In the post “The Basic DC Model for a NPN Bipolar Junction Transistor” we came up with a generic model for a transistor and in the last post “Reading a Datasheet for our DC Transistor Model Parameters” we assigned values to the parts of the model based on a real transistor. In this post we will use the model for to choose values for the external components. We will also see some of the difficulty designing with transistors. The circuit we will work with is not a good one for real use, but it is a good one to learn with.

I needed to create a set of specifications for our circuit shown in the first picture, In real life the specifications would be set by the signals driving the circuit and by the load we have connected to the circuit. However, this is just and example so I am free to choose. I first had to choose a voltage for the power supply, usually called Vcc. Randomly, I chose 12 Volts, because it is a common voltage used for power supplies and it is well below the maximum value on the datasheet. Next I have to choose a current flowing through the transistor. 100 mA for Ic sounds like a good number. It is one half the maximum current and finally I need to set the collector voltage. I set it at 6 volts or one-half of the power supply. This means half of the voltage will be dropped across the resistor and half will be dropped through the transistor. This also will create the power dissipated by the transistor to be 100 mA X 6 V or 600 mW. That is very close to the maximum of 625 mW but we will go ahead and use it for this circuit since we have no intentions of actually building it.

We are now ready to start calculating. To calculate Rc we use Ohms law. Rc = 6 Volts drop / 0.1 A = 60 Ohms. The power disippated by this resistor would also be 0.6 watts. We would want to use a 1 watt resistor.

Calculating Rb is slightly harder. First we need the current to flow through Rb. We know Ic = Beta * Ib. So, Ib = Ic/Beta. We know Ic and Beta and Ib = 100 mA / 200 = 0.5 mA. Now we need to calculate the voltage to be dropped across Rb. That is 12V-0.65 V = 11.35 V. Now finally Ohms law give us: Rb = 11.35V/0.5 mA = 22,700 Ohms. Power for this resistor would be 11.35 X .0005 or 0.005675 Watts.

To check the calculations and as practice using Qucs I ran a DC simulation on the circuit. This is shown in the second picture. So far so good… or is it? Now is when we come to the interesting part. ** (Note: if you click on a drawing it will enlarge. Use the browser back button to return to this page.)**

When I was assigning values to the various parameters of the model in the last post it seemed as if I was making a lot of assumptions. I was pretty much just slightly better than a wild guess.

What would happen if I am off on my assignment of Vo as 0.65 Volts? I ran two simulations, one with Vo = 0.7 V and one with Vo =0.6 V. I recommend you do these by hand to develop the skills, I used Qucs to provide the pictures. Conclusion: It did not affect things much. The reason why is Rb is doing most of the voltage drop and the current limiting so the tolerance of Vo is not affecting us much. However, if Vcc was much lower this would have a greater effect.

Now we are going to get derailed and things are going to make a fast trip down the sewer. The datasheet had a minimum value of Beta of 100 and a maximum of 300. We split the difference and called it 200 when we created the model. Real life, real devices, can have those values of 100 to 300 so what will that do to our calculations? In the next two simulations I held the values of Rb, Rc and Vo but changed Beta to see what it would do to our circuit currents and voltages.

As you can see things changed a lot. We probably should have expected there would be big changes with a 300% tolerance value in a key parameter. With Beta = 100 we ended up with only 50 mA of Ic current and 9 Volts drop across the transistor. With Beta = 300 we have 150 mA of Ic and only 3 Volts drop across the transistor.

We now have a few choices: Either we buy an whole big bunch of these transistors and check Beta until we find one that suits us or we test Beta before we design the circuit or we design a circuit that is more forgiving. We will choose the last choice.

We will use negative feedback to make the circuit more fogiving of the tolerances and that will be the next post.

However, if you want some examples of negative feedback we have already used it in several posts, but I am not sure I was explicit in using the term. The first example is the resistor Rf in Op-amps. Rf was used between the output and the inverting input to reduce the gain of the Op-amp. Examples are: “The Non-Inverting Op-Amp Amplifier” & “The Inverting Op-Amp Amplifier“.

We also used negative feedback when we talked about a speed controller on an automobile. Even if the control is completely human we keep an eye on the speedometer as we determine how much to depress the accelerator pedal. And if that negative feedback is not enough, there is always policemen with speeding tickets to really provide some negative feedback. An example of the speed controller and a generic negative feedback loop is: “Controlling the Speed of Our Rocket Car“.

Speaking of cars… my guess is this road feels a little bumpy to you about now. Although I showed you the real problem of tolerances, I hopefully have pointed a way out of this mess with negative feedback. We will get into the details in a few up-coming posts. Also, if anyone is concerned. I am focusing only on bipolar junction transistors at the moment. We will eventually get to the field effect transistors. It is best to learn these in detail and then compare and contrast to grow our knowledge.

Gary

“Using the Transistor D.C. Model – Our first circuit.” by Create-and-Make.com is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License.

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