Electronic – Single Transistor Amplifier Analysis

Finding the peak input voltage that still results in linear output drive is not too hard.

Think of how either Rl is projected back to the base of the transistor, or how Rb is projected to the emitter. You should be able to do that knowing the transistor gain. After that, it's a voltage divider to find what either the drop on Rl or Rb are. Remember that the B-E junction takes away a little from the overal available voltage. Then accounting for the 700 mV B-E drop, you get one from the other.

Once you know the maximum base voltage, you note that everything above applies to the other side, and the P-P voltage is twice the maximum.

Task 1 - find Ib.

We know that VBE is 0.7V (given).{quick check - is this possible with the given values? - without the base connected R2 would produce a drop of about 0.75V - this is a good result because we can always drop this voltage by taking current into the base}.

For VBE = 0.7 and R2 = 6.3k by Ohm's law I(R2) = 0.7/R2 = 0.11111 mA

The 11.3 volt drop (12 - 0.7) across R1 gives us a current of I(R1) = 11.3/93.7k = 0.120598mA

By Kirchoff's law the difference in current must be the base current 0.12 - 0.11 mA (2dp) = 0.01mA

Task 2 - the collector current

We know that current gain is 100 (given)

So Ic = 100Ib = 100 * 0.01mA = 1mA

Task 3 - find the voltage at the collector.

By Ohms law the voltage drop across Rc = Ic*Rc = 1(mA) * 6k = 6V

So the voltage at the collector (with respect to ground is 12 - 6 = 6V