: Treated as a perfect switch. It is a short circuit (0V drop) when forward-biased ( ) and an open circuit (0A current) when reverse-biased. Constant Voltage Drop (CVD) Model
VL=VZ=5.1 Vcap V sub cap L equals cap V sub cap Z equals 5.1 V Clipping and Clamping Circuits
Replace the diode with its equivalent circuit (e.g., a 0.7V source for ON, an open break for OFF). diode circuit analysis problems and solutions pdf
source pushes current into the anode. Assume the diode is . Step 2 (Substitute Model): Replace the Silicon diode with a DC source opposing the main battery. Step 3 (Apply KVL): Loop equation around the circuit:
Diode: V_F = 0.7V, r_d = 10Ω. Source 10V, R = 1kΩ. : Treated as a perfect switch
: Use KVL and KCL to solve for the diode current ( IDcap I sub cap D ) and diode voltage ( VDcap V sub cap D Verify assumptions : If you assumed ON : The calculated IDcap I sub cap D must be positive (flowing from anode to cathode). If you assumed OFF : The calculated VDcap V sub cap D must be less than the turn-on voltage (e.g.,
[ r_d = \fracnV_TI_D = \frac1 \times 25mV1mA = 25\Omega ] The AC output voltage is found by treating (r_d) as a resistor in a voltage divider. source pushes current into the anode
and a Silicon diode. The diode anode faces the positive terminal of the source. Find the circuit current and the output voltage Voutcap V sub o u t end-sub across the resistor. The positive terminal of the
Acts as a constant voltage source equal to the turn-on voltage ( Vγcap V sub gamma VDcap V sub cap D ). For Silicon, this is typically . For Germanium, it is Reverse Bias: Acts as an open circuit ( Piecewise Linear (PWL) Model
ID=VRR=9.3 V1 kΩ=9.3 mAcap I sub cap D equals the fraction with numerator cap V sub cap R and denominator cap R end-fraction equals the fraction with numerator 9.3 V and denominator 1 k cap omega end-fraction equals 9.3 mA , which is greater than . The assumption is Valid . Problem 2: Two-Diode Logic Circuit (Ideal Model) Circuit: Two ideal diodes, D1cap D sub 1 D2cap D sub 2 , have their anodes connected to a shared node via a resistor. The cathode of D1cap D sub 1 is connected to (GND). The cathode of D2cap D sub 2 is connected to . Find the output voltage ( Voutcap V sub o u t end-sub ) at the shared anode node. Step 1 (Assumption): D1cap D sub 1 sees a larger potential difference than D2cap D sub 2 D1cap D sub 1 is ON and D2cap D sub 2 is OFF . Step 2 (Model): Replace D1cap D sub 1 with a short circuit and D2cap D sub 2 with an open circuit. Step 3 (Analyze): Because D1cap D sub 1
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