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Bjt amplififer

Discussion in 'Homework Help' started by merk24, Sep 3, 2016.

  1. Tony Stewart

    Tony Stewart Well-Known Member Most Helpful Member

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    Install https://www.windows10download.com/ltspice-iv/
    then .asc files open in this.
    click stickman icon to run scope.
    click any node(s) for voltage or components for current.
    repeat after editing schematic values or adding, parts.
    use move or drag hand icon (rubbers band mode keeps nodes connected)
    voila!

    Car 12V 4W audio amps use many transistors in final stage in discrete or IC (>10k current gain, symmetry of bipolar gain, hence low distortion and maximum swing.) such that output impedance is <<1% of load R.
     
    Last edited: Sep 5, 2016
  2. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    So at what point can we try for an answer(s)?
     
  3. MikeMl

    MikeMl Well-Known Member Most Helpful Member

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    Most respondents to this thread seem to be all about showing off how smart and experienced they are instead of showing the TS what principle he/she needs to learn to solve the problem...
     
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  4. dave

    Dave New Member

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  5. EvilGenius

    EvilGenius Member

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    Here is a tutorial on Class A, Class B, and Class AB amplifiers that I found useful. Hope it helps the OP with the design aspect and modification of the circuit to his/her needs.
    http://www.electronics-tutorials.ws/amplifier/class-ab-amplifier.html

    With my limited knowledge I recommend working backwards. Separate two phases of input/output (when input goes positive work the top transistor and when input is negative work the bottom transistor). Find out what output current and voltage is needed, set Vbe of complimentary transistors to 0.7v to obtain base, collector, emitter voltages and currents.
    It also helps to utilize some transistor basics and shortcuts such as Ic/Ib=hfe (set to 100 to begin) and Ie=0.99 x Ic. Also set current below base to be 10 times Ib, and current above base to 11 times Ib. One can quickly calculate everything like a ladder. Then the currents and voltages for the third transistor which provides bias for the complimentary transistors can be calculated. At the final stage set the decoupling capacitors to create the bandwidth (or specific frequency operating window to isolate the biased transistor) and to create the smooth sine shape. From there on is making adjustment to obtain the optimum output gain, voltage, current and power rating. Just a few thoughts...
     
    Last edited: Sep 5, 2016
  6. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The original posting person said, "our project" and had specifications for it so it is probably a school assignment. I do not know why the teacher did not teach the basics first.
    The required gain is very low so 3 transistors should be able to do it.
    The small 2N4401 and 2N4403 transistors will get extremely hot but should be able to produce 1.24W into an 8 ohm speaker with the 12V supply.

    I show simulations of the original circuit with its 30 ohms load and with an 8 ohm load.
    I added bootstrapping and reduced the resistor values and show the simulation.
     

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  7. EvilGenius

    EvilGenius Member

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    I thought he needed 3w output?!
    I must have read that wrong.
     
    Last edited: Sep 5, 2016
  8. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    I changed the load to 8 ohms so we can get 3 watts from a 12V supply.
    problem: We need more current. The little output transistors can not with stand the heat and the current is very close to the max the part can with withstand. I chose 5A transistors not 600mA parts. These parts can connect to a heat sink to get rid of the heat.
    upload_2016-9-5_14-28-46.png
    Below is a graph of the old transistors. You can see the current gain is about 200 with current below 100mA. At 500mA the gain drops to 20. By choosing a bigger transistor there will be good gain at the desired current. So start at the load. Find what voltage, current and power you need. Choose transistors designed to work at that level.
    upload_2016-9-5_14-36-7.png
    To get 10x more current in the output you need (about) 10x the current in the bases. I changed the resistors to get about 10x more current through out the amplifier.

