Small Signal BJT Amplifier

Discussion in 'The Projects Forum' started by kaname08, Mar 1, 2012.

  1. kaname08

    Thread Starter Member

    Sep 22, 2010
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    I have a project at school,
    I already read the previous post about small signal ampli but haven't seen what i'm looking for,hope you understand.

    We need to make an amplifier using small signal input and using BJT for 2 stage,
    We are the one to choose the Av, and I chose 23k because its a two stage. I searched in the internet for the signal output of a typical mp3 player or vcd player. So my input is Virms=300mV and Zi=100Ω,I guessed 100Ω is the Zi because from what I searched it says the impedance of the mp3 is 100Ω.
    I used voltage divider,and here attached the diagram.


    My questions are as follows:
    1) I my Vi1 and Zi1 correct? and also what frequency to be used
    2) I chose 2N2222A (here's the datasheet http://pdf1.alldatasheet.com/datasheet-pdf/view/50091/FAIRCHILD/2N2222A.html ), what Icq,B and Vceq to be used here?or atleast how to compute it or determine the correct parameter to use?.
    3) For the Vcc,what should I consider to set my Vcc,but I was planning to use 12V(is it ok?).Also what to consider in choosing current source?
    4) Is it right that I started in setting my Av?and my Av,is it alright that I chose 23k? and I'm trying to start from Ac analysis going to DC,is that right?
    5) For Ce,Co and Ci , what is the proper computation for it? I already know the formula f= 1/(2∏RC), but I'm not really sure how to get it.


    hope someone would help.
     
  2. kaname08

    Thread Starter Member

    Sep 22, 2010
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  3. jimkeith

    Active Member

    Oct 26, 2011
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    Your Av seems out of order--23k or 23,000 is not possible with a 2stage bipolar amp--typ max Av per stage is 100. Then what is all this gain for? if attempting to get a 100mV output signal, this puts the input at 4.3uV which is really in the mud (noise region)--a low Z dynamic microphone puts out roughly 1mV.

    The schem looks workable otherwise.
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    Since the input signal is 300mV RMS which is 0.85V peak-to-peak and the battery is 12V. The maximum output swing will be about 8V peak-to-peak then the voltage gain must be no more than 8V/0.85V= 9.4, not 23,000.
    For a gain of only 9.4 then the emitter resistors do not need bypass capacitors.

    Every amplifier has a specified load resistance. What is it for your amplifier?
     
  5. kaname08

    Thread Starter Member

    Sep 22, 2010
    37
    0
    my load resistance is 16 ohms.

    how did you get 8V?
    so in computing here I need to use peak-to-peak and not just the peak?
    Is the 9.4 AvLT or AvNLT for the 2 stage?
     
  6. kaname08

    Thread Starter Member

    Sep 22, 2010
    37
    0
    The gain here is just to have a loud sound,well i'm not so sure about it,its for a 16Ω load,,
     
  7. Veracohr

    Well-Known Member

    Jan 3, 2011
    552
    76
    16Ω load sounds like a speaker - are you designing a power amp or a small signal amp?
     
  8. jimkeith

    Active Member

    Oct 26, 2011
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    99
    Do what Audioguru recommends or your audio output will be SEVERELY distorted.
     
  9. kaname08

    Thread Starter Member

    Sep 22, 2010
    37
    0
    I'm doing small signal amplifier.Why? Should I use 8Ω or 4Ω,,
     
  10. jimkeith

    Active Member

    Oct 26, 2011
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    99
    A small signal amplifier normally has a load impedance of 300Ω or greater while a power amplifier normally has a load impedance of 16Ω or lower. Class A amplifiers usually do poorly at driving speakers.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    I have made many audio power amplifiers but I have NEVER made a class-A heater. It wastes a lot of power making heat even when it is not playing.

    Class-A heating amplifiers are made (very expensive) for people who are "different".
    Everybody else uses class-AB which is much more efficient and has excellent spec's.
     
  12. Audioguru

    New Member

    Dec 20, 2007
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    You are making a class-A POWER AMPLIFIER, not a small signal amplifier.
    I have made many audio amplifiers but I have never made an obsolete old class-A amplifier.

    Most audio amplifiers are class-AB today and have been for the last 40 or 50 years.
     
  13. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    I recall working on class AB push-pull amplifiers that were manufactured in the 30's. Virtually the same as 'modern' tube type guitar amps.
     
  14. kaname08

    Thread Starter Member

    Sep 22, 2010
    37
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    So I should use 300 load impedance, so what other things should I revise so that I may be able to make the small signal amplifier?
     
  15. Audioguru

    New Member

    Dec 20, 2007
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    We do not know why you are making a small signal amplifier. An MP3 player can drive headphones very well with its little power amplifier.

    We do not know how you calculated that you need a voltage gain of 23,000.
    Don't you know that an input of 300mV times 23,000 is an output of 6,900V?
    6,900V into an 8 ohm speaker is a power of 6 million Watts which is ridiculous.
     
