common emitter amp problem

Discussion in 'Homework Help' started by ninjaman, Sep 14, 2016.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
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    Hello,

    this is not for a course or anything, i am trying to learn about amplifiers and have hit a wall with this. i am using this:
    http://www.learnabout-electronics.org/Amplifiers/amplifiers20.php
    a site that shows you how to design a common emitter amplifier. i want an amplifier with a gain of 20, however this will come later. for now, i want to get this straight and understand it. i have followed along and got this circuit which i built on multisim.
    upload_2016-9-14_23-4-21.png
    it said to use 1uf-10uf for C1 and C2, so i chose 1uf. then it said to choose a value of C3 that would have a low reactance compared to Re.
    so i thought that 1 ohm would be low compared to 1k ohm. so i used the formula for capacitive reactance and got 7957 uf at 20Hz.
    upload_2016-9-14_23-8-12.png
    the output is really distorted as well.
    i chose these values
    vcc = 9 v
    quiescent current = 1mA
    load resistor = half vcc divided by 1mA = 4500 ohms, real value 4700 ohms
    value of Re= 12% of vcc = 1.08 volts, this divided by 1mA to get 1080 but i chose 1kohm because it gave more accurate voltages when measured compared to calculated results
    base current = 6.6uA this from Q current over 150 which is the current gain of 2n3904 at 1mA
    base boltage = 1.78volts
    r1 and r2 bias current = base current * 10, 6.6 * 10 = 66 uA
    r1 value= vcc - vb / 66uA, 9 - 1.78 / 66uA = 109394 ohms i chose 110k ohms
    r2 value vb / 66uA , 1.78 / 66uA = 26970 ohms i chose 27k ohms

    any help would be great
    thanks
    simon
     
  2. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    It works OK in LTspice. Below is with a 1mV input. What input voltage gave you lots of distortion?

    Edit: should add this is with a 1kHz input.
    upload_2016-9-15_0-0-21.png
     
  3. crutschow

    Expert

    Mar 14, 2008
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    You will get distortion at higher input signal voltages due to the base-emitter non-linear (logarithmic) relation between current and voltage.
    For example, the output is noticeably distorted with a 10mV input.
    To reduce that distortion you can add a low value resistor in series with the emitter to provide AC negative feedback (with a corresponding reduction in gain).
    For example 100Ω in series with C1 greatly reduces this distortion.
     
  4. crutschow

    Expert

    Mar 14, 2008
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    Below is the amp simulation for a sinewave with a 100Ω resistor switched in series with the bypass capacitor for the second parameter step (yellow traces).
    As you can see the distortion is quite noticeable with no resistor (green traces),
    The input voltage was increased for the second parameter step to compensate for the reduction in gain and give a similar output voltage for both steps so you can see that the reduction in distortion is not due to a reduction in output amplitude.

    upload_2016-9-15_0-32-19.png
     
    Last edited: Sep 15, 2016
  5. ninjaman

    Thread Starter Member

    May 18, 2013
    306
    1
    Thanks for that, unfortunately the amplifier that I have to design will only have the components that I have mentioned above, no extra resistors or components involved.
    I am not sure about the capacitor for C3. I calculated 7957uF at 20Hz, is this a typical value. I know that probably sounds a bit odd, it is not a real life value for capacitor. The design record that I am following wanted the value in uF so I left it at 7957, which is 795mF. I think this value is very wrong. I used a value of 1 ohm in place of Xc. should I have chosen a frequency of middle of bandwidth. My thinking is 20kHz - 20Hz / 2, to get the middle of the bandwidth and use this...? Though the design record specifies keeping the reactance low at 20Hz and not middle of the bandwidth.
    Also, does it matter if the capacitor is electrolytic or ceramic. I am thinking that the main reason is due to power going through the capacitors. I have read that capacitors have an issue with non-linearity, varying capacitance with voltage.
     
  6. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    For the capacitor to have an impedance of 100Ω (one tenth of the emitter resistor value) at 20Hz, I make the value 79uF.
     
  7. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    If you are going for the kind of hi-fi where only those with 'golden ears' can hear the difference then electrolytic capacitors are fine. The common kinds of ceramic capacitors also change capacitance with voltage. For the case in question, 79uF, you would be best to use 100uF electrolytic. I don't think you would get that value in ceramic though the available values have been increasing.
     
  8. ninjaman

    Thread Starter Member

    May 18, 2013
    306
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    AlbertHall, is that "one tenth" a rule of thumb. I think I must of missed it, is that what would be used, one tenth of the emitter resistor for the capacitive reactance?

    Thanks for you help

    Simon
     
  9. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    Yes, and you would do the same for the input and output coupling capacitors.
     
    ninjaman likes this.
  10. crutschow

    Expert

    Mar 14, 2008
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    It's not the external emitter resistance that you need to bypass, it's the internal Re of the transistor which is much smaller (≈ 25 (1mA / Ic ohms).
    Thus you need about 330μF to get a -3dB point of 20Hz (giving an Re of about 23Ω for the simulation model} as shown below.
    Notice I also had to increase the value of C2 to get to that low rolloff.
    1μF for C2 gives a LF rolloff of about 60Hz due to the transistor input impedance.

    upload_2016-9-15_9-30-20.png
     
    Last edited: Sep 16, 2016
  11. ninjaman

    Thread Starter Member

    May 18, 2013
    306
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    So I think I have this now. The input and output capacitors start at 1uF and are to be adjusted after initial testing. I simulated the circuit with these results:
    upload_2016-9-17_16-52-33.png
    the input is 2mV and the output is 320mV (approx on both), so to find the gain I put the output over the input and got 160 (approx). Using the Rl over Re method would get 4700 / 1000 = 4.7 gain. So I am guessing that the gain I calculated from these simulated result is because of that capacitor across Re, it is meant to block DC and gain is increased also?
    Is this correct?
    I am going to do some research on transconductance and try to learn about the internal Re mentioned in the Crutschow post above.
     
