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. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Hey Guys,

    I need some help with a single stage BJT amp. I need to get some considerable current gain (18.5 A/A) from this stage as well as deliver 5 mA peak-to-peak to a load resistance.

    I wanted to initially go for a common-collector configuration, but I have this 2 kΩ resistor at the collector that has to be there (I need an output resistance of exactly 2 kΩ). What other configurations are there that I could consider that could give me high current gain (as well as some stability to changes β)?

    I really appreciate the help. I'll post a schematic in a minute. I am having some trouble posting it with this thread.

    Some important parameters:
    Vcc = 20 V
    Source resistance = 15 kilo ohms
    Beta = 225
    Capacitors are "large"
     
  2. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
    104
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    This was the general layout I was working with, but it just won't click. This circuit should be easy since I had to do one similar to it last week with a significantly lower gain trying to deliver 16 mA peak to peak.

    Also the load resistance is 7.58 kΩ.
     
  3. panic mode

    Senior Member

    Oct 10, 2011
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    the only configurations for high current gain are common emitter and common collector.
     
  4. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
    104
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    Yes, I've realized that common emitter is really the only way to go at this point. It's just that I'm having so much difficulty trying to get anything that comes remotely close to what I want. Do you have any recommendations as to where I should start to get something working?

    For the other circuit that I made that also had current gain, I used common collector, but the gain was only around 3 A/A.
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Have you tried to calculate the peak to peak voltage across 7568 ohms at 16 mA peak to peak?
     
  6. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    The circuit that required the 16 mA peak-to-peak current was another project, different from (but related to) this project. In this project, I need 5 mA peak-to-peak roughly at the 7580 ohm load.

    I'm not entirely sure what you are suggesting. Could you please clarify?
     
  7. CDRIVE

    Senior Member

    Jul 1, 2008
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    Please restate your requirements, as I don't get it. Also, why are R4 & R5 0Ω ?
     
  8. PaulEE

    Member

    Dec 23, 2011
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    If your supply voltage is +20v, I don't understand how you think you're getting 5 mA through 7.6K. That requires V = IR = 37.9v...also, current usually isn't "at a node", but rather, it flows "through a node". What you mean to say is that you need +37.9v "at the node" to cause the resistor to sink 0.005 amps, assuming it is with reference to ground...
     
  9. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    If the limit is a single device and not a single BJT
    Might I suggest a darlington in a three leg TO package.
     
  10. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Sorry to hear you didn't get my requirements. In what part exactly did I lose you?

    For the resistors: I just couldn't figure them out so I just left them 0. The purpose of the diagram was just to show the configuration I was looking at, not the actual values of resistors.
     
    Last edited: Apr 8, 2012
  11. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    The Vcc is not the only source of excitation in the circuit. The current source is varied until I get a 5 mA peak-to-peak swing through the load and Vcc provides that extra current which causes the amplification. If Vcc was the only source in the circuit, then it would be just a DC biasing problem.
     
  12. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Interesting. I only recently learned about the Darlington pair in class, and I recall that it gives extremely high gain (β^2 approximately). Are you suggesting just replacing the single transistor in the schematic with one pair (but otherwise keeping everything else the same)?
     
  13. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Does anyone have a suggestion for any specific kind of configuration (i.e., where certain resistors should be oriented)? If so, please post a schematic if you would be so kind. I'm really beginning to tear my hair out with this.
     
  14. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Whilst I am sorry I misread your circuit in respect of the load resistance value, Paul is making the same point I referred to.

    You cannot beat Ohm's law; what you are proposing in not what you have posted in your circuit.
     
  15. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    How about this. This is another circuit that I made last week, different from the one I am having trouble with now, and it provides the 16 mA peak-to-peak swing that I referred to earlier in this thread (it has a current gain of 3.3 A/A).

    I'm just trying to do a similar thing with the circuit I am working on now, except the gain I need is 18.5 A/A with a 5 mA peak-to-peak swing.

    The 7580 ohm load resistance that I use in the problematic circuit is the input resistance of this circuit I am posting now, by the way.

    Does this make sense or have I just confused you more? :(
     
  16. studiot

    AAC Fanatic!

    Nov 9, 2007
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    I think you are suffering from some fundamental misconceptions and need to start at the beginning.

    The circuit with a 470 ohm load resistor has 470 * 16/1000 = 7.5 volts swing across it.

    This is clearly possible from a 20 volts power supply.

    As Paul has pointed out, and I tried to get you to calculate and thereby understand, a 7580 load resistor with a 5 mA swing requires a voltage swing of 37.9 volts. This is impossible from a 20 volt supply. You will need something greater than 45 volts to achieve this, (which is larger than the 2n2222 rating.)

    What do you think the function of the transistor is and how do you think it works?
     
    Last edited: Apr 8, 2012
  17. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Nevermind, I see what you are saying now. This whole time, I was confused because I thought you were talking about something else. Yes now I see, it would require more voltage at the load than I have a available. I guess I'll have to contact my professor about this.
     
  18. studiot

    AAC Fanatic!

    Nov 9, 2007
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    We are trying to help here, but are not here to design your circuits for you.

    So if you would like to answer the questions asked they should help, they are designed to lead you in the right direction.

    Here is a simple explanation of how the transistor works.

    The emitter is fixed at ground potential.

    The transistor tries to force a value of collector current through its collector resistor given by beta times its base current.
    That is it is like a variable tap (faucet) that can be turned on more and more by cranking up the base current.

    The top end of the collector resistor is fixed at Vcc.

    So as a result of the collector current it has a voltage drop across it. The larger the collector current the larger this drop.
    So the other end of the collector resistor - the bit that is joined to the collector - falls in potential as its current increases.

    So as this point is driven up and down in voltage it can drive the voltage across a load resistor up and down, thus generating a current in the load.

    A similar argument can be made for an emitter resistor.

    The transistor draws power from the supply rail and delivers it as current and voltage to a load. It does not produce power of itself. So you can only draw the current and (in this case) voltage that is available at the power rail.
     
  19. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
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    Thanks. I'm sorry if I've been difficult to work with here, it's just that I am so frustrated that everything that I've tried has just blown up in my face (figuratively). Maybe I need to take a breather and come back with a clearer head.
     
  20. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Sure when you are ready

    :cool:
     
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