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AC motors
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[quote]QUOTE: <i>Originally posted by flee307</i> <br /><br />AC is a sign wave. DCC is pulsed DC that has a positive and negative component that is phase shifted. It's like a pulsed DC power pack that has pos and neg leads and ground. The pos is +12v from ground, the neg is -12v from ground, and they are 24v apart from each other. But they are still DC. <br />[/quote] <br />Now we get into similarities and differences between these terms, and pulsed DC and AC can look very similar. AC is not necessarily sine wave, any waveform (including rectangular pulses) that is one polarity for 1/2 its cycle and then reverses polarity for the other 1/2 cycle is AC. This is the type of pulses used for LANs (Ethernet etc) and is chosen so it will pass through transformers at computer interfaces without distortion. Pulsed DC alternates between one polarity only, then completely off, and has an average DC value that makes DC motors run in one or the other direction. DC pulses can also be sine shaped (MRC power packs) as well as square. It is also the type of signalling used for digital logic (TTL, CMOS etc) that does not have to pass through transformers. The bridge rectifier in a DCC decoder converts the AC pulses to DC to operate the decoder logic. <br /> <br />But a 50% pulsed DC wave will pass through a capacitor with some distortion, and if no DC offset is added afterwards it becomes AC since it becomes centered about 0 volts and has true positive and negative reversals. If a DC offset is added it stays pulsed DC. If it is not 50% duty cycle, it becomes AC with a DC offset (one polarity is a higher voltage than the other) so in effect it becomes both together. <br /> <br />The main difference between AC and DC is that if you filter pulsed DC with a large capacitor, you will still get a DC voltage at the average value of the pulses. But if you filter AC the same way, you get 0 volts. <br /> <br />If you look at AC or DC pulses on an oscilloscope they can look identical. The only way to know what you are seeing is if you know where ground (0 volts) is in relation to the waveform. <br /> <br />
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