Capacitor Charging Time: Calculating 5uF Capacitor Full Charge with 1V Power Supply

Hello, I have the above-mentioned problem
I have a 1V power supply and the C capacity is 5uF. What's the full charge time?

You still need to know the resistance through which it is charging, then the charging time is:

2.3 * R * C to 0.9 * Ucc

4.6 * R * C to 0.99 * Ucc

6.9 * R * C to 0.999 * Ucc

The general formula for the voltage of a capacitor charging from 0 to Ucc is as follows:

$$Uc=Ucc*(1-e^{\frac{t}{R*C}})$$

t - time

On the one hand, aiming at 0 and on the other hand, towards ?.
What do you understand "What's the full charge time?"
It is never full (? and since you did not mention that it would be charged by some R, it is close to 0 to the state when it is considered charged (but it is not fully charged but about 99.99%) for an ideal capacitor.
Rather, you mean the RC time constant read:
http://aneksy.pwn.pl/podstawy_fizyki/pdf/28.14.pdf

oh yeah .... I just had a physics exercise about RC and RLC .... thanks for the reminder, actually it's kind of stupid that I forgot about this simple and obvious formula.

thank you all for help .....


I hope that I will not ask such stupid questions anymore: /

There are no stupid questions, there are only stupid answers. Anyone can forget even seemingly basic things.

I remember that the capacitor constant in the experiment was about 66%, which is according to the assumptions ..

and I forgot the pattern ... eh ...

And one more question: why if I hook the led to a 1mF capacitor, it lights up brightly, after a fraction of a second it gets darker, and then slowly goes out ...
what to do to make it go out evenly (some special capacitor?)

mat17a wrote:

I have a 1V power supply and the C capacity is 5uF. What's the full charge time?

Of course, if we charge a capacitor through resistances, we will get the steady state (in our case, the charging of the capacitor) after about t = 5 * R * C. If, however, it is assumed that there is no resistance in the circuit, the charging time of the capacitor is infinitely short. (In practice, "very" short resistance always exists - eg of capacitor legs). The laws of commutation state that the voltage across the capacitance must be a continuous function. If this is not the case (we have 0 V and then in one moment we give e.g. 1 V - disregarding the resistance on which the supply voltage could be deposited in the initial moment, we force a voltage change to C). The current flowing in the circuit from C, being the derivative of the voltage over time (rate of change i = dUc / dt), reaches chile (infinitely short in theory) huge values (infinitely large in theory). The functions describing the behavior of the voltage and current C over time are no longer the exponents but the unit pulse (for voltage) and the Dirac pulse (also called the Dirac delta). Dirac delta charging of the electrolyte, eg 1000 uF, may "weld" its pin with the power supply cable. In serious applications, it is sometimes undesirable, e.g. due to the limited current strength of the diodes. In such a case, chokes must be used in series with C.

mat17a wrote:

why if I hook the led to a 1mF capacitor, it lights up brightly, after a fraction of a second it gets darker, and then slowly goes out ...
what to do to make it go out evenly (some special capacitor?)

Put a resistor between the diode and the capacitor, e.g. 220? -1k?.
Without a resistor, you will damage the diode very quickly!


The current through the LED depends exponentially (function $$e^x$$ )
from the voltage on it, so at the beginning a very large current flows limited only by the resistance of the diode structure and the lead wire inside it. The capacitor discharges very quickly until the structure resistance (U / I) begins to outweigh the resistance of the leads. Then, when the voltage on the capacitor drops below the diode forward voltage (a different value depending on the diode type), the diode turns off.

so as not to post a new topic when searching, this will refresh this. I have a question, what does the charging and discharging time of the capacitor depend on?

It depends on the charging or discharging current. The current may come from a current source, then the voltage across the capacitor changes linearly:

I * t = U * C

$$U(t)=Uc_0+\frac{I}{C}*t$$

Uc0 - initial voltage across the capacitor
I - charge or discharge current (then we give it with a minus)
t - time in seconds

The second case is charging and discharging through a resistor, then the charging / discharging current depends on the instantaneous voltage difference across the resistor, which causes the voltage across the capacitor to change exponentially.