BJT Circuit Analysis

Discussion in 'Homework Help' started by nDever, Feb 20, 2015.

  1. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    Hey,

    So I need to analyze a few designs that include BJTs, but all of my textbooks only cover MOSFET analysis. Can I model the BJTs as dependent sources (current-controlled)? How can I approach this?
     
  2. #12

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    Nov 30, 2010
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    Your post will be moved to the Homework forum.
     
  3. WBahn

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    Mar 31, 2012
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    Whether you model them as current-controlled or voltage-controlled depends on the model you use which depends on what is most appropriate for your circuit. It also depends on whether you are talking about large signal or small signal behavior.
     
  4. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    Sorry, I'm using them for a switching application. Could I assume them either saturated or in the active region and then use their β-α characteristics for the current/voltage?
     
  5. WBahn

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    If you are using them for a switching application, then you want them to either be saturated or off. You want to keep them out of the active region except during the brief transient switching times.
     
  6. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    Here is a schematic. I'm a little green when it comes to using transistors in this way, I've only really done logic circuits. Circuit.png
     
  7. WBahn

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    How can we possibly tell if that circuit will exhibit the behavior you want unless you describe what behavior you want it to exhibit?
     
  8. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    Behvavior...right! You might need that.
    So when SW is open, Q1 should be driven saturated which should cause C to charge through R5. When SW is closed, Q1 should be grounded and turned off while Q2 is driven, causing C to discharge through R4.
     
  9. WBahn

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    Your circuit won't do that. But it is not too far off.

    It's hard to interpret your schematic because it is drawn in a very unconventional manner. Here is a more conventional drawing that is (or is at least meant to be) exactly the same as your circuit:

    Qsw_wrong.png

    To use a BJT as a switch, you normally want to use a PNP to pull high and an NPN to pull low. You also want to connect the emitter of each transistor directly to the power rail (Vcc for PNP and COM for NPN) so that you can drive them into hard saturation. With that in mind, give it another go.
     
  10. #12

    Expert

    Nov 30, 2010
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    I want you to notice that this is a switching circuit and it works well with the emitters facing the power lines. This configuration is horrible for an audio amplifier or op-amp output because there is no valid way to stop with the output voltage in the middle of the power supply voltage. When you get to that part and see the emitters facing the other way, be aware of why it looks backwards.
     
  11. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    OK, let's try this again.
    Circuit2.png
     
  12. WBahn

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    Much closer and this may do what you need, but not quite what you described.

    When the switch is closed, Q1 will be ON and Q2 will be OFF. What is the effective resistance that the capacitor discharges through.

    When the switch is open, Q1 will be OFF and Q2 will be ON. What is the effective resistance that the capacitor charges through.

    Now compare that with what you want and, if necessary, make adjustments.

    Once you have a good topology, we can talk about resistor sizing for R1 through R3.
     
  13. WBahn

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    BTW: What tool are you using to make your schematics -- they look very clean.
     
  14. nDever

    Thread Starter Active Member

    Jan 13, 2011
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  15. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    You're right WBahn, my second circuit doesn't give me exactly what I want, but I can't figure out the correct structure. I can't have C discharging through the same branch that is charges through.
     
  16. WBahn

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    You are very close. Ask yourself why it discharges through R5 but it doesn't charge though it. Then ask yourself if you can do something similar with R4 except on the charging side.
     
  17. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    OK, I feel pretty dumb. I feel like I'm looking right at the problem... :confused:
    Does it have anything to do with R4 being connected to the collector?
     
  18. WBahn

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    Nope.

    Think of the discharging path. Ignoring R4 (remove it for now) you have a path that goes from the capacitor through R5 and then through Q1 to the negative supply rail (COM).

    Now think of what you want for the charging path. You want a path the goes from the capacitor through R4 and then through Q2 to the positive supply rail (VCC).
     
  19. nDever

    Thread Starter Active Member

    Jan 13, 2011
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    I can't imagine this thing working if I leave the RC branch where it is. When the switch opens, the voltage on R5 will go to Vcc which will charge the cap.
     
  20. WBahn

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    I don't know what you are referring to by "the RC branch" since you have two Rs. With the current circuit, when switch is open Q2 will turn on taking the top of R5 to Vcc. But the other end of R5 will be open and no current will flow through it. The cap will then charge through R4. Notice that R5 is NOT involved in the charging process. Why is that? You want the same thing to be the case for R4 during the discharging process so that it is NOT involved in the discharging process.

    Again -- you want:

    Charging: Vcc -> transistor -> resistor -> capacitor.
    Discharging: capacitor -> resistor -> transistor -> common.

    If you want different resistors in each path, then each resistor needs to go in the part of the path controlled by the respective transistor and not in any part of the path that is common to both.
     
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