Trying to create single stage BJT amp with high current gain. Stuck and need help.

Discussion in 'The Projects Forum' started by HunterDX77M, Apr 7, 2012.

  1. CDRIVE

    AAC Fanatic!

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    When component values are unknown the accepted practice is to state them as Rx, Cx, Lx, Zx, or, if you prefer, R?, C?, L?, Z?, etc. As for what part (exactly) that you lost me? ... Well, all of it because you can't re-write Ohms Law. Not even Congress or the Supreme Court can do that. Maybe Al Gore can but not quite sure about that. :rolleyes:

    The Vcc is not a source of excitation. It's your power source. This includes oscillators, which have no external source of excitation.
     
  2. panic mode

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    Oct 10, 2011
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    how about simple fresh start - can you post entire question exactly as stated in your book? that would eliminate any ambiguity and guessing...
     
  3. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Okay, I've looked over things and have a rebuttal for the voltage beyond the Vcc.

    7.58 kΩ * 5 mA = 37.9 V This is true.

    But what I mean by 5 mA peak-to-peak is a sinusoidal current with an amplitude of (5 mA)/2 = 2.5 mA. The current swings between -2.5 mA and +2.5 mA. It should never have a magnitude greater than 2.5 mA. I'm sorry that wasn't clear (my fault). The maximum voltage required at the load would then be:

    7.58 kΩ * 2.5 mA = 18.95 V < 20 V

    Do you agree with me?
     
  4. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Okay here it is. Hope this clears all the smoke.
     
  5. studiot

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    No.


    Whilst we do have (somtimes good) debates here, I thought you were seeking help, not a debate.

    Along with several others I have asked some questions, trying to probe your level of understanding in order to frame suitable responses.

    So far you haven't answered one that I can see.

    What did you make of my description of how the transistor operates?

    Edit I see another post whilst I was writing this one so here is another question.

    What do you understand by 'clipping' as referred to in your question?
     
    Last edited: Apr 8, 2012
  6. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    I understand it to be when the transistor is saturated and the output current (or voltage) levels off and won't increase any further with an increase in the input. Please correct me if I am wrong.

    I'm sorry if this turned into a debate, it's just that some of what people are telling me here confuses me when I try to relate it to what I learned in class. I'm really sorry. I came here looking for help and ended up agitating/irritating everyone. :(

    Can you please repeat the questions that you need my answers to?
     
  7. studiot

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    Clipping and saturation are different things entirely.

    I need to draw some diagrams and start at the beginning.

    I'll post them shortly.
     
  8. studiot

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    OK,

    Let us assume we have a +20 volt supply and let us create a potential divider by connecting a fixed resistor R1 in series with a variable resistor R2 across the supply.

    Now let us consider the voltage Va at their junction point A.

    I have drawn this in Fig1(a).

    If we adjust R2 down to zero then the voltage at A is also zero (Fig1 c)

    If we make R1 zero the voltage at A is +20 regardless of the value we adjust R2 to be. (Fig1 b).

    No matter what we do we cannot make Va greater than 20 or less than zero.

    So we have set the bias of point A to some value Va

    Now let us set Va to different values and inject a 20 volt peak to peak alternating signal at A.

    In Fig2 Va is set to halfway ie 10 volts. You can see that the alternating signal take Va ap to just touch the +20volts and down to just reach zero volts - a peak to peak of 20 volts.

    In fig3 Va is set to +15, however as previously noted Va can never exceed +20.
    You can clearly see the result.

    This is called positive clipping

    Note that Va no longer reaches the zero rail in its excursions and the bottom half of the sine wave is not clipped

    In Fig4 the opposite happens where Va is now set to +5 volts.
    The result is clipping of the negative half cycles as Va can never fall below zero. Needless to say

    This is called negative clipping

    If we injected a 30 volt peak to peak sine wave we we see simultaneous clipping of both positive and negative half cycles ie a 30 volt pk to pk signal will not fit between 20 volt rails.

    Please confirm that you understood all this.

    In the next installment we can look at how this appertains to a transistor circuit, however I will observe that the transistor behaves like the variable resistor R2.
     
  9. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Crystal clear. :)
     
  10. studiot

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    So if we return to transistors and make R1 the collector resistor and R2 the transistor resistance between collector and emitter.

    R2 is of course a variable resistance, which we vary electronically by varying the drive to the base. The greater the drive the lower the 'value' of R2 and thus the greater the value of the collector current.

    Can you now see the relationship to my earlier description of transistor action?

    And can you see why that means to maximise the possible AC output we must set (bias) the transistor so that the collector (point A) is halfway between the rails?

    Have you heard of the Q point? This is what is meant by the Q point.
     
  11. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Ah, okay. My textbook did a poor job of explaining quiescent points, which is why I didn't get what you were saying before.
     
  12. studiot

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    So where does that take us to with your requirement to drive 5mA pk to pk through 7580 ohms?

    As a matter of interest it is just possible from a supply of 20 volts, but not with a single transistor or even a pair of them.

    To achieve this magic feat you need to use a full bridge amplifier which alternately switches the load across the supply first one way then the other.
     
  13. HunterDX77M

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    Sep 28, 2011
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    A full bridge amplifier? That's certainly beyond the scope of anything I've learned, yet. I can't and won't give up on this, but based on all the awful failures I've had today (on paper and PSpice) and everyone's advice so far, it seems like what I want to accomplish is impossible.
     
  14. studiot

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    I don't know why you want to achieve this but a full bridge amp (often called an H bridge amp these days) is not so difficult. It will certainly be on your syllabus at some point.

    Did you understand what I meant about reversing the polarity of the load each half cycle?
     
  15. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    1) Why I want to achieve the whole thing? Well, it's an assignment. I have to do it.

    2) I'm afraid it's not on our syllabus. After looking it again, it looks like differential amplifiers are the last thing we'll be learning (and the semester is almost over).

    3) No, I didn't really understand what you meant. Is that like the push-pull that goes on in a Class B or AB configuration?
     
  16. studiot

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    Before I go tonight, here is more.

    For the first half cycle Transistors TR1 & TR4 are on and the other two are off.

    Thus end A of the load is connected to the positive and end B to the zero.

    Obviously the transistors are driven appropriately to provide the waveform.

    For the second half cycle the roles reverse with TR1 &TR4 off and TR2 & TR3 on.

    Now end B is at +V and end A at zero.

    The allows the 20 volts supply to be used two ways round to provide the nearly 20 volts peak each way needed for your load.

    Does this cheer you up?

    Goodnight.
     
  17. CDRIVE

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    Curious,.. where does this put the resolution of this topic? Has the professor restated the requirements of this assignment? Was there a misunderstanding of the parameters? Was a decimal point inadvertently dislocated? Are we still committed to a single BJT stage?

    Any of the above? :confused:
     
  18. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    I spent three hours today with the professor and he helped me understand what needed to be done. I'll work on the design I have now and post sometime tomorrow what I did for some resolution to this unbelievably long thread.
     
  19. studiot

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    I look forward to seeing your explanation.

    Meanwhile, I suppose you did not expressly exclude a transformer from your circuit so I suppose you could place the primary of a suitable transformer in the collector circuit and drive your load from the secondary.
     
  20. CDRIVE

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    Ha, believe me kid, this thread length doesn't come close to an AAC record. Don't despair though. If we all work together we can make it happen. :rolleyes:
     
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