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

Hi!

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

removed it 

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.

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.

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)

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)

Great explanation. I don't have the book so I'm not sure what the purpose/aim of the experiment is.
 You can also increase the DC voltage in small steps of maybe 0.1 or 0.5 volts. And you can see at what input voltage the LED just flashes for the first time. 

Thanks for the great explanation I think I'm almost there. Does this mean that the LED fades out completely once the capacitor is fully charged? For some reason I thought it was just until the capacitor reaches 67 percent. Is it that the capacitor only blocks DC when its fully charged?

As usual with these things, it isn't that simple but experimentation will help.  Take a look at figure 2-89 which shows an Oscilloscope measurement of what happens when the capacitor is discharged and then button A is pressed: you see a pulse followed by a tail-off.  The LED will light at the point that pulse is greater than its forward voltage and it will stay lit as long as the tail-off is greater than its forward voltage.  Put in a smaller capacitor and that pulse/tail-off effect will be shorter; put in a larger capacitor and it will be longer.  Of course, we're talking very fast times here which is why you only see a flash: shorter flashes will just appear as a dim light and longer flashes will appear brighter - and these are effected by the value of the LED series resistor which is current limiting.

So the capacitor lets the DC through when there is a sudden change (i.e. button A is pressed and 9V suddenly appears on the +ve side) and this will tail off rapidly based on the size of the capacitor/resistor combination.  I've been a bit misleading tie-ing this to the Time Constant, sorry about that, but I was trying to get the idea over that the effect will change based on the Capacitor/Resistor combination which affects the charging rate and thus that tail-off effect.  It seemed to be the simplest way to explain it but your question tells me I was over simplifying and confusing you, sorry. 

If you have an oscilloscope, or can get access to one temporarily, you could measure the impact of changing the capacitor and resistor sizes and relate what the scope shows you to the forward voltage of the LED but I guess you won't have gone that far yet; also whilst a simple enough experiment, it may be a little confusing at the moment.  I'm in danger here of patronising you of course which I definitely don't mean to do!

Did you mean 67% or 63% which is the percentage used in respect to the Time Constant?