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| Line Transformer (L1) Selection
My primary consideration here was that I did not want to use my house electric service for fear of RF energy backfeeding and destroying delicate computer, audio, video, telephone, and alarm equipment. Fortunately, since I live in the boonies and have frequent weather related power outages, I own a 5KW generator which I can use for totally isolated power. Not knowing much about Power Factor, I will equate 5KVA with 5KW to determine the minimum rating for the power transformer (L1). Rectifier (D1) Selection For current rating of the rectifier we can use either of two methods: First, we can calculate the turns ratio on the transformer: 13,800/220 = 62.73. This means that the secondary voltage will be 62.73 times the primary voltage. It also means that the secondary current will be primary current/62.73 (yes, I know that the transformer is not 100% efficient). I=W/V so the generator will deliver 5000/220 or 22.7 Amps. This current divided by the turns ratio (22.7 / 62.73 = .362 amps) yields 362 ma of output current. Or we can take our KVA and divide by the output voltage (5000 / 13,800 = .362) or 362 MA. This is the amount we can safely draw from the pole pig without overloading the pole transformer or the generator. Now onto
the voltage rating: We know that the transformer output is 13,800 Volts AC. Since it is not stated otherwise, this is RMS voltage. To convert to peak voltage we simply multiply by 1.414 which gives us 19,513 volts Peak. So you see Capacitor C1 will charge to almost 20,000 volts. This means that, at the bottom of the sine wave (0 Volt point), there will be a reverse voltage of almost 20,000 volts across D1. Now we have our absolute minimum ratings for the diode bridge: 362 ma 20,000 volts. This is the same as 362 ma and 10,000 per internal diode. Allowing for what I consider minimum safety margin, I would look for a 1 Amp 40,000 volt bridge rectifier.
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-------------------------- More Later --------------------------- |
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