Question: there 30V RMS (Vpeak = 42V), 2 A across the SCR. To control the firing angle of that SCR I am using Phase Control - Diode, Resistor, Potentiometer, Capacitor. If the trigger voltage for the SCR is 0.65V and the holding current is 9mA, what are the formulas to calculate Rt (total resistance of the Firing Circuit) and C needed in order to control the firing angle of the SCR between 30 and 150 degrees? Thank you for reading my messege.
Sorry -- mised the diagram the first time. Lets start with the basics. The output of the transformer will be 60V RMS or 84.84 Volts Peak. The sin(30) = 0.5 and the sin(150) = 0.5. With pots R2 or R4 at one end of their travel, C1 and C2 will be charged through R1 or R3. You did not label the two diodes but we'll call them D1 and D2. Assume they have a forward drop of 0.65 Volts. So to fire the SCR we need (0.65V + 0.65V) on C1 or C2 and we want this to happen at 30 degrees. Code ( (Unknown Language)): 84.84 * sin(30) = 84.84 * 0.5 = 42.42 42.42 - (0.65 + 0.65) = 42.42 - 1.3 = 41.12 This will happen if the RC time constant is about 0.1 times the time equivalent to pi/6 or 30 degrees. This is 1.388 mseconds. So Code ( (Unknown Language)): RC = 0.1 * 1.388e-3 = 139 microseconds So 1 uF and 139 ohms should do the job, this might need to be a 1W resistor Any resistance added by R2 will delay the time it takes to get the capacitor up to the firing voltage of 1.3 volts. This analysis is an approximation so don't hold me to it. I'd make R2 a 5K linear taper pot.