Editing Basic Circuit Building Blocks
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Latest revision | Your text | ||
Line 115: | Line 115: | ||
*D2 is a diode, it lets current pass only in the direction of the arrow. | *D2 is a diode, it lets current pass only in the direction of the arrow. | ||
*R resistor | *R resistor | ||
− | *Input a source of alternating current ( | + | *Input a source of alternating current ( ocasionally DC in which case the whole circuit serves only to protect against a reverse connection. |
C1 the first, main, filter capacitor. | C1 the first, main, filter capacitor. | ||
− | C2 the second filter | + | C2 the second filter capicator. |
Discussion: | Discussion: | ||
− | In this circuit C1 is a classic filter capacitor it charges while the diode conducts, it discharges and supplies current when the diode does not. R and C2 are a second stage filter. With R set to 0, it simply adds to the value of C1. With R in the | + | In this circuit C1 is a classic filter capacitor it charges while the diode conducts, it discharges and supplies current when the diode does not. R and C2 are a second stage filter. With R set to 0, it simply adds to the value of C1. With R in the circit it forms a low pass filter which helps remove the ripple from the power ( at the cost of some voltage drop ). In the old days R would often be a low value inductor which had a similar effect without the voltage drop. A capacitor alone is often put across a circuit component that uses power to supply bursts of current and stop noise from being propigated through the power supply. |
The amount of ripple in a simple circuit like this can be determined from the supply frequency voltage, output current, and the capacitance. The amount of time without any input voltage is 1/2f. Given an output current I, the charge transferred is is I/2f. The voltage sag is then just the charge divided by the capacitance, or I/2fC. An inductor added to this circuit will compensate for voltage sag by inducing a voltage if the current starts to drop. | The amount of ripple in a simple circuit like this can be determined from the supply frequency voltage, output current, and the capacitance. The amount of time without any input voltage is 1/2f. Given an output current I, the charge transferred is is I/2f. The voltage sag is then just the charge divided by the capacitance, or I/2fC. An inductor added to this circuit will compensate for voltage sag by inducing a voltage if the current starts to drop. |