THIS IS GEARED TO BEGINNERS ONLY.
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.
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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.
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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
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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.
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.
----------------------------------------------
-------------------------------------------------
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.
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----------------------------------------------------------------
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
-------------------------------------------------------------
---------------------------------------------------------------
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.