Russell McMahon wrote: >- All versions of this sort of system can and will kill people. >- All parts of the circuit are potentially at mains potential. >- Component failure of several sorts can change the "potential" to "actual". >- Doing this on a one-off or amateur basis lacks the controls needed to give >you half a chance of not killing someone. Your implication is that amateurs can not competently build safe systems using dangerous voltages. I disagree. I do agree that any discussion about these types of systems should include frequent warnings about possible dangers, but I think it's quite possible for a hobbyist to build completely safe systems this way. To make a system safe, you must ensure that all conductive (metal) surfaces that are exposed or could be exposed through wear, breakage, or curious fingers are grounded. Everything that isn't grounded (which means it isn't and can't be exposed) must be insulated by an insulation that cannot be removed, worn, or damaged, and will not allow such things as fingers, paper clips, etc, to penetrate and reach dangerous voltages. This is usually not a small task, which is what Russel is getting at. However, in the case of something that can be completely enclosed in a rigid plastic or metal box, it is pretty easy. When you add switches and indicators, it gets tricky. For example, your switch or lamp must meet the criteria above EVEN WHEN BROKEN by dropping it or having something dropped on it. If you choose to ground some parts, you must do it so that the ground cannot fail or be broken, etc... I'm sure that others will add to my requirements. ***************************************************************** Olin Lathrop wrote: >> The available current is restricted by the reactance of the capacitor. >> Xc = 1/(2 x pi x f x c) >> >> example for 50Hz mains at 230VAC. >> >> ie. 1uF -> Xc = 1 / (2 x pi x 50 x 1e-6) = 3183 Ohms >> Maximum current = V/Xc = 230/3183 = 72mA > >You can't just model capcitors as resistors that vary as a function of >frequency for this purpose. To do this correctly, you would have to deal >with the complex reactance and frequency. The reactance (which is the imaginary part of a complex impedance) is 3183 ohms. The complex impedance is 0+j3183 ohms. The previous writer was in fact doing exactly what you suggested should be done. >A much simpler way to do this is to compute the charge transferred each >cycle. The current is the charge transferred each cycle times the frequency >of this transfer: > > I = q * f > >where I is the average current, q the charge transferred per power line >cycle, and f the power line frequency. The charge on a capacitor is q = V * >C, where V is the voltage and C the capacitance. > >Assume a simple circuit where you connect one end of a capacitor to the >power line, the other end to ground via two parallel diodes, one in each >direction. In the real circuit if you wanted a small positive supply, the >diode with cathode to ground would actually be a little higher dumping its >current into the unregulated voltage instead. For now let's ignore this >extra voltage drop to see what the current is in the limiting case. We >therefore only care about the current in one of the diodes. In this >configuration, the voltage on the cap swings from about the negative power >line peak voltage to the positive. At 230V RMS, the peaks are >+- sqrt(2)*230 = 325V. Therefore a charge of 650V * C is being transferred >thru the capacitor every power line cycle. So to use the numbers from the >above example: > > I = 650V * 1uF * 50Hz = 32.5mA > >You can now also see that a supply voltage of 5 or 12 volts will have little >effect on the current since that amount will only reduce the 650V value in >the above equation. In other words, this kind of capacitive charge pump >will look like a current source. This implies you probably want to control >the initial unregulated voltage with a shunt. This would likely be a zener >or a zener together with a power transistor. The example above could dump >almost 1/2 watt into a 15V load. I don't know that your way is simpler (that's a matter of opinion), but it is more accurate. The reason for the large discrepancy is that the original poster was proposing using a bridge rectifier to capture both the positive and negative charging cycles on the capacitor. Using your calculations, it should be I = 650V * 1uF * 50Hz * 2 = 65 mA for a bridge rectified circuit. For the more common half-wave rectifier circuits, your calculations are correct. ***************************************************************** Jinx wrote: >"The reactance of a capacitor C is approximately X= 1/(6fC). Under an >rms voltage V, the rms current through the capacitor is I=V/X=6fCV. >For a 230V 50Hz mains input supply, the rms current Irms through a >1uF capacitor would be 6 x 50 x 0.000001 x 230 = 0.069A = 69mA > >With live to one side of C, and the other side of C 1/2-wave rectified >by two diodes + reservoir cap Cr, the Irms through C converts to a mean >DC current of 0.9 x Irms. Output voltage is smoothed by Cr as approx >0.9 x Rload x Irms" The value of 69 mA is not really different, just the result of round-off. If Jinx had used the more precise value of 2*pi = 6.28 instead of rounding off to 6, he would have calculated 72 mA as in the previous messages. His source also points out that this is a calculation of RMS current, which is not exactly what is available in a rectified circuit. The correction factor of 0.9 quoted here is the difference between 72 and 65 mA. 72 mArms * 0.9 = 65 mAavg Jinx's source fails to note that this amount of current is available only if you bridge rectify the input. Using the half-wave rectifier they propose, the correction factor should be 0.45 (which is half of 0.9). Now everybody's numbers will agree: 230V * sqrt(2) * 1uF * 50Hz = 32.5 mA or 230V / (1/ (2 * pi * 1uF * 50Hz) )* 0.45 = 32.5 mA If you look carefully, you will see that these are in fact the same equations, but with the constants gathered together differently. We have arrived at the same conclusion through two different paths. ***************************************************************** Now I write: What's the drawback to the bridge-rectified circuit? It seems offer twice the current for the same capacitor value. The bridge-rectified circuit truly does offer twice the current. Its major drawback is that the output voltage is not referenced to either of the incoming lines. In some cases this is not a problem, while in others it is. In phase-control work, where you need to fire a triac, the gate voltage should (generally) be referenced to the neutral side. Thus the half-wave rectified circuit is more appropriate in this case. The half-wave circuit is also marginally safer in applications where the output voltages can be referenced to mains neutral instead of hot. The bridge rectified circuit voltages are referenced to neutral for half the cycle, then to hot for the other half. I hope this clears up a few things! Don -- http://www.piclist.com hint: PICList Posts must start with ONE topic: "[PIC]:" PIC only "[EE]:" engineering "[OT]:" off topic "[AD]:" ad's