Capacitor charging

[Peter Bullock]

If you connect a capacitor to a battery without a resistor how quickly does it charge ?

Pete.

 

[Bluejets]

Find the circuit resistance and use that to calculate Time.
If I remember correctly 1/5T is normally used for most equations because of the nature of the charge curve.

 

[(*steve*)]

Included in the circuit resistance will be the internal resistance of your power supply. This internal resistance works a bit like a fixed resistance n series with a voltage source.

The time constant (R x C) it the time taken for the capacitor to reach about 60% of the source voltage. After another time constant it's 84% (that's 60% + 60% of 40%). Another time constant takes it to almost 94%. For most practical purposes, after 5 time constants it's all over.

That "about 60%" is actually 63.2%. After 5 time constants the voltage on the capacitor is greater than 99% of the final value.

 

[Peter Bullock]

Thanks for the help. I have another question regarding capacitors. In my textbook there is a circuit using two 555 circuits. The first circuit is a monostable. The output from the first 555 is fed through a capacitor and then into the trigger of the second 555. In my textbook it says: The capacitor linking the output of the first 555 to the trigger of the second is essential. It ensures that only the change from high to low gets passed through - otherwise the second time period would never stop. Can someone shed some light on this for me ?

 

[Bluejets]

Timing on the 555 starts with a pulse on 2 which has to be shorter than the output.

 

[Hopup]

Capacitor is not essential at least I have built working circuits with 2 555's without it.

 

[Audioguru]

Signetics invented the 555. Then Philips bought Signetics. Here is the article about why a coupling capacitor is required:

 

[AnalogKid]

A 555 monostable circuit is not a "true" monostable because it does not have positive feedback to make the output pulse width independent of the input state after it is triggered. It also is not a normal retriggerable monostable, because if it receives a 2nd trigger input before the first timing period is finished, the timing period does not immediately restart. IOW, the 555 is weird.

If the trigger input is held low longer than the monostable pulse width, the output stays high until the trigger input goes high. The coupling capacitor forms a differentiator circuit that "extracts" the leading negative edge from whatever is driving the trigger input. Shortly after the driving signal goes low, the trigger input goes high even though the driving signal still is low. Now when the 555 times out, the output goes low at the correct time.

ak