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

minimaltom

Hi, im having a small problem with the -  part of the chapter, that is experiment 9 and the 2nd edition. In figure 2-85 and 2-86 there is an LED and a 470 ohm resistor added beneath the 10k resistor. Why have the 10k resistor in this circuit at all? Won't the capacitor charge itself fine through the 470 resistor? You don't need to measure the time the capacitor charges in this experiment and if you did wouldn't the 470 resistor alter the time it took to charge?

Thanks

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  • Hi!

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

  • in reply to minimaltom

    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.

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