# bjt transistor circuit finding the resistances

Posted on Feb 7, 2014

I have been given the following circuit diagram and have been asked to compute R2 and RC such that, at the Q-point, the circuit will behave as follows: VCE=5V VBE=0.7V IC=2mA β=100(where β=ic/ib)

There are many places we can begin looking at this. We know that the collector current is 2 mA, and that VCE is 5V. Since the emitter is at ground, the collector must be at 5V, and so the voltage across RC is therefore 10V: the difference between 5V and 15V. So RC must be 10V/.002A=5KΩ. Next, R2 and R1 which span a voltage from −15V to +5V must form a voltage divider such that the top of R1 is at 0.7V (VBE). Hint: the voltage divider spans a range of 20V, and the 0.7V transistor base voltage is 15.7V above the bottom of the voltage divider. This approach assumes that we can ignore the base current because it is small. Often when analyzing transistor circuits we can do that, but not in this case because R1 is such a high resistor. The voltage divider is not "stiff" at all with regard to the resistance in the base circuit of the transistor (which has no emitter resistor at all). A more exact answer requires that we account for the base current. The transistor is carrying only 2mA of current, and so is nowhere near hard saturation, and so the base current is only 0.02 mA, or 20 micro-Amperes (2mA divided by β). Determine how much current is flowing through R1 from its resistance, and 15.7 voltage. The current flowing through R2 is the sum of the transistor base current (.02 mA) and the current through R1. Knowing the current through R2 and the voltage across it, we can calculate its resistance.

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