Applications of Semiconductor Devices — Practice Quiz

A_V = –β × R_L / r_i

Step 1: In CE amplifier, the change in collector current is Δi_c = β × Δi_b. Step 2: Output voltage change is Δv_0 = –Δi_c × R_L (negative because increase in collector current reduces V_CE). Step 3: Input voltage v_s = Δi_b × r_i, so A_V = v_0/v_s = –β × R_L / r_i. The negative sign indicates 180° phase shift between input and output. β × r_i / R_L reverses the fraction. R_L / r_i misses the β factor.

200

Step 1: Use the formula |A_V| = β × R_L / r_i. Step 2: Substitute values: |A_V| = 50 × 2000 / 500. Step 3: |A_V| = 100000 / 500 = 200. So the voltage gain magnitude is 200. Option 25 results from dividing β by R_L/r_i instead of multiplying. Option 100 is half the correct value (common error of using R_L/r_i without β). Option 50 is just the value of β.

NOR gate

Step 1: NOR = NOT + OR. Step 2: OR gate gives '1' when at least one input is '1'. Step 3: Inverting the OR output: NOR gives '1' only when OR output is '0', which happens only when BOTH inputs are '0'. AND gate gives '1' only when both inputs are '1'. OR gate gives '1' when at least one input is '1'. NAND gate gives '0' only when both inputs are '1' (i.e., output is '1' for all other combinations).

Transistor is cut off; V_0 = V_CC

Step 1: When V_BB = 0, the base current I_B = –V_BE / R_B, which is negative (no forward bias). Step 2: Since I_B < 0, the transistor does not conduct and is in the CUT-OFF region. Step 3: Since no collector current flows, there is no voltage drop across R_L, so the entire supply voltage V_CC appears at the output. This is like an open switch – no current means full voltage at output. Saturation occurs only when V_BB is high enough to drive sufficient base current.