| 摘要 |
PROBLEM TO BE SOLVED: To realize efficient and speedy charging by a simple and economical method. SOLUTION: When a transistor Tr1 goes into an OFF state, voltage is impressed to the positive terminal of an amplifier AMP1, which raises the output voltage and turns on a FET Tr3. This causes charging current I1 to flow through a resistor R18, creating a potential difference across the resistor. This potential difference, detected by an amplifier AMP2, raises the output voltage of the amplifier. When the voltage of the negative terminal of the amplifier AMP1 becomes larger than that of the positive terminal, the output voltage decreases to 0V, which turns off the FET Tr3. This prevents the charging current I1 from flowing, which decreases the output voltage of the amplifier AMP2. Consequently, when the voltage of the negative terminal of the amplifier AMP1 becomes smaller than that of the positive terminal, the output voltage rises again, which turns on the FET Tr3. A capacitor C is charged by the repetition of this process. |
