The diode used in the circuit has a fixed voltage drop of 0.6 V when forward biased. A signal \( v_s \) is given to the ideal OpAmp as shown. When \( v_s \) is at its positive peak, the output \( (v_{OA}) \) of the OpAmp in volts is \(\underline{\hspace{2cm}}\).
The transistor Q1 has a current gain \( \beta_1 = 99 \) and the transistor Q2 has a current gain \( \beta_2 = 49 \). The current \( I_{B2} \) in microampere is \(\underline{\hspace{2cm}}\).
The output \( V_o \) of the ideal OpAmp used in the circuit shown below is 5 V. Then the value of resistor \( R_L \)in kilo ohm (kΩ) is: