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Chapter 3: Diode Approximations There are three (3) Diodes Approximations: • • • Ideal (1st-Approximation) Second approximation Third Approximation • For the 1st-approx. assume the diode drop voltage is zero (Perfect closed switch) For the 2nd-approx. assume the diode drop voltage of 0.7 volts For the 3rd –approx. assume the diode drop voltage of 0.7 volt and consider the forward bulk resistance of the diode: Vd = 0.7 V + Id x Rb (3-5) Ignoire bulk resistance of the diode if Rb < 0.01 Rth • • • • Forward Bulk Resistance Calculate the Bulk Resistance (Rb) from the formula given below: Rb = (V2 – V1)/(I2 – I1) Forward bulk resistance (Rb) is very small less than 1 Ohms or 10 Ohms. The bulk resistance is the dynamic resistance while current pass through the diode, but static resistance that usually measured by meter is relatively small compare to the bulk resistance. The reverse resistance of the diode is very large greater than 1 M-ohms. Example of 2nd-approximation Thevenin’s Equvalect Circuit Second Approximation Example: In this Example the diode is forward-biased to act as a closed switch to turn ON the 5 V @ 5W Light Bulb. Calculat the Lamp current and compare that to the measured lamp current by MultiSim. The load current can be calculated by using the 2nd-approximation and Ohms' Law: Since this is a complex circuit (more than one loop) we must apply Thevenin's Theorem first to convert that to a single loop: Vth = [6/ (10+6)] (18V) = 6.75 V Rth = 6|| 10 = (6 x 10)/(6+10) = 3.75 Ohms Lamp Resistance (RL) can be calculated from: IL = P/V = 1W / 5V = 0.2 A IL = 0.2 A = 200 mA ===> Lamp resistance (RL) = VL/RL = 5V/0.2 A = 25 Ohms From the Thevenin's Equivalent circuit diode current (load current) can be calculated by Ohm's Law: Id = IL = (Vth - Vd)/(Rth + RL) = (6.75V - 0.7 V) / ( 3.75 +25) Id = IL= 6.05 / 28.75 = 0.21 A = 210 mA (calculated Lamp current) IL(Measured) = 0.207 A = 207 mA ====> They are very close !