  12. crutschow

    Expert

    Mar 14, 2008
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    Yes.
    The capacitor bypasses the emitter resistor for AC so the gain is approximately the collector resistor divided by re.

    Here's a discussion on gm which is used to calculate re, and here's a discussion on re.
     
  13. ninjaman

    Thread Starter Member

    May 18, 2013
    306
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    i am getting very confused with this. my coursework is to design a common emitter amp with -20 voltage gain.
    i have created this amp:
    upload_2016-9-18_19-50-40.png
    I tried to create an AC analysis and managed to get the low end to around 24Hz at -3db. but the high end is not working. i have read the course notes which seem to disagree with things i have found on the internet and what is mentioned on here. this is the coursework example for designing a common emitter amp
    upload_2016-9-18_19-53-3.png
    upload_2016-9-18_19-53-36.png
    upload_2016-9-18_19-54-5.png
    on another site the current through R2 is found by 10 * base current instead of the 25 * base current shown above. also, the capacitor values do not seem all that important here.
    i don't think that i need a capacitor across Re, it is not shown on the coursework and would increase the voltage gain above 20.
    and my value for Re is less than what is mentioned above. it says that the current through the emitter resistor is roughly the same as collector current. and that the emitter resistor voltage is 10% of VCC. i worked this out to be 0.9 volts giving a emitter resistor value of 900 or 910 (real value), this changes the voltage gain though. so that 10% rule doesn't apply.
    i have created some differences with my values and thinking to get the voltage gain of 20, would these differences cause and issue?
    i understand that the rules of thumb mean that the values could vary considerably but how much is too much. as long as the bandwidth is right and voltage gain correct with no distortion, that should be all that matters...yes?
     
  14. AlbertHall

    Well-Known Member

    Jun 4, 2014
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    With 220Ω Re and 4.7k Rc you have your x20 gain but you need to adjust the base bias to get back to half supply at the collector.
    Or use 1K Re (to get the DC correct) and then across that a capacitor with a series resistor of 300Ω to get the correct AC gain.
     
  15. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    What other site were you looking at?

    I can tell you that this site, http://www.electronics-tutorials.ws/amplifier/amp_2.html uses 11 x Ib when computing R1 and 10 x Ib when computing R2.

    Then there's this site, http://www.rason.org/Projects/bipolamp/bipolamp.htm which recommends 5 to 10 times Ib for what we are calling R2.

    You stated you need a x20 amplification, which I assume is the minimum value. You haven't stated your desired output voltage or input voltage. If you were using line level audio input that's one standard, microphone level audio input would be much lower.

    Looking at Vc in your circuit, your placed the Vcc/2 at 4.5V. That certainly is true, but should you not consider the Ve DC level? That brings up the topic of distortion. Do you have a distortion figure in mind?

    Sometimes the nearest value is the incorrect choice.
     
    Last edited: Sep 18, 2016
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    It is funny that you are trying to use all those "rule of thumb" without understanding from where those rules come from in the first place and do not understand their limitations. Do you even understand how BJT amplifier the voltage ?
     
  17. ninjaman

    Thread Starter Member

    May 18, 2013
    306
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    Hello JoeJester,
    Thanks for your reply. I have made a number of assumptions based on the course work question which is:
    upload_2016-9-20_11-59-0.png
    I have assumed the input voltage to be 0.4V peak at 5kHz. This is the value used in the example. There are a few different things to look at, one is transient analysis. I assumed that the things I am required to cover are covered in the coursework and that is all. I have not spoke to the tutor yet due to how I am choosing to study the course. From what I have gathered I am required to answer just the question without any extras. I assumed that the distortion subject would come up in the analysis of the circuit and all that would be required is: look at the sine wave, is it distorted? if so change values in amplifier.
    There is no mention of lines or mics and the methods for finding values, these "rules-of-thumb" are written as such in the course work and in the book that I am using. The book is called "Success In Electronics" by Tom Duncan.
    https://www.amazon.co.uk/Success-El...369539&sr=8-2&keywords=success+in+electronics
    I have looked in another book that I have and it doesn't mention multiplying the base current. It mentions finding the emitter voltage by calculating R2 voltage, subtract from this the base-emitter voltage and you have the emitter voltage. The resistor values are chosen already in this circuit. My point is, there seems to be many ways to find the values of components, voltages and currents and I am not sure which method to follow, which is frustrating.
    I imagine that the gain remains the same, -20. There isn't any mention of a distortion figure. There isn't a lot of mention of Ve DC level and distortion in the coursework, or either of the books that I am reading. I have given up with the art of electronics books.
     
  18. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Your input voltage is too high 0.4Vpeak * 20[V/V] = 8Vpeak ---> 16V peak to peak. So your supply voltage (Vcc) must be higher than 16V.
    And LvW put it very nicely.
     
    Last edited: Sep 20, 2016
  19. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    All I am saying is you needed to know something about what you were trying to design.

    A gain of -20 was a good beginning. Your choice of 9 volts and a collector resistor limited your peak to peak swing of 8 volts. That limited your input to 0.4 volts pk to pk.

    You knew your input was 0.4 volts peak or 0.8 pk to pk.

    It's good to write your basic design specifications in the problem statement. If you stated a class A amplifier, then little distortion is assumed.

    Well now you can revisit your design and repost.
     
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