current of -hfe*ib (minus because it is entering the node). But, ib is (V3-V2)/hie, so the current entering node 3 from the current source is given by -hfe*(V3-V2)/hie.

    We add these 3 currents and equate the sum to zero, giving us the 3rd equation of the 4 we need:

    (V3-V2)/hie + (V3-0)/R4 - (V3-V2)*hfe/hie = 0

    At some point you may need to rearrange this equation (and the others) so that the coefficients of each unknown (the unknowns are V1, V2, V3 and V4) are gathered together, with zeros for place holders as needed, like this:

    (0)*V1 + (hfe/hie - 1/hie)*V2 + (1/hie + 1/R4 - hfe/hie)*V3 + (0)*V4 = 0

    I hope I haven't made any sign errors (or other dumb errors); it's easy to do, especially on an exam when you're under time constraints! For this homework problem you can just take your time and double check.

    Now you will have to write the other 3 equations, and solve them as a system of simultaneous linear equations. You should probably substitute numeric values for the various symbolic variables before solving the system. Otherwise, you are going to get painfully complicated expressions.

    When you write the equations for nodes 1 and 2, don't forget that at some point you will substitute Vs for V1. You can use 1 for the value of Vs if you want. It will vanish when you calculate Av.
     
  7. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    thanks for pointing out the typo errors.

    i have learnt network analysis including nodal, KCL,KVL etc..
    i use to hand calculate simultaneous eq.

    btw, there is a slight twist to the orginal problem now that the emitter R is dropped. So the equivalent cct became as attached.

    at node2,
    V2/1.1K + (v2-Vs)/5k + (V2-V3)/100k=0

    at node 1,
    (Vs-V2)/5k= is

    so, V2/1.1K -is + (V2-V3)/100k=0

    at node 3,
    (V3-V2)/Rf + hfe*ib + V3/Rc + V3/RL=0
    (V3-V2)/100k +50ib + V3/10k + V3/10k=0
    (V3-V2)/100k +50ib + V3/5k =0
    (21V3-V2)/100k = -50ib

    iL=V3/10k

    but how do work out these 3 equations?

    thanks
    stupid
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    After thinking about this problem, I think it might be better to take V1 = Vs as a known, and have only 2 equations. Then rearrange them with coefficients for each variable (V2, V3) collected together. Then they are in a proper format to use a simultaneous equation solver such as this one:

    http://math.cowpi.com/systemsolver/2x2.html

    When writing equation 2, note that Ib is the current through hie, which is V2/hie.

    Your 2 equations should be:

    Eq1: V2/1.1K + (v2-Vs)/5k + (V2-V3)/100k = 0

    Eq2: (V3-V2)/Rf + hfe*V2/hie + V3/Rc + V3/RL = 0 which becomes
    Eq2: (V3-V2)/100k + 50*V2/1.1k + V3/10k + V3/10k = 0

    Now you have to collect coefficients for each variable, in each equation, leaving zero as a place holder where needed (not needed here).

    Eq1: (1/1100+1/5000+1/100000)*V2 + (-1/100000)*V3 = Vs/5000

    Let's set Vs = 1, so equation 1 becomes:


    Eq1: (1/1100+1/5000+1/100000)*V2 + (-1/100000)*V3 = 1/5000

    Eq2: (50/1100-1/100000)*V2 + (1/100000+1/10000+1/10000)*V3 = 0

    Use the solver at the link I gave above. They use X and Y for the unknowns, instead of V2 and V3. Type the four coefficients multiplying X and y in the left four slots, and the two constant terms (1/5000 and 0) on the right and the click "solve". Be sure to use enough decimal places (5 would be good) when you convert an expression such as (1/1100+1/5000+1/100000) to a decimal number such as .0011191

    Now you have the two voltages at V2 and v3. From these you can calculate your three problem requirements. For example, since we took Vs to be 1, the voltage gain, Av, is just V3/Vs = V3/1 = V3; the voltage at V3 is Av.

