BJT CC Amplifier

For a BJT CC Amplifier, you generally know what the values of your input parameters are such as, voltage,frequency, and impedance. Also, you generally know what the the values of your output parameters are such as, resistance and current. The output voltage is not as important because for a CC, the output voltage follows the input, that's why we call it a voltage/emitter follower. In this example, our input is supposed to mimic a signal generator with a signal thats 1VAC at 1khz. (50 ohms is a typical value for signal generators) With this input, we want to drive a 1k-ohm load at 1mA. There are two parts, we start with the dc analysis and then move onto the ac analysis. DC Analysis: The primarily goal is about establishing the correct operating point to ensure the transistor works in the active region. Start by choosing the voltage divider resistor values so that we have around 5-10 times more current at the output(Emitter), compared to load. The current throught the voltage divider should be at least 10 times greater than the base current. Since our load requires 1mA, we should have either 5mA or 10mA of current at the output, which is the emitter resistor. We choose this equation to find the base and emitter voltages: VE = VB - 0.7 The emitter voltage is 0.7 volts less than the base voltage, so choosing a value like 3.7-volts at the base, would lead to 3-volts at the emitter. Since we know the emitter voltage is 3-volts, and we want 5-10 times of current for the emitter current(5-10mA), compared to the load current which is 1mA,lets calculate the emitter resistance: RE = 3V / 10mA = 300-ohms. Now lets calculate the voltage divider resistor values by choosing a current of 0.5-1mA for the divider to keep power dissipation low. We can use the voltage divider formula to solve for the resistor values: VB = VCC * R1 / (R1 + R2), where VB=3V and VCC=12V. Or we can use KVL and ohms law to get the values of R1 and R2 directly as seen below. R2 = VB / I_divider = 3.7V / 1mA = 3.7k-ohms | R1 = (VCC - VB) / I_divider = (12V - 3.7V) / 1mA = 8.3K-ohms. The BJT is acting as a amplifier, operating in the active region, so we can use this equation: IE = IC + IB = IB*Beta + IB = IB(Beta + 1). Since we know IE, solving for IB we get: IB = IE / (Beta + 1) = 10mA / 101 = 99-micro-amps. Now we need to check if I_divider is greater or equal to IB times 10 using this equation: I_divider >= Ib * 10 = 1mA >= 0.99mA. As you can see, 1mA is greater than 0.99mA, even if you round 0.99mA to 1mA and get 1mA >= 1mA, equation still holds because of the equal sign. If you want to have I_divider be more larger than IB*10, you can scale down the divider resistors to have a I_divider of 1.5mA, instead of 1mA. It's wiser to scale down the divider resistors as opposed to changing the base voltage, because we already calculated some values with our chosen VB. At this point, we are done with DC Analysis, and can move onto the AC Small Signal Analysis.