32

A pretty thorough debunking of these ‘power saving’ devices can be found here: http://www.nlcpr.com/Deceptions1.php Besides this, the box appears to be shoddily made and has no safety approvals. You should unplug it at once, it’s a hazard. Like many scams, these boxes are based on a small kernel of truth. And that kernel is, applying a capacitor across the ...


32

All I find is this giant capacitor. Is this literally a capacitor with a switch that turns on a light? Would this capacitor provide any surge protection in either direction? Wondering if it might prevent the dryer from creating a surge? This device is NOT a surge protector. While it may have been advertised as such on Amazon (there is not really any ...


17

Short answer: an ideal battery has constant voltage \$U_{nom}\$ (nominal voltage) until it's empty, so energy stored is: \begin{aligned} E &= U_{nom} \times I \times T\\ &= U_{nom} \times \text{capacity} \end{aligned} an ideal capacitor has voltage proportional to charge, so it decreases linearly to 0: \begin{aligned} E &= \int_0^T U(t)I\ dt\\ ...


5

It's for mains power factor correction, which is what the capacitor is doing across the supply. The box is labelled as such. Correcting power factor is not a scam but a single box that can correct it for all loads is. In correcting the power factor, it is attempting to get as much as possible of the power drawn from the supply (apparent power) to be ...


3

When it comes to voltage dependence of capacitance, class II ceramic capacitors are considered exceptionally bad, while tantalum, polymer and class I ceramics are considered very stable (source 1, source 2). I won't comment on the other types since I couldn't find authoritative sources, but I believe nothing is remotely as bad as class II ceramics. Class II ...


3

The 5V from the USB should be stable. I'm not sure why you think you need a boost to be honest. Even if you did need 5V, I thought you would want a SEPIC or buck-boost or something that can bring both higher and lower voltages to 5V. Is the boost converter only there to make up for the Schottky diode's voltage? If so, don't use a Schottky; Use a PMOS circuit....


3

The difference is that a perfect battery is a constant voltage source. If you discharge the battery the voltage will not change until it will suddenly drop to zero when it is completely discharged. A capacitor, even an ideal one is not. If you discharge a capacitor, the voltage will drop along the way to zero when all energy is extracted. To get the energy ...


2

Yes, having a 22 uF capacitor will violate the specs that say max 10 uF. However, if you limit the inrush energy to below the allowed energy vs time curve, then you can fill a larger capacitor slowly. But you will be instantly powering the whole board and all the capacitors via schottky diodes, all while charging a battery, so total surge needs to be taken ...


2

if charge injected by S1 is q1, and input capacitance of opamp is ignored, the voltage at X and P jump to infinity. I did not quite understand how this occurs. Can you help me understand? The formula for voltage produced by charge q on a capacitor of value C is V = q/C. If C = 0 then any charge injected into the capacitor (op amp input) will produce ...


2

input capacitance of opamp is ignored, the voltage at X and P jump to infinity For an ideal Op Amp... If \$ Z_{in-} = ∞\$ and \$I_C=C\cdot dV/dt = V_{in-}/Z_{in-}\$ then \$ V_{In-} = Z_{in-}\cdot CdV/dt= ∞\$


2

Important difference: The voltage of capacitor rises when current is pushed into it and voltage drops when current is drawn from it. So the voltage is not constant as it is proportional to the charge stored in the capacitor. Q=I×t=C×U. Batteries can be approximated to have constant voltage when current is drawn out or pushed in. You don't draw batteries ...


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