novice capacitance question?

[alanzhao]

here is the simple diagram

--------||+------+|(--------

I have a 3v battery and 10uF capacitor connected in series, I want to know

1) the output voltage?
2) the time it takes to charge and discharge the capacitor?

I am new to electronics, please help.

Thanks!

 

[bmcculla]

You need to make sure you have a current path even with capacitors. So as it stands the circuit doesn't make any sense. If you have the two unconnected leads connected together and consider the + side of the cap as the output then:
1) The output voltage will be 3V after a small amount of time.
2) The time it takes to charge the cap depends on the internal resistance of the battery. This resistance isn't very predicatable; it depends on temperature the amount of current you are drawing and the how charged tha battery is among other things. To get a predictable charge time you need to add a resistor in series with the cap.

Hope this helps
Brent

 

[alanzhao]

sorry, I didn't make the circuit connected, two leads are actually connected.

Questions:
1) Why doesn't the capacitor increase the output voltage as it charges and stores voltages? you said output voltage will be 3v after a samll amount of time? how do I increase the voltage?

2) what if I add a 100ohm resistor in between the battery and capacitor, and the result? How do I apply T=RC? How should I pick a resistor value?

Sorry for bunch of questions

Thanks!

 

[Nigel Goodwin]

Questions:
1) Why doesn't the capacitor increase the output voltage as it charges and stores voltages? you said output voltage will be 3v after a samll amount of time? how do I increase the voltage?

The capacitor can't charge higher than the supply voltage. To increase the voltage, increase the battery voltage.

2) what if I add a 100ohm resistor in between the battery and capacitor, and the result? How do I apply T=RC? How should I pick a resistor value?

That would slow the charging time, T is in seconds, R in ohms, and C in Farads. Without an external resistor, R is the internal resistance of the battery.

 

[alanzhao]

but how can a 3v charged capacitor used to run a 6v application, for example?

 

[Nigel Goodwin]

but how can a 3v charged capacitor used to run a 6v application, for example?

It can't!.

To run a 6V application off a 3V source, you would need to generate a 6V supply. You can do this in various ways, a switch-mode supply or a charge pump (for low power) are two common ways.

 

[alanzhao]

I found this project **broken link removed**

How can an LED be illuminated by a 1.5V circuit, when the forward voltage of an LED is about 2V? The LM3909 uses the 100uF capacitor as a charge reservoir, building up a voltage of about 2V before discharging the cap through the LED.

It runs 2v application off a 1.5v source, can you explain this to me?

Thanks for help!

 

[Nigel Goodwin]

I found this project **broken link removed**

How can an LED be illuminated by a 1.5V circuit, when the forward voltage of an LED is about 2V? The LM3909 uses the 100uF capacitor as a charge reservoir, building up a voltage of about 2V before discharging the cap through the LED.

It runs 2v application off a 1.5v source, can you explain this to me?

Thanks for help!

It's a simple charge pump application. Effectively what it does is put the capacitor in parallel with the battery and charge it to 1.5V, it then switches the capacitor in series with the battery - where 1.5V + 1.5V gives 3V, enough to power the LED. This process repeats continuously, in this case used to flash an LED - but the same principle can be used to give a continuous supply (at low currents). Try looking up datasheets for the ICL7660 IC.