_{E}configuration as a current source to drive LED as shown in the following figure. In this circuit, laser diode ADL-65055TL from Laser Components was used. Its operating current is 25mA, maximum current is 35mA, and operating voltage is 2.2V.

For a 12V power supply, assuming V

_{CE(saturation)}is approximately zero, R

_{E}is calculated as $$ R_E=\frac{Vcc-V_L}{I_{max}}=\frac{12-2.2}{35m}=280 \Omega $$ I did not want large gain for this CE configuration and bypass capacitor was not used. I wanted the ac output current of this class A amplifier to swing between 23mA and 29 mA. Using the quiescent emitter current for transister 2, I

_{E2}=26mA (near operating current), emitter voltage is, V

_{E2}=R

_{E }x I

_{op}= 280 x 26m = 7.28V. To achieve high input impedance, a CC amplifier was cascaded in front as shown in the following figure. Assuming base-emitter junction voltage, V

_{BE}, as 0.7V, the quiescent base voltage of transister 1 is V

_{B1}= 7.28+0.7+0.7=8.68V. Neglecting the small base current, R

_{1}and R

_{2}are calculated as $$ V_{B1}=Vcc \times \frac{R_2}{R_1+ R_2}$$ $$ \frac{R_1}{R_2}+ 1=\frac{12}{8.68V}$$ Then, R

_{2}=2.6 R

_{1}. If we choose R

_{1}as 22k arbitrarily, R

_{2}will be 57k. The ac input voltage should be (29-23)m x 280= 1.68V peak to peak. The resulting prototype using available components in my lab is shown here.