Single Transistor amplifier Design HELP PLEASE

Discussion in 'Homework Help' started by stinky111, Apr 7, 2011.

  1. stinky111

    Thread Starter Member

    Apr 7, 2011
    39
    1
    Hi All,

    could anyone please help me design a single transistor amplifier to run from 24VDC Supply,
    for a assignment I have been given.

    Thanks in advance, Any help appreciated..
     
    Last edited: Apr 7, 2011
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    783
    The homework forum doesn't actually work like that. Familiarize yourself with the conditions of participation in the sticky at the top of the homework forum listings.

    By all means submit a design of your own (however flawed) and you will generally find help arrives in no time at all.

    This general topic has been revisited many times on the forum. Run a search on the site for similar threads. You might consider checking out the AAC e-book notes on amplifier design - Volume III Chapter 4 in particular..
     
  3. stinky111

    Thread Starter Member

    Apr 7, 2011
    39
    1
    Thx for that I will do, I will resubmit what I have so far & go from there,
    Sorry my question wording was not the best, was looking for help only not just someone to do my homework for me as you pointed out.
    Thx for the reply

    As attached what I have so far, Any advice??thx
     
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    Last edited: Apr 7, 2011
  4. acelectr

    Member

    Aug 28, 2010
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    0
    Well what is the exact purpos? What are the limitations? What amount of gain is wanted & also what about the limitations for the lower and upper 3dB points? To start from somewhere one would need some reference information so to the desing considering that reference.
    First advice would be to simplify the circuit as much as possible. For instance start with some thevenins and begin to form current equations.
    good luck!
     
  5. hobbyist

    Distinguished Member

    Aug 10, 2008
    774
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    Here are equations you can study to bias your given circuit.

    Voltage at the collector terminal ref. to ground, is called (VC).
    Volt. at the base term. ref. to ground is called (VB).
    Volt at the emitter term. ref. to ground is called (VE).
    Current flow from the supply into the collector is called (IC).
    Current flow through the base into the emitter is called (IB).
    Current flow from the emitter to ground is called (IE).
    Supply voltage is called VCC.

    REdesignating your schematic:

    R2 will be called (RC).
    (RE) will be chosen, ..(it originally was R4.)
    R1 and R3 will be calculated.

    Equations:
    1). VC = (VCC / 2)
    2). IC = (VC / RC)
    3). RE = (2 to 20) times less than RC. (choose a value)
    4). VE = (IC x RE), where you made a choice for RE in above step.
    5). VB = (VE + 0.65v)
    6). R3 = (10 x RE)
    7). ID = (VB / R3)
    8). R1 = [(VCC - VB) / ID]
     
    Last edited: Apr 7, 2011
  6. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The gain of your "amplifier" is 1. So it is not an amplifier, it is a piece of wire.
    Don't you know which ratio of resistors (plus an internal transistor resistance) sets the gain of a transistor amplifier?
    The load impedance is also very important since it reduces the gain.
     
  7. stinky111

    Thread Starter Member

    Apr 7, 2011
    39
    1
    Thanks so much, for all your replys, the equations are a real big help, I now understand how I arrive at the values I need to design my circut now.. I will post results, thx
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    300
    I wonder if this has done you so much of a favour. Someone has given you a tool-kit of formulae (or at any rate rules-of-thumb) which may let you make a working circuit. Do you actually understand why these rules are as stated?

    One could also argue over whether these given rules are really good practice, for instance in the arbitrary choice of RE, but maybe it's best not to confuse matters further.
     
  9. hobbyist

    Distinguished Member

    Aug 10, 2008
    774
    64
    Hi,

    Electronics can be very confusing learning, when you are just new to the field,
    sometimes the best encouragement to stick with it comes by hands'on, training, actually getting a circuit to work. (That's why I'll show pics. or a video of it working).

    That's my experiance with these forums are abouit, since I'm not a professional as most are on these boards,
    I can help the newcomers get a chance to be encouraged, by designing a circuit that works, enough for them to want to continue on in there learning on the subject.

    Since I am not a professional in this, I can relate to the students on here as I am a circuit designer on the most BASIC level of electronics, learning curve.

    That's why my circuits are so simplistic in design, it all comes from working at the beginners level in this field.

    Then as these students progress, they come into you guys' level, for more professional detail in a design.
     
    Last edited: Apr 8, 2011
    stinky111 likes this.
  10. stinky111

    Thread Starter Member

    Apr 7, 2011
    39
    1
    Thx hobbyist,

    Just to confirm to all I am very much a newcomer to electronics, (Its obvious I guess), I have just done a complete 180 career path change.

    I have only started studying again this year to become an electrician...
    After being out of school for over 20 years & the oldest in my class never that great at maths I'm struggling with the theory, but am top of my class at all practical assessments.

    I realize that some of my questions may seem like no brainers for most, I can assure u all that they are genuine & I appreciate all help given as I'm trying my best to absorb all I can, to pass my cert & progress with my new career.
    Then hopefully one day be on the electronics level as most posters on this forum..
     
  11. hobbyist

    Distinguished Member

    Aug 10, 2008
    774
    64
    Hi,
    Thanks for replying back,
    The equations I gave are your basic 'run of the mill', type to get a transistor stage biased so it can act as an amplifier, however, in a real world design, there will be parameters given by the application it is used for, such as input and output impedances, and output signal voltages, input signals, voltage or current gains, ect...

    So then the design will need to be acted upon by the constraints given, in the application, that would mean a change in some of the equations given, by adding different values to have a more particular amplifier,
    but these things you will pick up on as your course goes on,
    but for now, experiment with the equations given, to see how the transistor stage works, when various signal amplitudes is applied to its input.

    Later you will learn how to adjust the AC gain by using a bypass capacitor across RE,
    there is the first order approx. equation for that, than there is also the more detailed calculations, which involve more math values to be included.