    You can calculate the current IL by dividing V3 by 10000. You can calculate Is and Ib similarly and get the current gains as ratios of IL to Is and to Ib.

    See if you can get it to work and report back your results.
     
    Last edited: Mar 22, 2010
  9. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    hi,
    the answers:
    V2=0.0609
    V3=-13.183

    assume Vs=1v
    is= (1-0.0609)/5k=0.18782mA
    iL= V3/10k=-13.183/10k=-1.384mA
    ib=0.0609/1.1k=0.0554mA

    gains:
    iL/ib=23.796
    iL/is=7.369
    VL/Vs=V3/1=-13.183

    if the voltage gain has a negative sign, what does that mean?

    thanks,
    stupid

     
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Your answers for v2 and V3 are exactly right but you've made a small numeric error. This computation:

    iL= V3/10k=-13.183/10k=-1.384mA

    should be:

    iL= V3/10k=-13.183/10k=-1.3183mA. This has led to some small errors in your current gains. You might want to recalculate before you hand in your answers.

    I get:

    iL/ib=23.804
    iL/is=7.0189

    The negative voltage gain is because a common emitter amplifier has a 180° phase shift from base to collector.

    If you apply a small signal increasing voltage at the base, the voltage at the collector will decrease; a decreasing voltage at the base leads to an increasing voltage at the collector. A small sine wave applied to the base will be 180° out of phase at the collector.
     
    Last edited: Mar 22, 2010
  11. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    i tried a current divider rule on hfeib on the RL & Rc for iL whose answer is 1.384mA.

    looks like the difference was quite substantial, doesnt it?

    about negative volt gain: is that the effect on small signal voltages or dc voltages?
    is the positive peak of small signal at input is negative peak at output?

    thanks
    stupid

     
  12. The Electrician

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    If you're going to do that you have to take into account that some of the current goes into the 100k resistor. If you calculate the current in the 100k by knowing the voltage across it, and subtract that from hfe*ib, then divide by 2, you'll get 1.31827 mA. But, it's easier to just recognize that the voltage across RL is -13.183, so the IL is just -13.183/10k.

    It's because when you change the base voltage in such a way as to increase the base current, you also increase the collector current. If the transistor is an NPN, this means increasing positive voltage at the base. It doesn't matter if this increase in voltage is DC due to changes in the bias circuitry, or if it's due to the input signal.

    In a common emitter topology, when the collector current increases, the collector voltage decreases because of increased voltage drop across the collector resistance.

    So, if the input signal is a sine wave, the positive peaks at the input correspond to negative peaks at the collector. That's what is meant when we say there is a 180° phase shift. It's just an "inversion" of the input signal. It's what is meant by a voltage gain with a negative sign.
     
  13. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    This problem as you originally posted it had an emitter resistor. Did your instructor delete that because he thought it was too hard?

    You could probably solve that one now that you've done the simpler one. Maybe get extra credit?
     
  14. stupid

    Thread Starter Active Member

    Oct 18, 2009
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    i thought so. the solution would be complicated. dont u think so?

    btw, had the emitter resistor included the gain would suffer, right?

    thanks
    stupid

     
  15. The Electrician

    AAC Fanatic!

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    Yes, the voltage gain is about 1/4 what is is without the emitter resistor.
     
  16. peck

    New Member

    Mar 21, 2010
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    Hi all,

    For the previous KCL node equation. I think there is some problem....

    it should be (V3-V2)/Rf = hfe*V2/hie + V3/Rc + V3/RL right??

    Please correct me if i was wrong.....

    Peck.
     
  17. The Electrician

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    Oct 9, 2007
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    This would be equivalent to:

    (V3-V2)/Rf - hfe*V2/hie - V3/Rc - V3/RL = 0

    As I explained earlier in this thread, the convention I use is that currents leaving a node are taken to be positive.

    When using KCL, all of the currents are equated to zero, so it appears that the last 3 currents, - hfe*V2/hie - V3/Rc - V3/RL, have the wrong sign.
     
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