Charging time of capacitor [closed]

What will be the charging time of the capacitor and how to calculate it?

closed as off-topic by Olin Lathrop, Voltage Spike, Dmitry Grigoryev, Dave Tweed♦Jul 14 '18 at 23:10

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• not enough research. – Marcus Müller Jul 9 '18 at 6:13
• didn't you get the image? – Sender Iot_thingworx Jul 9 '18 at 6:17
• Marcus means there is not enough research by you. You need to show some effort. The red circle is not a standard electronic schematic symbol so we don't know what it does. – Transistor Jul 9 '18 at 6:19
• I saw the image. It's just that "capacitor charging" is really not something that you can't research yourself on the internet; you'll get info about the voltage and current over time. If you don't understand that, it'd be good to ask here - but as you ask it right now, it's just that you're not showing any attempt of your own and expect us to copy wikipedia for you (very dramatically speaking) – Marcus Müller Jul 9 '18 at 6:20
• We don't just do your homework for you here. – Olin Lathrop Jul 9 '18 at 11:25

2 Answers

Both diode properties are available with a right click. But charge time will be almost instantaneous tracking input 40Hz peak since the ESR (diode) ~ a few Ohms with high current thus few hundred microsecond delay dV/dt is small compared to 40Hz

The 63% decay RC time constant is 170*100uF =17ms so depending on the input voltage it may stay illuminated with a half wave rectified sine current with a tail exponential decay then hold at 1.8V the threshold for a Red LED.

You may select the LED and right click add to scope and verify.

Since the diode conducts on negative peaks , the LED Anode (+) is known to be in the same polarity as the diode in series.

• Thank You Tony :D .The input voltage is 0.040v as it is from piezo plates that are connected in parallel. – Sender Iot_thingworx Jul 9 '18 at 6:48

If the diode is a standard silicon PN diode, and the 100$\mu$F capacitor is a standard electrolytic capacitor, and (as you said in a comment) the input voltage is 40mV, then the capacitor will never charge to a voltage that will light an LED.

The diode will never be forward biased so the current through it will be very, very small. The leakage current through the capacitor will probably be as large as the current through the diode, so you will never obtain a significant voltage across the diode.