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Or the capacitor will just charge to peak and current won't flow ?
This really depends on what your power supply is. Normal voltage source will not cause what you descirbed. A current source surely will.An open LED, say D2 will cause full supply (divided) Vr to D5, D6, D7, D8.
An open LED, say D2 will cause full supply (divided) Vr to D5, D6, D7, D8. If supply voltage is beyond a point, will there be a chance for smoke, as in a mains/capacitive current limited circuit ? Or the capacitor will just charge to peak and current won't flow ?
As in reliability, safer, better design, and the rationale ? All diodes are LED.
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