    I use the first order approx. for most of my circuits, unless the circuit requires a more detailed calculation, then I go deeper with the math involved.

    But you will see as you first start out, a transistor amp. stage can be successfully designed using first order (ballpark) figures, for the calculating of bias components.
    When the constraints are within reason, of such a design.
     
  12. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    @ stinky : I must apologise to you - I had assumed that you were one of the many young students (typically from certain eastern countries) who come seeking answers to assignments, seemingly without making much effort of their own. In some cases the aim appears to be to get answers for on-line assessments, with no aim other than to get better grades. In my opinion these students often have little real understanding of their subject and do not seem to wish to acquire any. The imperative is to get answers so as to avoid failing their course. This forum attracts a large number of enquiries of that kind, which may lead to some of us developing rather negative attitudes on the subject.

    It is of course more challenging to begin studying a new subject as a mature person, especially if this has little in common with what you were doing before. All the same, I would urge you to make the greatest possible effort to understand at least the basics of theory. This may be difficult, but otherwise much of what you are trying to learn may remain a series of disconnected facts, leaving them much harder to remember, let alone understand and apply.

    As for the best way to proceed, there is no substitute for actually working through examples. Reading through the subject matter is helpful, but there is no substitute for applying it, in theory as well as in practice. Unfortunately, the subjects requiring the most effort may be those that you least enjoy (Maths?), but that's just how it goes. Spending little time on a necessary subject because you do not like it is a recipe for failure.

    I will leave you with this thought: it is quite correct that the emitter resistor value should normally be smaller than that of the collector resistor in the sort of four-resistor bias you are considering. Why is this so? Does it help you to imagine a case where the emitter resistor is much bigger?
     
    hobbyist likes this.
  13. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Some engineers bias a transistor by formulas they memorized. They don't care to understand why.

    Other engineers know why and how a transistor should be biased and use simple arithmatic to calculate the resistors.
     
  14. stinky111

    Thread Starter Member

    Apr 7, 2011
    39
    1
    Guys, These are equations so far please comment,
    Im stuck at my calculation of R1, getting some weird answers using:

    R1=[VCC-VB)/ID]



    Given Q1= PN2222A, hFE= 300, VCC=24VDC


    Choose R3: R3=150-ohms

    R4= (R3x5) R4=750-ohms

    Choose IC: IC=15mA

    Ib= IC/hFE (15mA/300) IB=20uA

    IR1 & IR2={Ib x 10} or 200uA.

    Ve=R3 x Ic, (150 ohms x 15mA) VE=2.25v

    R2=(VR/IR2)=(2.25v/200uA) R2=11.25m ohms

    Vb=(VE+0.65v)=(2.25+0.65v) Vb=2.9v

    Id=(VB/R3), (2.9v/150 ohms) ID=19.33??????

    R1=[(VCC-VB)/ID] (24v-2.9v)/19.33????????
    Thx
     
    Last edited: Apr 10, 2011
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    783
    I think your Ic of 15mA is rather high for a typical single stage class A amplifier. I'd use something more modest - say 3mA. Unless you must have the 15mA of course.

    In any case given your hFE=300 your Ib=15mA/300=50uA - not 20uA. Check your calculations!

    Typical rule-of-thumb for a stable voltage divider control of Ic is to set the divider current [ID] to about 10Ib or 500uA in your case.

    Since you haven't specified a target [unloaded] voltage gain it's hard to proceed the design from this point. The voltage gain Av will be approximately Av=R2/R4. There's a negative sign there normally, but let's consider only the the gain magnitude for the moment.

    If you have a target Av then you can proceed to estimating values for collector & emitter resistors R2 & R4, using the additional information for Ic and your selected Vc.
     
  16. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    783
    Your calculations/method also let you down at a couple of other points ...

    I see you have set R2=750Ω and R4=150Ω giving you a target gain magnitude |Av| of 5. For reference, I have used the component ID's as you posted in your schematic.

    You correctly show Ve=2.25V at Ic=15mA.

    Vb would therefore be about 2.9V. Correct again.

    This would set R3=2.9/500uA=5.8kΩ

    The current in R1 would be about 550uA (Ib + R3 current). The voltage drop across R1 is 24-2.9=21.1V.

    So R1 would be given by

    R1=21.1/550uA=38.36kΩ

    A slightly different approach is to take the current in R1 as 500uA and the current in R3 as 450uA - which is what I think you intended. I just took the 500uA flowing in R3 rather than R1.

    So in your case, with the base current of 50uA and 500uA in R1.....

    R3=2.9/450uA=6.44kΩ

    And R1=21.1/500uA=42.2kΩ

    Hope that helps.
     
    Last edited: Apr 11, 2011
  17. Adjuster

    Well-Known Member

    Dec 26, 2010
    2,147
    300
    Be careful with orders of magnitude and prefixes. You have written

    Leaving aside whether the values you start with were the right ones, the correct result of 2.25V/200uA is 11 250Ω or 11.25kΩ.

    This might just be the result of a slip of the finger to the adjacent key, but the result is that the answer is out by a factor of a million!

    11.25m ohms (11.25mΩ) means 11.25 milliohms, that is 11.25*10^{-3}Ω or 0.01125Ω.

    For reference, here is a list of SI prefixes. There are quite a few that I've never used! http://en.wikipedia.org/wiki/SI_prefix
     
  18. stinky111

    Thread Starter Member

    Apr 7, 2011
    39
    1
    Wow, thanks guys,( T N K & Adjuster) appreciate the quick responses. Now I see where Ive gone wrong..
    Yes correctly one of my mistakes was a slip of the finger on calculator, i will have to be very careful or that in future..

    Once again, thx for your time, help appreciated.
     
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