How Capacitors Work -- The Learning Circuit 29

John, over the years, I'd hear a few examples relating voltage to the height of a water fall and current to the mass flow of water.

In diode/resistor circuit above, I can picture a throttle valve being a resistor in that it would restrict the water flow.  An initial empty being barrel plumbed in parallel could be the capacitor.

What would the diode be?  Or does the water analogy have its limitations?

Also, in an electric circuit - do the electrons preferentially go to the capacitor first or do some electrons go to the negative of the battery at the same time.  Does that depend on the capacitors construction?

I know I could google around for this, but I feel like old school conversation.

-Sean

Hi Karen,

Just a clarification. The resistor in the circuit will slow the charge as it limits the current that can flow at a specific voltage. If there were only a resistor in the circuit the capacitor would charge and approach the voltage of the battery. The diode on the other hand is what causes the inability of the power source to charge the capacitor to full voltage. The diode junction is an energy barrier and it takes a specific voltage to cross it. In the case of the Red LED the junction may present a 1.2 volt energy barrier and this energy barrier will take that much voltage away from the power source voltage so that the capacitor can only approach the source voltage less the diode junction voltage at full charge.

John

Yes. That's what Jon and John were trying to explain. In a practical application, since there is a resistor and diode in the circuit, causing a voltage drop, the capacitor voltage at full charge would be less than the supply voltage.

Hi Mey,

To be honest, it is more fun trying it out, this is a safe and easy experiment. You'll need a multimeter, but any newcomer to electronics will require this anyway (even a $5 multimeter is fine) and a few different resistors and capacitors to experiment with.

Or, you can simulate it. If you click here, you'll see a circuit with two resistors and a 5V battery. By using the red slider marked "Variable Resistance" on the right, you can vary the value of the bottom resistor. Notice that it is possible for the center voltage test point (marked TP) to vary, and it will reach as high as 4.5V which is very close to 5V. So you're right, the voltage will be lower, but it could be by an extremely tiny amount.

If the bottom resistor was of an even higher resistance then the TP voltage will get even closer to 5V. That shows that having a resistor in series doesn't necessarily mean there will be a lower voltage by any significant amount.

To read up on this in more detail, google 'potential divider'. The wikipedia page has the formula so you can play with trying different values and seeing its effects - or you can try it for real as mentioned, or in a simulator.

A quick question, since we have a resistor in series there will have a voltage drop and will this not result in charging the capacitor to a voltage less than the input voltage? Pls excuse me if this is silly question, I am a new to electronics

Oh, I see what you're saying. I churn these videos out so quick that they start to jumble together. I was thinking of a charge circuit I use in the next video. Yes, I see what you mean and the graphics in the video could be misleading. I'm not always sure what visuals are best for demonstrating lessons, so sometimes I air on the side of simple rather than perfectly accurate. The point was that the capacitor can charge up to match the supply voltage. I was focused more on getting that point across, so the circuit is more a theoretical than practical example. Though the circuit would work as shown, you are correct about the loss of voltage through the diode(LED).

As John says, I was just pointing out that it wouldn't reach 9V (by showing you the result in a simulator).

Here it is without the diode (I've shortened the timescale to make the shape of the curve more visible)

Charging through a resistor like that, and the form of the curve it produces, is where a textbook would go after introducing the capacitor.

Thinking a bit more about the circuit you showed, whoever designed it wanted to know when the charging was taking place and when it had finished (the LED lights during the charging and extinguishes at the end). You wouldn't do that for a small value of capacitance, where the charging would be over quite quickly, so perhaps it was intended as a charge circuit for a supercap where it might take several minutes to charge the part.

Hi Karen,

The symbol that Jon is using is a power source and the denotes that it is a DC power source the same as your battery. The point Jon is making is that the inclusion of the LED in your schematic will take the LED junction voltage from the voltage of the battery so that your 9 volt battery actually only charges the capacitor to 9 volts less the LED junction voltage or perhaps a total of 7.8 volts. The resistors in line will slow the charge of the capacitor but not keep it from getting close to full charge.

John

For clarity, are you saying that alternating current is more effective/efficient for charging a capacitor than DC?

That circuit isn't the best way to charge a capacitor to 9V.