# Capacitor charging time in a circuit which has resistor in its output side.

#### san514

Joined Jul 9, 2018
5

Here the input voltage is 0.040V and current is 0.013A what will be the charging time of the capacitor?

[There is a diode at the input side.The red circle is LED and it has a switch at the end.]

#### crutschow

Joined Mar 14, 2008
27,179
You can answer that yourself.
What is the normal forward drop voltage of the diode?

#### ericgibbs

Joined Jan 29, 2010
12,901
hi san
Welcome to AAC.
Post the charging time you calculate.? we can then check your work.
E

#### WBahn

Joined Mar 31, 2012
26,398
View attachment 155871

Here the input voltage is 0.040V and current is 0.013A what will be the charging time of the capacitor?

[There is a diode at the input side.The red circle is LED and it has a switch at the end.]
Is the input voltage 40 mV rms? or peak? or peak-to-peak?

Where is this 13 mA coming from? If the source is a 40 mV, then it is establishing the voltage across it. The current will be whatever the current is.

#### MrAl

Joined Jun 17, 2014
8,239
View attachment 155871

Here the input voltage is 0.040V and current is 0.013A what will be the charging time of the capacitor?

[There is a diode at the input side.The red circle is LED and it has a switch at the end.]
Hello,

Are you really saying the input voltage is 0.040 volts? It doesnt matter if that is RMS or Peak it is not high enough to make the diode work as a diode, and even if the diode did work as a diode the LED would not work because the LED forward voltage is far too high for that (1.5 to 3.5 volts approximately).
Thus there is no current flow.

Did you mean maybe 4v or even 40v or something?
I think you should double check that first.

#### san514

Joined Jul 9, 2018
5
Hello,

Are you really saying the input voltage is 0.040 volts? It doesnt matter if that is RMS or Peak it is not high enough to make the diode work as a diode, and even if the diode did work as a diode the LED would not work because the LED forward voltage is far too high for that (1.5 to 3.5 volts approximately).
Thus there is no current flow.

Did you mean maybe 4v or even 40v or something?
I think you should double check that first.
It's 4v

#### san514

Joined Jul 9, 2018
5
Is the input voltage 40 mV rms? or peak? or peak-to-peak?

Where is this 13 mA coming from? If the source is a 40 mV, then it is establishing the voltage across it. The current will be whatever the current is.
Its 4v exactly.sorry posted it wrong by mistake.

#### san514

Joined Jul 9, 2018
5
Hello,

Are you really saying the input voltage is 0.040 volts? It doesnt matter if that is RMS or Peak it is not high enough to make the diode work as a diode, and even if the diode did work as a diode the LED would not work because the LED forward voltage is far too high for that (1.5 to 3.5 volts approximately).
Thus there is no current flow.

Did you mean maybe 4v or even 40v or something?
I think you should double check that first.
4v

#### ericgibbs

Joined Jan 29, 2010
12,901
hi san,
Your duplicated Thread in the Projects Forum, on the same Topic, has been deleted.
Please do not Create duplicates.

MOD.

#### san514

Joined Jul 9, 2018
5
The charging time is 0.017sec

#### MrAl

Joined Jun 17, 2014
8,239
The charging time is 0.017sec
Hi,

How did you calculate that?

The charge time is related to the line frequency, i wont say anything more just yet.