  16. kaname08

    Thread Starter Member

    Sep 22, 2010
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    I am going to make a small signal amplifier for my project,and I'm just the one who set 23k gain,, but I can also use other Av.
    Now I get how you computed the output voltage.
     
  17. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Hi
    kaname08 ,

    I'm only a hobbyist in playing around with transistor circuits, so I can only help you on that level.

    You are getting a lot of response to this thread, that you are not going to be able to design a small signal amplifier with the values you constrained yourself to, such as 16 ohms Zout.

    However your not getting the full story of why this is so.

    So I've put together a little lesson in this principle, to help you grasp better of what is being told to you in this thread.

    I designed a circuit trying to keep to the reality constraints you chose for your design.

    These are
    VCC=12v.
    Rload=16 ohms
    Zin= > 100 ohms.

    Here is how I went about this design.

    Because the load is 16 ohms, I choase to match it with a RC value of 16 ohms. Reality I needed to use two 5W power resistors for RC and the load of 10 ohms and 7.2 ohms in series.

    The voltage across RC should be close to around 1/2 of VCC, around 6v.

    Calculating current flow through RC gives around 375mA, which is a lot of current for a small sig. transistor.

    However with your choice of transistor, the data sheet shows max. DC continuous current to be 600mA. So because you chose a transistor with a higher current rating, our calculated value is feasable for this transistor.

    Now at this stage to calculate the volt.gain, you need to take both the Rload and RC in parrallel with eachother, which is around 8 ohms, so to get any significant gain for this stage, RE needs to be much lower.

    However because RC//Rload, is already very low, then a very low RE of 4ohms would at least give a gain of 2, so make RE=4.2 ohms, but because of high current the need to parrallel 3 12 ohm 0.250 watt resistors must be done to attain around 4 ohms so as to not burn up RE.

    Now from there calculate the v. drop across RE, then add to that 0.7v. to get the Base resistor divider voltage, (VB). Then to get to a final Zin at the first stage of >100 ohms, make the base resistor to ground around 15 to 20 times RE. I chose 62 ohms.

    Then by calculating the current flow through this resistor, I was able to get the value needed for the top resistor, which came out to 270 ohms.

    Keeping in mind I need to get a >100ohms Zin, I need to choose a higher value RC for the first stage (input side), so I matched the bottom resistor of the second stage and chose 62 ohms.
    Now the gain of this stage is approximately on the order of Zin of second stage= {[(Bmin. + 1) x RE] // 62 ohms // 270 ohms,} ( this is ignoring any inherent resistance called (re))
    Then all this in parrallel with RC of this first stage of 62 ohms, using a Bmin of 35, gives a Zout of this stage of around 23 ohms.

    So again to get some kind of gain with this stage choose 10 ohms for RE, which gives a volt gain, for this stage (AV) of around 2, just as the first stage,

    So now to get the >100 ohms Zin of this amp, make the base resistor 15 to 20 times RE, which I chose 180 ohms. Then calculate the top resistor which is around 1.1K ohms, and build the circuit and check for voltages close to 1/2 VCC +/- 1v.at the colectors.

    Also do a calculating check on your Zin on this stage.
    The Zin of this stage is {[(Bmin +1) x 10 ohms // 180 ohms // 1.1K ohms.]} ~= 108 ohms using Bmin of 35 as per data sheet.
    OK total gain of amp calculates out to be first stage AV =2 and second stage AV = 2 so Atotal = (2 x 2) NOT 2+2 but always multiply stage gains, NOT add. AVtotal=4

    To check it apply a signal voltage into the amp and using a oscilloscope adjust the input voltage until the output waveform shows NO distortion, then you can calculate the Av by dividing the input waveform voltage into the output waveform voltage.

    Here is the final schematic of this circuit.

    I had to use 2N3904,s which has a lower current rating then your transistor.

    schem.jpg

    Here is what the amp waveforms look like when there is NO load on the output.

    s1 nl.JPG

    and the waveform with an 17.2 ohm Load.

    s1 fl.JPG

    Now you have a better understanding of why a Class A stage is good for small signal large output impedance matching, versus a low output impedance, such as a speaker. A class A amp would consume a lot of power to run a low impedance device. Thats where Class AB power amps come into the picture.

    However to show that all is not loss even though a Class A amp is not fitting for power amplification, I redid the first design, to use a lower transistor current, to run a 8 ohm speaker load, by changing the RC of the output stage to a 120 ohms which brings IC to around 50mA, and had to readjust the top base resistor to 750 ohms, to get this IC value.

    Here is the NO load waveforms, this has extremely high gain because of 120 ohms RC divided by 3.9 ohms RE, you can see the dial settings on the oscope as well. Both dials are on the same voltage settings.

    s2 nl.JPG

    Now watch how much this gain drops when the 8 ohms speaker load is applied.
    The gain would be around (8 ohms / 3.9 ohms) ~= 2.

    s2 fl.JPG

    However even though the output voltage is in the Millivolt region, all is not loss, here is a video showing the difference in loudness of the 8 ohm speaker when the signal generatoer is connected to it directly, then as the speaker is connected to the generator through the amplifier.

