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Bipolar Junction Transistor DC Biasing Calculations Example.

The model of the circuit we are biasing.

The model of the circuit we are biasing.

It is Number Crunching Time! Tonight we set the DC Bias of the circuit shown in the first diagram. Once we do that we will run several Qucs simulations of the circuit with the numbers we calculate. We will do those simulations with different values for Beta and Vo and see how well our negative feedback helps us tolerate the variations.

As a reminder:  In the post “Reading a Datasheet for our DC Transistor model parameters.“, we determined the following values:  Vo = 0.65 V;  iceo = 0.05μA;  Beta (β) = 200 but can actually be from 100 to 300.

Now that we have the model down the next step is to determine our goals.  This is where in real life things get hard because the devices feeding this circuit and the device we are feeding will determine our goals.  However, we are baby-stepping and we have quite a bit more to learn before we tackle real life.   So.. I pick some numbers.  We want an Ic of 50 mA (+/- 2 mA) and the voltage across the transistor (VCE) of 4 Volts.

Now that we have the specifications set we are ready to get out the calculator and have at it.

Step 1:  We have to choose a value for VCE. The suggested rules said 3 to 5 times VCE. 15 Volts sounds like a good number and that is 3.75 times VCE.

Step 2: The rule says VRe should be the same or a little less than VCE. So, I choose 3.5 V.

Step 3: We need to calculate the value of Rc.  First we must calculate VRc.  VRc = VCC – (VCE + VRe = 15-(3.75+3.5) = 15-7.25 = 7.75V.
Rc = VRc / Ic = 7.75 / 0.05 = 155 Ohms.
Note:  I got a little lazy in this because actually Ic = the planned Ic + Iceo, but Iceo is so small it will not matter.

Step 4:  Calculate the Value of Re:  First we must calculate Ie.  Ie = Ic + Ib. and Ic= β * Ib.  Using some algebra, Ib = Ic / β.  = 0.05 / 200 = 0.00025.   Now Ie = Ic + Ib = 0.05 + 0.00025 = 0.05025.
Re = VRe / Ie = 3.5 / 0.05025 = 69.65 Ohms.  (Yes, I know this is a crazy value because it is not made with this kind of accuracy.  But for right now we will use it.)

We have completed the output side of the transistor, and now we move on to the input side of it.

Step 5:  Calculate Rb1:  First we need IRb1.
The rule is IRb1 should be 10 to 100 times Ib. 50 X sounds good.
IRb1 = 50 X Ib = 50 X 0.00025 = 0.0125 A
Vb = Ve+Vo = 3.5 + 0.65 = 4.15 V
Rb1 = ( Vcc – Vb) / IRb1

Step 6: Calculate or one remaining Item Rb2:
First we must figure out the current through Rb2.  If 50 X Ib is going through Rb1 and Ib is going to the base, then 49 X Ib is going through the resistor Rb2.  IRb2 = 49 X 0.00025 = 0.01225.
Rb2 = Vb / IRb2 = 4.15 / 0.01225 = 339 Ohms

The "as calculated" model of the circuit model in Qucs.

The “as calculated” model of the circuit model in Qucs.

Now to check our work I created a Qucs model of the circuit and did a DC simulation.  Everything matched up exactly.  It should because it is the same model!

Now the moment of truth.  What happens if we use the same values for all the components but the value of β changes?  Is our negative feedback helping us?

 

 

 

The model with the calculate value for the components but the value of Beta is 100.

The model with the calculate value for the components but the value of Beta is 100.

This is where a program like Qucs helps a lot because several interacting values (Ic, Ib, Ve,Vb and Vc) would all change at the same time and it would require several calculation iterations to calculate all of those. To do this on Qucs I simply changed the value of Beta and clicked simulate. With Beta = 100 our Ic value dropped from 50 mA to 49 mA and our Vce dropped from 4 volts to 3.97 V. I would call that good!

 

 

 

 

I have ran several other simulations and have the results summarized in the table below.

 

Beta Vo Ib Ic Ie Vc Ve Vb Remarks
200 0.65 V 0.25 mA 50 mA 50.3 mA 7.25 V 3.5 V 4.15 V As Calculated
100 0.65 V 0.49 mA 49 mA 49.4 mA 7.41 V 3.44 V 4.09 V Beta 100% low, Ic 2% low
300 0.65 V 0.168 mA 50.4 mA 50.6 mA 7.19 V 3.53 V 4.17 V Beta 100% high, Ic 1% high
200 0.60 V 0.254 mA 50.7 mA 51 mA 7.14 V 3.55 V 4.15 V Vo 9.3% low, Ic 2% high
200 0.70 V 0.247 mA 49.3 mA 49.6 mA 7.35 V 3.45 V 4.15 V Vo 9.3% high, Ic 1.4% low

Summary: Our negative feedback worked and the circuit is much more tolerant to variations in manufacuting tolerances of Beta and Vo.

The next step will be to make the resistors reasonable values. In other words the standard values. I think at this point it will be helpful to create a video of me doing that in a Qucs simulation.

The next step after that will be to use Qucs with a SPICE simulation of the transistor. I have already done that with the exact calculated values of the resistors. I will create another version once we use standard values for the resistors.. When I created the SPICE version I ran into some problems and had to troubleshoot my typing. That will also be very helpful in a video format to show you how to troubleshoot it.

After we get through all that on to better things… actually running a singnal through an amplifier! Fun times are coming.
Gary

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Bipolar Junction Transistor DC Biasing Calculations Example.” by Create-and-Make.com is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.

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2 comments to Bipolar Junction Transistor DC Biasing Calculations Example.

  • Danel Jones

    Hi Gary,

    Really nice series on BJT modeling and design. I believe you have a typo in your commentary. In the Beta = 200 case, Vce = 3.75V instead of 4V.

    I found your site when I went looking for additional QUCs tutorials. I don’t know what my problem was, but I’ve had no problem duplicating your results.

    So, thanks and keep up the good work.

Leave a Reply to Danel Jones Cancel reply

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