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I am building a circuit like this:
enter image description here
From here.
This got me wondering how much current can I draw from a circuit like this? The question would seem to be what is the wattage of an capacitor?

Edit:
What I am trying to ask is can I draw 100mA or 1A? And how would that affect the components?

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  • \$\begingroup\$ Note that ALL portions of this circuit should ALWAYS be considered to be at full mains voltage for safety purposes when connected to mains. || The main "implication" of Vlad's answer, which he clearly states, is that maximum current is about 32 mA with values shown. He gets 31.63 mA by calculation and 33.6 mA by simulation. I max ~~~~= V/R_x = (Vin-Vout) / Xc. Rx = reactive "resistance" at mains frequency. Xc = capacitor reactance at main frequency. Also take off I x Rseries BUT this is smallish. You get ABOUT 60 mA per microfarad of series cap at 230VAC. Achieving death would be easy. \$\endgroup\$ – Russell McMahon Oct 3 '12 at 9:37
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I'll assume the frequency is 50Hz,as the link says. You have:

\$ (2 \cdot 120 \Omega)+(470k \Omega || 470nF)+(2 \cdot V_{FWD})+(5 \cdot LED) \$.

Let's say the LEDs have \$ V_{FWD}=2V \$

\$ X_C=\frac{1}{2 \pi f C} = \frac{1Meg}{2 \cdot \pi \cdot 50 \cdot 0.47} = 6.77k \Omega \$

\$ Z_{eq} = 470k \Omega || 6.77k \Omega = 6.67 k \Omega \$

The total (RMS) current will be:

\$ I_{total} = \frac{ V_{in} - 2 V_{diode} - 5 V_{LED} }{2 \cdot 120 \Omega + Z_{eq}} = \frac{230-1.4-10}{240+6670} = 31.63 mA \$

Still, due to the large capacitor's reactance compared to the total resistance, you will have a large displacement factor, plus islanding due to the forward drop voltage. The displacement will be (approximately, it doesn't include diode's/LED's resistance):

\$ Z_{tot} = 2 \cdot 120 \Omega + 6.67 k \Omega = 6.91 k \Omega \$

The two 120\$ \Omega \$ are too small so we can leave them aside, therefore leaving us with:

\$ \phi = arctan \frac{R}{X_C} = arctan \frac{470 k\Omega}{6.91 k\Omega} \approx 89^{\circ} \$

Which is almost pure reactive, therefore the power factor will be (considering the simplifications we made) \$ \approx \$ 1.5%.

A quick simulation with the following schematic:

schematic

with the following results:

values

Which are quite close to the calculations, save the power factor who counts the distortions and harmonics, as well.

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  • \$\begingroup\$ I am not sure exactly is the implication of you answer. It has been a good 15 or more years since I have done these kind of calculations. I edited the question to clarify my intent. \$\endgroup\$ – Gerhard Oct 3 '12 at 9:22
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  • Note that ALL portions of this circuit should ALWAYS be considered to be at full mains voltage for safety purposes when connected to mains.

  • Achieving death would be easy.

Vlad's answer clearly states the current achievable using both calculation and simulation. If you cannot follow his explanation then you should be exceedingly careful in trying to apply it and even more careful trying to extend it. This circuit kills.

The main "implication" of Vlad's answer, which he clearly states, is that maximum current is about 32 mA with values shown.
He gets 31.63 mA by calculation and 33.6 mA by simulation.

I max ~~~~= V/R_x = (Vin-Vout) / Xc.
Rx = reactive "resistance" at mains frequency.
Xc = capacitor reactance at main frequency.
Also take off I x Rseries BUT this is smallish.

You get ABOUT 60 mA per microfarad of series cap at 230VAC.
Extending this circuit to supply 1A is trivially easy in theory.
But - Sticking around 20 uF into the mains and expecting to draw low voltage out the other end constitutes an exercise in faith beyond what is reasonable.

If the load goes open circuit the output WILL BE AT MAINS VOLTAGE when the protection zener fails. At the post mortem they will probably established why it failed. Maybe from a mains spike, maybe bad heatsinking, maybe a wiring error, maybe ... .

"Some people who died here ..." as the charming signs at US national park read, will do so without any apparent failure having happened. Some may die because "Phase" was applied to the lower and not upper input in the diagram. The fact that the diagram does not show that phase should be the upper connection and or that there is no attached "this machine kills" notice attached indicates conceptual problems along the way.(ie the notice was never there OR somebody decided it was not worth copying.

Some people who dies here will do so even when phase is connected to the upper lead.

While this circuit can "safely" drive LEDs if "FULLY" isolated at all tines, the relatively low cost of a transformer or switching power supply. and doing it properly makes this a poor choice in most cases.

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    \$\begingroup\$ Russel, you're right, I assumed someone using this kind of circuit to know about the consequences, but the warnings should come first and foremost. I should have to make a habit out of this. \$\endgroup\$ – Vlad Oct 3 '12 at 10:34

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