For Beginners to practice biasing

Discussion in 'General Electronics Chat' started by hobbyist, Sep 28, 2009.

  1. hobbyist

    Thread Starter AAC Fanatic!

    Aug 10, 2008

    Here is a very good excersise for someone just starting out in learning how to bias transistors, and gaining some experiance in logicaly thinking through the process.

    First use a CC config. for the output because it is easier to grasp for a beginner.
    I'll explain my example circuit in detail to show how this exercise is done.
    I bias my CC amp to have a VCC = 18v. and a VE @ 11v. across 2k ohm load resistor.

    Now that will make the VB = 11.7v.
    I will continue to use a resistor from base to ground to be around 10 times greater than emitter resistor.
    So RB1 = 20K ohms.

    current IRB1 = (11.7v. / 20k ohms) = 585uA.
    and RB2 calculates to be around 10K ohms.
    Here is the scheme so far.
    Bias ex. 1.

    bias ex. 1.jpg

    Now replace the RB1 resistor with a CE and bias it to keep the 11v. output at the emitter of Q1.

    So I have the IRB1 established at 585uA. and the VC of Q2 established at 11.7v. so I will make the voltage at the emitter of Q2, to be around 1/2 of VC of Q2. make it 5v.

    Now RE2 = (5v. / 585uA.) = 8547 make it 8.2K ohms.
    again 10 times it = 82K ohms for RB3, and VBQ2 = 5.7v. so (5.7v. / 82k ohms) = 70uA.

    So RB4 = (18 - 5.7) / 70uA. = 175K approx. so make it 180K.
    Bias ex 2.
    bias ex.2.jpg

    Now I want to change only the Q2 stage to give an output at Q1 of 6v.
    Now that means I need to calculate the current flow through RB2 when VBQ1 = 6.7v. (remember VEQ1 = 6v. output).

    So VBQ1 will be 6.7v. and V. across RB2 = (18 - 6.7) = 11.3v.
    therefore current IRB2 will be ( 11.3v. / 10K ) = 1.13mA.

    Now to solve for the RE2 value I'll choose around 1/2 of VCQ2 which is 3v. and take (3v. / 1.13mA) = around 2.7K ohms for the new RE2.
    Now the new VBQ2 = 3.7v.

    again make RB3 to be around 10 times RE2 = 27K ohms. (RB3 = 27k)
    Now again calculate the current through this resistor, and then solve for the value of the new RB4 value which comes out to be around 100K ohms for RB4.

    Bias ex. 3
    bias ex.3.jpg


    Practice using different output voltages for Q1, write down your calculations, then before you go to check the voltage at the points, VEQ1, VBQ1, VEQ2, VBQ2, say out loud to yourself what the voltage should be then check that voltage point, to confirm your analysis.

    This could be done with CE amps as well youll need more calculations including the voltage across the collector resistor and VC of the output transistor,

    but just work at it and practice until you get good at predicting (calculating ) the voltages at a point in your circuit and verifying it with an actual measurement. It will help you gain confidence too.

    By all means have FUN with it.
  2. hgmjr

    Retired Moderator

    Jan 28, 2005
    Do you happen to have an assumed Vbe and Beta figure for the transistors used in the problems you have posted?

  3. Audioguru


    Dec 20, 2007
    He assumed 0.7V for Vbe. The base is at the fairly high voltage of 11.7V so if Vbe changes to 0.6V or to 0.8V there will not be much effect on the biasing.

    Since he used the value of the resistor from the base to ground to be 10 times the value of the resistor from the emitter to ground then the Beta of the transistor can be from about 50 to hundreds without much effect on the biasing.
  4. hgmjr

    Retired Moderator

    Jan 28, 2005
    I see that the Vbe is 0.7. I would like to have the OP confirm the Beta just for completeness.

  5. hobbyist

    Thread Starter AAC Fanatic!

    Aug 10, 2008
    I wasn't using any beta figure, I was implementing using a 2n3904 transistor, because it seems to be pretty common, and by establishing a base ground resistor, at around 10 times RE for this excersise, seems to be a good choice to keep a stiff enough divider current for this transistor.

    As well as for a beginner to learn how to keep a reasonably high input impedance to that stage.

    This excersise is not to establish the Q point for an amplifier, as normally 1/2VCC, but rather to learn how to calculate what voltages should be at a specific point in a circuit, and using basic equations and theorems, to establish the value resistors needed to achieve those voltage points.

    While also encountering the problems associated with working with nonlinear devices, and getting the practice of knowing how to establish chosen voltages, wether it be through resistors or even using diodes to establish voltage biasing,

    such as a zener diode being used to establish VE, that could further be implemented into building a series feedback voltage regulators, and things.

    I feel this excersise can build confidence in someone just learning, how to design a transistor circuit, because once you can calculate and emperically measure a voltage close to your calculations at a point in the circuit that seems formidable, by looking at it in schematic form, breaks the mindset, "I'll never understand this stuff",
    to a confidence that says,"wow I can actually calculate what this voltage and current should be right at this point", Tjat gives the beginng learner, a boost in the "I CAN understand this stuff a little more than I thought" direction..
  6. hgmjr

    Retired Moderator

    Jan 28, 2005
    I see now that you have a good reason for not including the details I mentioned.