Diode Circuit Analysis Problems And Solutions Pdf

RL(min)=VZIL(max)=12 V9 mA≈1.33 kΩcap R sub cap L open paren min close paren end-sub equals the fraction with numerator cap V sub cap Z and denominator cap I sub cap L open paren max close paren end-sub end-fraction equals the fraction with numerator 12 V and denominator 9 mA end-fraction is approximately equal to 1.33 k cap omega If the load resistance drops below , the load will draw more than , forcing the Zener current below IZKcap I sub cap Z cap K end-sub

A refined version that accounts for the real forward voltage drop. In this model: diode circuit analysis problems and solutions pdf

Three diodes (D1, D2, D3) are connected as follows: Anode of D1 to +5V, cathode of D1 to node X. Anode of D2 to node X, cathode of D2 to -5V through a 1kΩ resistor. Anode of D3 to node X, cathode of D3 to output (V_o). A load resistor 2kΩ from (V_o) to ground. All diodes are silicon (0.7V drop). Find (V_o). RL(min)=VZIL(max)=12 V9 mA≈1

. The maximum current available for the load occurs when the Zener draws its bare minimum current: Anode of D3 to node X, cathode of D3 to output (V_o)

IS=VS−VZRS=24 V−12 V1.2 kΩ=12 V1200 Ω=10 mAcap I sub cap S equals the fraction with numerator cap V sub cap S minus cap V sub cap Z and denominator cap R sub cap S end-fraction equals the fraction with numerator 24 V minus 12 V and denominator 1.2 k cap omega end-fraction equals the fraction with numerator 12 V and denominator 1200 cap omega end-fraction equals 10 mA Write the KCL equation at the output node: