Capacitor not charging?

Question

I am yet another software developer trying to step into the world of electronics... (so sorry for the basic questions...)

I am trying to understand how a capacitor works, but obviously I am missing a crucial part from all the helpful websites like sparkfun and adafruit.

I failed to solve an issue I currently have in a pretty basic project that reads a temperature from a thermocouple and displays the value on an LCD. An electrical engineer at work pointed me in the direction of adding a capacitor to filter the noise I introduce when I touch the thermocouple. Long story short; the capacitor doesn't seem to charge, no matter how I add the capacitor to the circuit, almost no voltage comes out; ever.

So I thought to be smart, and make a new circuit of:

• a battery supply (9.6V)
• an LED
• a capacitor "104"

The circuit is simple, all components are in series, so I expected that my LED would light up, and when I disconnect the battery, the LED would fade out. But again... nothing happens. The LED never lights up, and my multi meter measures 0.8V after the capacitor. But also, when I hold the multimeter probes to the leg of the capacitor, the measured voltage drops from 0.8V to close to zero in no time.

Please forgive me if this drawing is wrong (and please do tell me what is wrong so that I can learn from it):

• Why doesn't the capacitor charge up to the voltage of 9V (but seem to stop charging at 0.8V?)?
• Why does it discharge when I measure the voltage with a multi meter?

PS: there is no resistor between in the circuit limiting the current, and the batteries are 8x 1.2V rechargeables = 9.6V

Answer 1

Your circuit should look like this:

^{simulate this circuit – Schematic created using CircuitLab}

Also, if the value of your capacitor is small, when you disconnect the power source, the led might go out very quickly. Are you sure your capacitor is 104F?

edit: as G36 points out, you probably meant it was marked 104, which means 100nF (except for with aluminium capacitors). To see any afterglow, you'll probably need at least 1000uF

Answer 2

First of all you have to know the working voltage of your LED, if the LED requires a 5V input, applying 9.6V will damage your LED and you will get a no response result. So you have to add a resistor in series to drop the extra 4.6 Volts from the circuit.

Since your battery rated at 9.6V and assuming you have a LED that works on 5V as input and consumes 20mA of current, the required series resistance value can be calculated using kirchhoff's voltage law, which states that "for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero"

this means Vs ( Battery ) = V ( LED ) + V ( Resistor )

where V ( Resistor ) = I ( current ) x R ( resistor value )

By Substitution 4.6v = 20mA x R , so R = 230 Ohms. So you need a resistor of at least 230 Ohms. So the next higher popular standard resistor is 330 ohms.

Its not necessary to connect a capacitor to this circuit, the main purpose of a capacitor is to smooth the voltage generated from the power source and to eliminate voltage ripples. But if you intend to add a capacitor it should be connected in parallel with the LED. Because all the loads are designed to function optimally when connected to rated supply voltage. When capacitor is connected in parallel, full supply voltage is available to the load. However when it is connected in series, it drops the supply voltage, resulting in altered performance of the load.