Make: Electronics (2nd edition) experiment 9 - capacitive coupling

Hi!

Could you please post a screenshot of the circuit diagram for better understanding?

[Not posting an image of the circuit because it's copyright]

The answer is in the text:

• when the capacitor is discharged (initial state), and you press button A, current will flow through the capacitor, charging it, and through the 10K resistor, and the LED and 470 resistor in parallel with the 10K resistor.  If the LED did not have the 470 resistor in series then the current through it may well blow it.  The 470 is really in the circuit to protect the LED and has a coincidental small role to play in charging the capacitor - but that isn't its real purpose in this experiment.
• Once the capacitor is charged the LED will turn off because the capacitor will not pass DC current: the current's only path is via button B which is open thus nothing flows.
• In order to see the LED flash, the speed of charging the capacitor has to be slowed down.  This is achieved with the 10K resistor in series with it.  If you tried a small value resistor here, you may well not see the LED flash at all; conversely, trying a larger resistor you may see the LED light for longer.
• The 470 and 10K resistor are in parallel with each other so, yes it does somewhat impact the charging rate, and that rate can be calculated by determining the actual resistance (you should be able to calculate total resistance from resistors in parallel by this stage) and thus the time constant. 

What he's teaching you here is that even with DC, a capacitor will allow some current through whilst it is charging; what he's reinforcing though is that you should always current limit your LEDs to stop blowing them.  Note that when he moves on to replace the LED with the DMM, that 470 resistor isn't needed.

If you have plenty of LEDs to hand you could experiment to try and make it blow: remove the 470 resistor and see what happens; try substituting larger capacitors.  I expect at some time you'll see it flash and then no more.

Don't forget to discharge that capacitor using button B before removing it from the circuit!  Especially if you try larger ones.

Hope that helps but feel free to ask more questions.

removed it 

Are the bullet points yours or from the book? I can't find that written anywhere in it.

Thank you

As Andrew mentioned the copyright issue, you may delete the picture or post a hand drawn image of the circuit. Also, Andrew gave a good explanation of the working of the above circuit.

There are two cases here -

Switch A is on and Switch B is off

Switch A is off and Switch B is on

For the first case when you press switch A and switch B in off, the 1000uF capacitor will charge through the parallel combination of 10K and LED and 470 ohms resistor.

The 470 ohms resistor is for current limiting for the LED to prevent it from shorting. If 470ohms resistor is absent then a large amount of current will flow through the LED when the capacitor is charging and this will damage the LED.

Now to answer your questions, yes having two resistors and LED in the charging path of the capacitor will change the charging time of the capacitor. However, if you wish to charge the capacitor only using the 470 ohms resistor and the LED, the current won't flow unless and until the LED starts conducting. Think of the LED just as a diode that won't conduct unless the forward voltage of the diode has been achieved.

The second case when switch A is off and switch B is on is for discharging the capacitor assuming it was fully charged in the first case.

They are mine - just trying to rephrase what is written there to explain it in a different way.  

Good point about the diode.

Great that's makes it clearer thanks.

In that case tho wouldn't the capacitor only charge through the 10k resistor after all and not the 470 resistor since the led won't allow current to flow because it isn't conducting? (Except for the initial moment when the led flashes of course)

thanks, that's very helpful

You could test this by removing the 10k resistor and see what happens.  There has to be a certain amount of voltage for a diode to conduct so if that voltage doesn’t appear on the diode, the capacitor can’t charge.  Note that the LED and 470 resistor are in parallel to the 10k resistor so there is always a path for the capacitor to charge.  Because it can charge and you see the LED flash, then it clearly demonstrates that there is enough voltage across the LED (diode) for it to conduct as well.  You could try an measure this with a DMM across he LED terminals but it may be too quick to measure unfortunately.  Increasing the 10k resistor may help show this.  Actually, that is the purpose of the next experiment in the book.

The LED flashes because the capacitor charges up quickly and then stops conducting, thus the LED is only conducting for a very short time.  Increase the size of the 10k resistor to 100k or 1m and you should see the LED light longer.  You could determine this time using the formula to calculate the time constant given in the book in the same section and then see if reality matches theory.

The purpose of this experiment is to prove that (a) capacitors block DC; and (b) that until charged, DC will pass.  The easiest way to visualise this is by placing the LED in circuit, in parallel, so you can see this. The alternative way is to use a DMM in place of the LED (the next experiment) or just to believe the science!

You could also experiment with the 470 resistor and see how that affects both the brightness of the LED as well as the (small) impact it has on charging time.  I’d suggest that you don’t go much below 470ohms though as although it should be ok to a certain point, you risk blowing the LED.  Look at a datasheet for a typical LED for its current handling and you’ll see why, given a 9V input supply.

Don’t forget to discharge the capacitor(s) before removing them from the circuit (good habit forming)