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I'm a newbie working my way through Charles Platt's Make: Electronics (2nd Ed). I'm on Experiment 8 and I'm confused about something. The circuit looks like this: circuit schematic

What is supposed to happen is that when the button is pushed the capacitor charges before the power is cut to the button, it keeps the relay switched until the capacitor discharges. So the right-hand LED stays on for a few seconds before the relay switched back and the left-hand LED comes back on. Platt notes that if the right-hand LED is removed, when holding down the button the left-hand LED will fade out over about a second before popping back on and cycling again, since some of the power from the capacitor flows through it. What I don't understand is why the same thing doesn't happen when the right-hand LED is present. In other words, in the original circuit why does the left-hand LED wink out immediately instead of fading when the push button is held down. As an experiment, I modified the circuit so that each LED had its own 470-ohm resistor. Once I did this, the left-hand LED would indeed fade instead of winking out. Can anyone explain this behavior to me? Thanks!

If it helps, here's the modified circuit on my breadboard, sorry for the crummy image quality: modified circuit

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  • \$\begingroup\$ The schematic in the image is too hard to understand.Try symplifying it to make it clear. \$\endgroup\$ – Daniel Tork Jan 6 '16 at 20:01
  • \$\begingroup\$ I'm not really sure how I can simplify it and still keep all the components. I copied it from the book except the book doesn't have a schematic with the capacitor included. Like I said I'm pretty new at this... \$\endgroup\$ – oh_the_werewolf Jan 6 '16 at 20:15
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schematic

simulate this circuit – Schematic created using CircuitLab

Circuit 1. Full circuit.

  • Initially D1 is on. V1 = 9V.
  • Press SW1 the capacitor charges and the relay turns on, disconnecting the supply. V2 = 9 V. V1 = capacitor voltage which is falling. D1 turns off - probably by the time V1 reaches 8.5 V because it's anode is held at about 7 V by the current flowing through D2 and R1.
  • The capacitor holds the relay on for a few seconds and the relay drops out.
  • If SW1 is still pressed the cycle repeats.

Clarification: For current to flow in D1 the anode (top) will need to be about 1.5 - 2.0 V higher than the cathode (bottom). When the relay initially picks up V1 = V2 = 9 V so both LEDs light. Then C1 starts to discharge and V1 falls to 8.9, 8.8, 8.7 V. Meanwhile D2 is being powered directly from the 9 V battery. If, say, 15 mA flows through D2 and R1 the voltage at the top of R1 will be \$I \cdot R = 0.015 \cdot 470 = 7 V\$. But V1 is falling and at 8.5 V there will be (8.5 - 7) = 1.5 V across D1 and it will be very dim and shortly after go out completely.

schematic

simulate this circuit

Circuit 2. Left LED only.

  • Initially D1 is on.
  • Press SW1 the capacitor charges and the relay turns on, disconnecting the supply.
  • If SW1 is held the V1 = capacitor voltage. D1 gradually decays with C1 voltage until the relay drops out.
  • If SW1 is still pressed the cycle repeats.

schematic

simulate this circuit

Circuit 3. Right LED only.

  • This time D2 turns on and gets full 9V.
  • Relay hold-up lasts longer because C1 is only discharging through the relay coil. (D1 is missing.)

schematic

simulate this circuit

Circuit 4. Independent LED resistors.

  • This time V2 goes to 9 V as before and lights fully.
  • V1 starts at 9 V and follows C1's voltage. In Circuit 1 D1 turned off quickly because it's anode was held at about 7 V by D2. This time it isn't so it remains forward biased with current flowing and light emitting!

Good first question. Some tips for future: There's a schematic button on the editor toolbar. It's quite easy to use. Draw your circuits with + at the top and - at the bottom. It's easier to see what's happening then.

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  • \$\begingroup\$ Thank you! I'm still not sure I quite understand what you mean when you say in Circuit 1"D1 turned off quickly because it's anode was held at about 7 V by D2". I will make sure and use the schematic in the editor next time. \$\endgroup\$ – oh_the_werewolf Jan 6 '16 at 21:09
  • \$\begingroup\$ See my clarification. Ask again if it's still not clear. \$\endgroup\$ – Transistor Jan 6 '16 at 21:30
  • \$\begingroup\$ So the current flowing through D2 from the battery creates a voltage drop across the resistor, which means much less voltage drop across D1, and therefore no current flowing through it? This has to do with the LEDs being non-linear, am I right? \$\endgroup\$ – oh_the_werewolf Jan 7 '16 at 2:17
  • \$\begingroup\$ Yes to first sentence. Kind-of to the second. It's not so much that they're non-linear because even if they were linear diodes (there's no such thing) the effect would be similar once the voltage across D1 starts to fall. I think you've got it though. \$\endgroup\$ – Transistor Jan 7 '16 at 7:46
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Note that when the right LED is present,you have more loads being fed by the capacitor,thus the discharge rate is faster.The LEDs have their own resistance.Know that when you put two resistors in parallel,the resistance will decrease,thus the capacitor will be diacharged faster through the LEDs and the 470 ohm resistor(the time it will take for a cap to discharge is 5xtime constant.The time constant is:RxC,where R is resistance in ohms and C is capacitance of the cap in farads).The same things happens when you place a rsistor for each led:total resistance decreases,the diacharge time decreases.

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