# Using capacitors for flashing lights

I am a beginner!

I've been trying to devise a very simple circuit to confirm in my mind how capacitors work. The goal is to make an LED flash on and off using a capacitor or two and one or more transistors.

I've tried various things but I'm getting nowhere fast. I've seen circuits online but they tend to involve two LEDs, inductors or other strange setups.

Can anyone explain the theory behind what I'd need to do (without delving into the maths too much please)?

As far as I can tell a capacitor could be used to control a transistor to make it switch off after a second or so.. my problem is getting it to turn back on!

If anyone can describe what needs to happen (logically) in order to make the LED turn on and off that'd be really good!

Cheers, John.

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See perhaps the simplest LED flasher circuit: cappels.org/dproj/simplest_LED_flasher/… – Alfred Centauri May 13 '14 at 13:19
@AlfredCentauri: You've completely missed the point of the question, which is about capacitors, not flashing an LED. I'm pretty sure that someone who's struggling with how capacitors work is going to be completely baffled by the concept of "negative differential resistance". – Dave Tweed May 13 '14 at 13:25
@DaveTweed, the link in my comment above is not an attempt to answer his question but rather, additional information for the OP to consider in his quest for a "very simple circuit" with capacitor. Regardless, your opinion on this is of no interest to me. – Alfred Centauri May 13 '14 at 13:28
@AlfredCentauri: It's obviously of some interest, or you wouldn't have responded. But I'm not here to get into a p*ssing contest with you; I'm here to make sure that visitors to this site get the answers they're seeking. – Dave Tweed May 13 '14 at 14:43
For what it's worth, one of the classic light-blinkers was essentially a largeish capacitor, a current-limiting resistor so it would charge slowly, a sufficiently high DC voltage source (typically a 90V "B" battery), and an NE2 bulb (which would ionize when the voltage got high enough, lighting up and conducting until it had discharged the cap enough that it could no longer stay ionized and then stop conducting for another charging cycle). – keshlam May 13 '14 at 16:36

What you've seen with two LEDs it is called an astable multivibrator.

It has two states that change whenever one of the capacitors is discharged.

Assume Q1 is ON. This means the positive side of C1 gets connected to ground. This leads to reverse charging of it (discharging), thus modifying the voltage at the negative side - it becomes positive and biases Q2 which turns ON. Q2 starts discharging C2 and it turns ON Q1 again. And this cycle repeats forever.

While either of the capacitors is charged, the base voltage of the transistor is negative, thus the transistor remains OFF. The frequency of this is related to the time needed for the capacitor to discharge (when it is connected in reverse polarity to ground via a resistor - e.g. C1-R2).

Source: Wikipedia - Astable Multivibrator

LED Flasher circuits usually work on the same principle.

Source: next.gr

When they have one LED, the other is usually replaced by a resistor.

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Why don't both 'halves' of this circuit operate at exactly the same time? Shouldn't current flow into both C1 and C2 at the same time? Shouldn't both transistors then turn off at the same time? – John Hunt May 13 '14 at 13:48
@JohnHunt good question. In and ideal application (perfectly matched resistors and capacitors) this circuit wouldn't work. On power-up both transistors have the tendency to turn ON. Due to slight differences in components values, one will be the first. Thus it will block the other one from switching ON. – Cornelius May 13 '14 at 13:52
Thought that might be the case. Coming from a programming background that sort of thing doesn't come to mind! – John Hunt May 13 '14 at 14:46
@johnhunt in programming, that is called a race condition. – Passerby May 14 '14 at 2:21

As far as I can tell a capacitor could be used to control a transistor to make it switch off after a second or so.. my problem is getting it to turn back on!

That's correct. Once the capacitor in your circuit charges enough to turn the transistor on, there must be a way to start the capacitor discharging until the transistor turns off.

A very simple circuit, using a unijunction transistor is below:

If anyone can describe what needs to happen (logically) in order to make the LED turn on and off that'd be really good!

Initially, the capacitor $C_2$ has zero volts across and the UJT is off (an open circuit).

Thus, the capacitor charges through $R_1$ and, after some time, the voltage across $C_2$ is large enough to turn on the UJT.

Once the UJT is turned on, $C_2$ rapidly discharges through the UJT and LED, lighting the LED, until the UJT turns off and then the cycle repeats.

This is a type of relaxation oscillator. Here is an introductory tutorial on the UJT relaxation oscillator that you might find helpful.

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Unijunction transistors are strangely different from normal transistors, and also a little hard to get hold of. – pjc50 May 13 '14 at 14:17
@pjc50, see, for example, digikey.com/product-search/… – Alfred Centauri May 13 '14 at 20:22
@AlfredCentauri Actually, there are no UJTs available from Digikey, only dusty stock from odd suppliers. The ones you found are PUTs, which are quite different devices (they require a couple resistors to work similarly to a UJT but they are completely different inside). Here, Farnell/Element 14 has some: uk.farnell.com/multicomp/2n2646/transistor-unijunction/dp/… – Spehro Pefhany May 13 '14 at 21:34
@SpehroPefhany, shame on me, I didn't catch that when I did the UJT search at Digikey. Thanks for pointing that out. Looks like Newark carries the 2N4870 in stock: newark.com/multicomp/2n4870/unijunc-transistor-0-3w-50ma-to/dp/… – Alfred Centauri May 13 '14 at 21:44