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wanted to ask if it's possible to make a blinking led using only a capacitor and a resistor, and using a normal 1.5v AA battery as a power supply, like the image, *pd: knowing how my teacher is, it wouldn't surprise me if it's not possible.

enter image description here

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5 Answers 5

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The circuit diagram is horribly confused. The point of this puzzle presumably is that an LED has a voltage drop of 1.6V minimum. So you cannot light it with a 1.5V battery alone. Now the arrow drawn at the image of the battery presumably means not the direction of the current (which would be pretty pointless given the polarity of the LED) but rather means "please rotate the battery between its connectors". So the idea is to create some pseudo-AC voltage. One phase of the voltage is used to charge the capacitor, and the next phase puts the capacitor in series with resistor and battery in order to light the LED. This circuit will not work for that purpose and I don't see how one can get by without adding more diodes or using a more complex commutator on the capacitor rather than the battery.

However, it may also be that this is intended to be a baffle-the-student device. In that case, either the LED is a blinking LED to start with and/or the battery is not actually a battery but rather a pulse generator and a button battery stuffed into an AA battery housing. Since the circuit diagram for no good reason at all supplants a picture of the device rather than its circuit diagram, presumably out of some perverse sense of accuracy, I consider this option the more likely one.

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  • \$\begingroup\$ It is possible to light a red LED from a brand new 1.5V battery, in practice. \$\endgroup\$
    – Simon B
    Feb 16, 2016 at 15:35
  • \$\begingroup\$ So, is there a way to make it work? it doesnt matter how is it placed, as long as only 1 capacitor, 1 resistor and 1 battery is used. \$\endgroup\$ Feb 17, 2016 at 1:09
  • \$\begingroup\$ I suspect that a sort of relaxation oscillator was intended, in which the LED conducts (i.e., blinks) when the capacitor reaches a sufficient charge/voltage level. The conducting LED discharges the capacitor and the cycle repeats. As noted, this won't work with the circuit as shown. \$\endgroup\$ Apr 22, 2022 at 23:00
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Not only is it not possible, but the LED won't even light up- it is reverse biased. So the circuit does nothing (the capacitor charges to ~1.5V and that's that).

Battery life should be excellent, however.

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    \$\begingroup\$ For the "battery life should be excellent", almost a +1. Oh, what the heck. +1 \$\endgroup\$
    – Marla
    Feb 15, 2016 at 23:24
  • \$\begingroup\$ Wellll.... if the wire is long and you happen to wrap it around your finger, you don't even need the resistor or the capacitor to flash the LED. Would that qualify? :-P \$\endgroup\$
    – Asmyldof
    Feb 15, 2016 at 23:33
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    \$\begingroup\$ Could you explain why does this happen? I have some thoughts, but i'm unsure. \$\endgroup\$ Feb 16, 2016 at 0:12
  • \$\begingroup\$ And if it's a white LED, which needs about 3 volts, the LED won't light up even if you do reverse the battery. Won't work for green or blue, either. \$\endgroup\$ Feb 16, 2016 at 16:55
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Yes. But you will need to remove and reinsert the battery for every blink. This is a trick answer to a trick question.

The capacitor will provide fading as it discharges, Depending on its value, but it may be too quick for human sight to see.

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  • \$\begingroup\$ In that case adding a capacitor with more capacity could work? But the voltage would be higher, wouldnt that fry the LED? or what about using some sort of resistor so the capacitor will charge in more time, making the LED blinks more visible? \$\endgroup\$ Feb 15, 2016 at 23:44
  • \$\begingroup\$ The fading will only happen when the battery is removed. It's capacity is not the same as its voltage. This is not really a practical solution, as you are blinking by cutting power. It's a trick answer for a trick question. \$\endgroup\$
    – Passerby
    Feb 15, 2016 at 23:47
  • \$\begingroup\$ I don't see how the capacitor in this circuit ever provides power to the LED, charging or discharging. And the only way it discharges is through the LED reverse leakage current. \$\endgroup\$
    – Paul
    Feb 16, 2016 at 23:47
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It is not possible with a normal LED.
There are, however, blinking LEDs that look just like normal ones but have a blinking circuit integrated. They need no other components, just a power supply, to blink.

E.g. here is a datasheet.

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you need at least 3V to light up most of LED.

here is good example of LED blinking with capacitor, resistor and transistor : http://blog.jongallant.com/2015/01/simple-blinking-led.html

LED

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    \$\begingroup\$ A wiring diagramm is not useful at all, especially if the type of transistor is not given. BTW: is the center pin of the transitor not connected? \$\endgroup\$
    – Curd
    Feb 16, 2016 at 11:31
  • \$\begingroup\$ PN2222 Transistor. I'm also not sure why the center pin is not connected. :( \$\endgroup\$ Feb 16, 2016 at 11:42
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    \$\begingroup\$ Here's the schematics: cappels.org/dproj/simplest_LED_flasher/… \$\endgroup\$ Feb 16, 2016 at 11:51

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