Timeline for Trouble understanding charging capacitor on bridge rectifier
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2016 at 23:00 | comment | added | Anthony X | Consider that in real life, the AC source to the bridge rectifier is typically a transformer. The transformer windings have resistance which will cause the input voltage to drop with increased load (e.g. capacitor inrush). Also, the AC source to the transformer will also be imperfect and cause a drop in voltage. Real-world example: older tube TVs. Turn one on and you may see lights on the same circuit dim momentarily due to inrush currents inducing voltage drops on the supply side of power transformers. | |
| Oct 15, 2016 at 21:50 | comment | added | João Pedro | Thanks alephzero. The thing I'm having trouble understanding right now is this: If I have Vi=10V, 0V diode forward drop, how could I have a peak voltage in Vo less than 10V? Assuming everything ideal ofc... | |
| Oct 15, 2016 at 21:44 | comment | added | alephzero | Don't forget that iIn real life, if C is very large the "inrush current" charging up C in the very first half-cycle after switch-on can be big enough to fry the diodes! | |
| Oct 15, 2016 at 21:09 | history | edited | João Pedro | CC BY-SA 3.0 |
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| Oct 15, 2016 at 20:33 | comment | added | João Pedro | Thanks guys, that makes sense now. If possible I would like to ask something else. Supposing a forward voltage drop of 0.6 in each diode, the peak voltage on Vo would be V1 - 0.6 - 0.6 = V1 - 1.2. If diodes are considered as ideal with 0V drop in forward bias, then Vo = V1 in this case. There is one exercise here made in my classroom where the peak voltage in Vo is different V1, assuming all diodes ideal with 0V drop, ideal sources, etc. Does it make sense? Could the capacitor not be fully charged by the end of a half-cycle? Thanks again. | |
| Oct 15, 2016 at 19:59 | comment | added | Russell McMahon♦ | Add to what they said: as Vmains rises above Vcap and diode starts to conduct Icap ~= V/r = (Vmains-Vcap) / (Rmains + Rwiring + ESR cap + ) = very large current peaks. Adding a small sreading resistor on series with HV out of bridge or in mains input helps. Without this in real world RFI peaks occur due to diode pulse currents. | |
| Oct 15, 2016 at 19:55 | answer | added | D.A.S. | timeline score: 1 | |
| Oct 15, 2016 at 18:37 | comment | added | FiddyOhm | You should also understand that "charges with the value of V1" means the uni-directional peak value of the sine wave minus two forward diode drops. (Many books will tell you that the forward drop of an ideal diode is 0 volts; in My Book the forward voltage drop of an ideal diode is 0.6 volts, which is a lot closer to the ideal real world than 0 volts.) Uni-directional peak value for a perfect sine wave input (most line voltages are not) would be 1/2 the peak-to-peak voltage you see on an O-scope and definitely not the RMS voltage, which is what you read on a DVM connected to the AC source. | |
| Oct 15, 2016 at 18:31 | answer | added | EM Fields | timeline score: 4 | |
| Oct 15, 2016 at 18:22 | comment | added | user16324 | There is still a time constant, and the C is still part of it, but the R isn't. Instead there is a different (usually much smaller) R composed of the sine wave source impedance, the forward resistance of the diodes, and the interconnecting wire. Whether that's "instantaneous" or not depends on the details. | |
| Oct 15, 2016 at 18:09 | history | asked | João Pedro | CC BY-SA 3.0 |