Common Emitter Amp Design

Discussion in 'Homework Help' started by hydra504, Nov 15, 2007.

  1. hydra504

    Thread Starter Member

    Oct 18, 2007
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    I am having the most difficult time trying to slove this design question for one of my labs, I need HELP!!! And please anyone correct me if I am wrong. The actuall design question is on one of the pdf files that I uploaded. I have been working on one of the sheets with the Title as Wink. Any one can check my answers to see if it's right? Thank you any help is truely appreciated.:confused::(:eek:
     
  2. chesart1

    Senior Member

    Jan 23, 2006
    269
    1
    I've examined your calculations and I think they are correct. What exactly is the problem you are having?
     
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I was just looking at the handwritten stuff you have on the first page, under section 8.1

    Over on the right side, you calculated ie as -.0667 mA; this is correct.

    You calculated ic as .931 ma; this is correct.

    You calculated ib as -46.9 mA; this is incorrect.

    If you calculate(-47 mV - (-0.4 mV))/10k you should get -4.66 uA, not -46.9 mA. Maybe your calculator hiccupped? But, in addition the sign is wrong. Currents into the transistor are taken as positive. If you calculate beta as .931 mA/4.66uA you will get a beta of 199.8 (beta should always be positive). Then your caclulation for alpha becomes 199.8/(199.8+1) = .995, a much more reasonable number. Alpha of a junction transistor should always be less than 1.

    Under section 8.2, it says to calculate some things for the circuit of figure 6.3, but that page shows only figure 6.1. What does the circuit of figure 6.3 look like? Are you calculating things for the circuit of figure 6.1 when you should be using figure 6.3?
     
  4. hydra504

    Thread Starter Member

    Oct 18, 2007
    16
    0
    Thank Electrician for the help, yeah I realized that I made some errors on the first page of the lab. Actually that first circuit is what we built in lab and took some measurements. Figure 6.3 has been actually cut off so thats a typo from the teacher himself.

    the Problem where I am having is the second circuit on the second page. The Common Emitter Amp problem. I am trying to figure out the two resistors R1 and R2 on the amp design. and I am having the hardest time ever.

    I did some homework last night and ALOT of reading....I need to find the Q point of the transistor with the given values. I know the Q point is determined by Ic and Vce. But since there is only one volatge source 9V connected to two resistors a 1K and R1, I can't seem to figure out where do I start.

    I think my calculations for the Wink Companies Page, on the Emiter Amp are wrong because I do not include in my calculations R1. So there fore I need to find Q and R1 and R2, which is where i'm stuck...
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    First, I notice on the Wink page, bottom of the right hand column, in a little box, you have Vb = 2.278 V. You've transposed a couple of digits when you copied the result from your calculator; it should be 2.287 V. Just make sure you don't transpose digits that are more significant! :-(

    Ok, to start, calculate ic when the transistor is saturated. You got the right answer, but it can be gotten more simply. If the transistor is saturated you have 1k in series with 220 ohms for a total of 1220 ohms in series with (9 volts - .2 volts). Thus ic is 1220/8.8 = 7.213 mA. This ignores the fact that the emitter current is slightly larger than the collector current by an amount equal to the base current; the base current ib = ic/beta. And this ignores the fact that beta probably isn't 150 when the transistor is saturated.

    But this is a good first approximation, so forging ahead we want to find the Q point so we will have symmetrical collector current excursions. To a first approximation this is obviously the case when the collector current is 1/2 of what it is when the transistor is saturated. So let ic = 7.213/2 = 3.61 mA; then ib = ic/150 = 24.1 uA.

    Since (to a first approximation) ie = ic, we have Ve = 3.61 mA * 220 ohms = .794 volts. Then Vb = .794 + .7 = 1.49 volts.

    Now, the bias network. There are multiple solutions for R1 and R2 that will work. For example, you could delete R2 and let R1 be (9 volts - 1.49 volts)/24.1 uA for a value of 311.6k ohms. But there is an important secondary consideration. Since production lots of transistors don't all have a beta of exactly 150, we have to accommodate the production spread of beta. Looking up the specs of a 2N2222 transistor, we find that only a minimum beta is given at a collector current of 10 mA; the value given is 75. However, there is a min-max range given at another value of collector current, and it is a 1 to 3 range. So, let's assume that the beta may range from 75 to 225. I notice that the beta given in your problem is exactly halfway between this min and max.

    The goal is to select the values of R1 and R2 so that the base current extracted from the junction of the two resistors will not upset the voltage division ratio of the R1-R2 divider more than a certain amount as the transistor beta varies from min to max. To make this the case we want the current in R1 and R2 to be substantially larger than the min to max base current variation due to beta variation. I don't know if your instructor has discussed this, or given you a value for acceptable variation, so I'll pick a divider current about 10 times the median base current. That base current will be the current we've already calculated since the given beta is the median beta.

    I just noticed that you didn't use the same designations for the divider resistors on your Wink page as those used in figure 6.1 of the problem sheet. I'll use the ones on your Wink sheet, as I did in an earlier paragraph.

    Let the current in R1 be 11 times the base current (since it has to carry the base current in addition to the divider current). Since Vb = 1.49 volts and ib = 24.1 uA, the voltage across R1 will be 9 - 1.49 = 7.51 volts and the value of R1 will be 7.51 volts / (11*24.1 uA) = 28.3k ohms. The current in R2 will be 10 times the base current. Then R2 = Vb/(10*ib) = 1.49 volts/241 uA = 6.18k ohms. I think your problem wants you to select some standard values for R1 and R2 and calculate how much the Q point is different from what it is with the theoretically perfect R1 and R2. I'll leave that up to you.

    If you wanted less variation in the Q point due to variation in beta, then you will have to make the current in the divider a larger multiple of the base current.

    Remembering that the current when the transistor is saturated is an approximation because it doesn't take into account the fact that ie is slightly larger than ic (by the base current), a more accurate solution can be had by solving:

    1000*ic + 220*(ic + ic/150) = 8.8

    Solving this gives ic = 7.204 mA and ie = 7.25 mA. All the rest can be derived from these values as before.

    I don't know how to answer part d, e and f because figure 6.1 doesn't show RL and I don't what current is being referred to in part d and f. Anyway, you can probably do the rest yourself.
     
  6. hydra504

    Thread Starter Member

    Oct 18, 2007
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    The Electrican....

    :: bows ::

    Thank you so much. I realize now what I did and have to do. You def made this problem so much easier and explain everything so much better than my instructor. I dunno what to say....thank you soo much!!!!!!!:D:D:D And I will make sure I won't make any more calulator mistakes anymore....:cool:
     
  7. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You're welcome. Good luck!
     
  8. kierra

    New Member

    Jun 30, 2009
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    disgusting subject
     
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