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If 60Hz, the 470Ω resistor plus the 5643Ω impedance of the 0.47μF results in 6113Ω total opposition to alternating current.
Calculate as if it was only a 6113Ω resistor in the circuit.
The unmarked capacitor to the right also presents a impedance.
Both in series to power applied at S1 and S2 does the division as if it were resistors.
Calculate as if it was only a 6113Ω resistor in the circuit.
The unmarked capacitor to the right also presents a impedance.
Both in series to power applied at S1 and S2 does the division as if it were resistors.
There is no way of determining the input voltage, it could be 240V and 50Hz or something very different.
The CR circuit will control the current (see #2) and the Zeners will clip the voltage to approximately +/-15V peak, passing excess current.
It's been about 50 years since I learned about calculating total impedance(Z) so I might be wrong, but I'm pretty sure you cannot simply add resistance and capacitive reactance. Rather you must use vector summation that accounts for the lead factor in capacitors (or lag factors in inductors). By my very rusty calculations the vector summation of a 470 ohm resistor and a 5643 ohm capacitor reactance is 5662 ohms.
Frank
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That is correct, but doesn't get you any closer to determining what the AC applied voltage and frequency is since you don't know how much current is allowed in the zener.
Certainly. But it is unsafe to do so to provide a low voltage DC from the mains supply. Better to spend a few bux for a transformer, or use a transformer-isolated, high-frequency switching, power supply. Component failures in the type of supply you posted are typically accompanied by exploding capacitors, melted conductors and fire.
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Not to forget deadly elecrical shock to persons!
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