    There is a significant change in loudness, even though it is small signal output.

    http://www.youtube.com/v/ayhmPw47gKY?version=3&hl=en_US&rel=0"></param><param

    Hope this helps you better understand transistor design constraints, and the importance of choosing proper parameter values.
     
    kaname08 likes this.
  18. Audioguru

    New Member

    Dec 20, 2007
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    A 2N2222A output transistor will burn up because it must dissipate 1.5W.
    The output power into 16 ohms of this amplifier is only 0.12W RMS.
     
  19. kaname08

    Thread Starter Member

    Sep 22, 2010
    37
    0
    Hobbyist thanks for the things you discussed, i have understand lots of things, but I have few questions.
    About the Re that you used for solving the gain, you used (RC//RL)/RE, but isn't it Av=(Rc//RL)/re ? or you used re as RE?

    Also what current rating should I use for my 12V Vcc, I mean for supply voltages there is also a supply current.

    And I have an Additional question, I will also need to put volume control. Should I compute the resistance of the potentiometer now or I can compute it after finishing the computation of the amplifier? I dont even know how to compute the potentiometer to use.
     
  20. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    Small (re) should be used when a small collector current is present,
    if I remember correctly re~= to 26/IE where IE refers to milliamps.

    With a minimum of 375mA flowing then re ~= (26/375) which is very small compared to RE.

    However if RE was bypassed in anyway fully or partially then re would need ro be included into the equation as you stated.
    When RE is not capacitively bypassed, then if a large enough current is flowing then re could be ignored for calculating Av. as well as Zin in some cases.

    In answer to your current consumption, a good design practice would be to use the data sheet and design for a collector current in a midrange, that shows the maximum and typical Beta value.
    However don't use the beta values in your design calculations, you want to eliminate as much as possible the beta factor by using RE and the voltage divider configuration at the base term.

    Volume controle: That could be elaborate , or just a pot in series with the input signal.
    I think the easiest way to start out with a volume controle addon, would be too design and build the input stage first, then test it out too see how loud it is with direct coupling from signal source, then test it out by putting resitors in series and checking output volume, then you could use that value to give the range of pot needed. Otherwise you could do voltage divider equations, and calculate everything ahead of time.


    If you want to design a simple cascaded 2 stage amp for experimental purposes, then choose a RC value that will allow y7our chosen current , and then work the calculations to solve for RE based upon AV of that stage.

    Then build it and check to see if your getting close tio calculated voltage values for VC and VB.

    If VB is too small compared to calculations, remove the transistor from the circuit, and check to see if VB at the voltage divider goes back up to normal value you calculated.
    If it does, than your RE is too small compared to your base resitor to ground, (NPN assumption)
    whats happening is there is excessive base current flowing causing the divider to upset.

    Either change RE to make it larger, then recalciulate the base resistors needed to reflect this change, OR lower the base to ground resistor, and calculate the top resistor needed to reflect that change.

    Once your satisfied with voltage measurements, you can then do some fun calculations on this stage.
    Solve for the gain with no load, then decide on what is the smallest load resistor that would have the smallest effect on the stage. This could be done empirically through voltage measurements with osciloscope or multimeter. This is AC gain (hfe) if you have some sort of input signal to inject into the amp.

    When you find a load value with satisfactory results, then calculate what the AV will be now with this stage loaded.
    Once that is done you could now design a second stage at the input to increazse AV overall.

    First calculate Zin of the output stage by {(Beta min. +1) x RE +re)} // the voltage divider system}
    Then try to use a RC value much lower than that for this input stage your now working on.

    Then do all your calculations for ICQ and AV and choose resistor values that would work for you.
    Then calculate what the AV would be for that stage, by first knowing Zin on the output stage, (use Beta min, for worst case calc. then multiply both stage gains and see how close it comes to a real measurement of the amp.

    From there you could then start choosing load values first, then try to design 1 stage to work into it with a chosen gain, and keep practicing this until you can understand in better detail how to choose the proper constraints, for a given design.

    Then begin to study up on other configurations such as emitter followers, these have no voltage gain, but large current gain, this kind of stage is used for matching a low load impedance to a inherently high output impedance of a preceding transistor stage.

    Then you could start combining these different configurations for unique design requirements, such as low value loads driven by high output stages, once you tackle this you can then learn the benefits of Class B amplifiers, for driving more current with more efficiency, then that would lead you to Class AB amps that take care of unherent distortion effect with Class B amps.

    Class AB amps are for power amplification, driving low value load resistances (speakers ect..)

    while class A amps are for small signal amplification. Which requires more voltage gain then current gain.

    Then eventually yoyu could put all this together to make your own audio amplifier for your personal devices, but it all starts out as one stage at a time, decide what is the requirement and what best configuration would meet that requirement, and every stage needs to work in unison with eachother, so as to work to a common cause of amplifying the signal more and more until a desired output is attained.

    Thats a basic analog system.

    Digital systems are even more easier and interesting to build using all transistors too, but thats a whole different ballgame for playing around with transistor circuits.



     
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