This model helps in the design of simple DC power supplies. In the example variable sheet below, the model is being used to predict the average dc output voltage Vdc and the ripple voltage that would be obtained with a specified transformer secondary voltage Vrms, half-wave rectifier, filter capacitor C, and load current Idc. Set wave = 0.5 for a half-wave rectifier, or wave=1 for a full-wave (bridge) rectifier. The output voltage regulation can be determined by calculating Vdc as a function of Idc). (Note: The equations used in this model are approximate and are reliable only for small ripple factors < 0.2. The model assumes a constant voltage drop across the rectifier diodes of 0.6 volts per diode).
Variable Sheet _________________________________________________________________________________ St Input Name Output Unit Comment Vdc 10.206667 volts average dc output voltage (V,mV) 10 Vrms V rms transformer sec voltage (V,mV) .5 wave .5 for half-wave, 1 for full-wave rectif 40 Idc mA average output current (A,mA,µA) 60 f Hz Line frequency (Hz) 100 C µF filter capacitance (F, µF) r .18855332 ripple factor (unitless ratio) Iac 7.5421328 mA rms value of ripple (A,mA,µA) RL 255.16667 Ohms load resistance (Ohms, KOhms) Vripple 1924.5009 mV rms ripple voltage (V,mV) PL .40826667 watts Power dissipated in load resistance (wat Rule Sheet _________________________________________________________________________________ S Rule Vdc=1.414*Vrms-Idc/(4*wave*f*C)-1.2*wave "Approx eqn for average output voltage Iac=r*Idc "definition of ripple factor (current) r=1/(4*sqrt(3)*wave*f*C*RL) "Approximate equation for ripple factor Idc=Vdc/RL "Ohms Law or load resistance Vripple=r*Vdc "definition of ripple factor (voltage) PL=Vdc*Idc "Joules Law for load resistance