    The goal is not clear to me. I think your teacher wants 10x more power out. So change the output transistors to parts that can work well at 10x the current.
    -------------------------------------------------------
    Another option: (every thing in this picture functions as one, very high gain transistor)
    Increase the current and power capability of the output transistors AND (increase the current gain) so Q1 does not need to be changed. Q5= the old low current transistor. Q4= the new high current transistor.
    Question: Why do I need two diodes? D3, D4
    Question: What is Q5+Q6 called?
    upload_2016-9-5_14-52-27.png
     
  9. EvilGenius

    EvilGenius Member

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    Ron
    What is the purpose of C3 in your circuit? More transient output current?
     
  10. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    With out C3:
    When you turn on Q2 hard; there is little voltage across Q2 C-E. This will cause there to be very little voltage across R3. Q2 will not turn on well because you can not get base current. (no voltage - no current)
    upload_2016-9-5_16-18-27.png
    At steady state; (No signal) The output is at 6V or 1/2 supply. There is 3V of charge on C3.
    Transistor on hard with on C3. Q2E=11V Q2B=11.7V So the current into the base is only 0.3V and (70+70) ohms. About 2mA (base current) is not enough current to get much emitter current.
    Transistor on hard with C3! Because there is 3V stored on C3, when Q2E=11V the bottom end of C3=11V and the top end of C3=14V. Q2B=11.7V. Across R3 there will be 2.3V. 2.3V/70 ohms = 33mA. By adding C3 we increase the base current from 2 to 33mA. So the emitter current can be 16 times more.
    upload_2016-9-5_16-22-56.png
    This is called boot strapping.
    Reach down, grab your shoe strings and pull hard. If you had gain like the transistor you could pull your self off the floor.
     

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  11. EvilGenius

    EvilGenius Member

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    Thank you. I read a little about bootstrap and understand what you are describing. I also understand you want the discharge rate to be longer than transient input switching period. How did you come up with value? Based on longer than 1khz time period using 1/2pixfxc formula? How did you decide that the top of C3 is applied to the bottom 70 ohm and not, let's say, to the base directly? You need a resistor to slowly discharge it?
     
  12. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Yes,
    An audio amp should work to 20hz. (more or less) But right, if you are only amplifying 1khz then the cap should not discharge much in 1/1000.

    What we are trying to do is make a little supply that is greater than 12V to pull up the base. We could have removed the resistor from +12V and connected it to a 15V supply and had similar results.
     
  13. EvilGenius

    EvilGenius Member

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    What are the cons to your bootstrapping in brief and should we be worries about cap discharging and transistor to shut off before the down cycle and even worse the next up cycle?

    Thank you in advance for your answers. It helps me understanding designing them better.
     
  14. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Since audio frequencies go down to 20Hz then you want an audio coupling capacitor to have its reactance equal to or less than its load resistance. But this amplifier has three coupling capacitors so the reactance should be calculated at 20Hz/3= 6.7Hz.
    The calculation is simply C= 1 divided by (2 x pi x R x f).

    With a 12V supply the maximum output swing might be 8.5V p-p which is 3.0V RMS. Then for an output of 3W the load must be 3 ohms which is almost a dead short and the peak current will be 1.42A.
     
  15. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    4 ohm load
     

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  16. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Ron, your amplifier has clipping at the bottom of the waveform. With your 4 ohm load its output power is only (10.4V - 2.0V)/2 divided by 2.828, squared, divided by 4 ohms= 2.2W. With a load that is 3 ohms then the output is 3W.
     
  17. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Ron, I zoomed your schematic so its details are visible:
     

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  18. EvilGenius

    EvilGenius Member

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    I do not see any comments from OP. The least he could have said was thank you.
    Thank you for your expert comments. It helped me learn more even though I find this type of amplifier not very practical.
     
  19. EvilGenius

    EvilGenius Member

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    Humor side: D3 and D4 are used so the circuit has another leg to stand on the work station.
    Combination of Q5+Q6 is called "Transistors in Heat" or humping T's?!
     
  20. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    That is why you are called "Evil".
     
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  21. audioguru

    audioguru Well-Known Member Most Helpful Member

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    I am a dirty old man and am surprised that I never noticed humping from darlington transistors.
     

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