A detailed explanation would be greatly appreciated 
This circuit
Thanks again
Looks like a (combination) voltage multiplier / astable multivibrator. I’ll do an analysis when I have time.
II just took another look at it. I don’t think it employs a voltage multiplier (there are no diodes). It just uses RC time constants. More coming…
The quick answer is that it uses a decaying input voltage to turn the system on and off. When the capacitors discharge (through the light and resistors), the value of the input changes. This input controls the output voltage, raising it. By altering the values of the resistors/capacitors, you increase or decrease the interval of the flashes. The LM 317 regulator is a very simple circuit, and uses the input voltages to regulate the output. IIRC 0Volts at the input produces full voltage at the output… and the inverse.
I think… but it’s been a while since I’ve done that sort of work.
-Butler
O.K., here we go…
The LM317 maintains 1.2 V (nominal) between its “OUT” and “ADJ” pins.
When you first fire up the circuit, 1.2 V instantly shows up between the LM317’s “OUT” and “ADJ” pins. (All capacitors are still “dead.”) Because there’s no resistor between the LM317 and C2, the 1.2 V instantly impresses itself across C2. In fact, as long as the circuit is powered 1.2 V remains across C2 all the time. I’m not even sure why he included this capacitor, other than to decrease the effective output impedance of the LM317. So for analysis, C2 can be ignored.
We have two charging circuits going on here:
- R1 & C3
- (R1 + R2) & C4
Both are charging at the same time. Circuit 1 (obviously) has a shorter time constant than circuit 2.
Now lets leapfrog to the light bulb. What’s the voltage across it? Simple… it’s the voltage across all three resistors (R1 + R2 + R3) plus the voltage across C2 (which is always 1.2 V).
Now back to the RC circuits… In circuit 1 we have C3 charging through R1. In the beginning C3 is “dead,” i.e. it looks somewhat like a short circuit. This means it’s “gobbling up lots of current” through R1. Well, if we have current through R1, then we have a voltage across R1 (Ohm’s Law). And it doesn’t take much current to produce a sizable voltage across R1; as little as 1 mA will produce 10 V (which means there’s 11.2 V across the bulb). Alas, the bulb lights at the beginning of circuit 1’s charging cycle.
At the same time circuit 1 is charging, circuit 2 is also charging, albeit at a slower rate. But keep in mind that C4 is also gobbling current through R1. This increases (and prolongs) the bulb voltage even more.
But all good things must come to an end. Why? Because the capacitors (C3 and C4) eventually charge up. As they do, they consume less and less current through R1 and R2, which means the voltage across them also decreases (Ohm’s Law again). This, of course, means the voltage across the bulb goes down, which makes it go off.
So what happens when the bulb goes off? C3 and C4 discharge. And then the process starts over again.
Disclaimer: I didn’t spend hours and hours analyzing the circuit (perhaps only 10 minutes), so I could be missing something significant. If anyone else wants to take a stab at it please feel free to do so.
Definitely not a voltage multiplier. It seems to just be a squirrely way to build an oscillator that can drive a twelve volt lamp directly.
The usual way to do this would be to build an oscillator of the desired frequency and duty cycle and have it drive a power transistor that the drives the lamp. Someone got really clever and found a way to make an LM317 (an adjustable voltage regulator) oscillate and used the IC itself as the driver.
The LM317 comes in TO-3 and TO-220 housings. The TO-3 is a big honker that can handle a lot of current. The TO-220 unit can handle up to 1.5A - which is enough for the 10 watt bulb mentioned in the description.
The LM317 is an amazingly flexible device. It can regulate current or voltage, and as this circuit shows, it can oscillate controllably. I’ve even got a schematic that uses an LM317 as an audio amplifier.
Agree. It is certainly not the way I would do it. (Too many damn electrolytic capacitors.)
A better approach would be to use a 555 configured as an astable multivibrator. The 555’s output would drive a FET or bipolar transistor. If Jukeball uses this approach he should keep two things in mind:
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An incandescent light bulb is slightly inductive. This means that, when the bulb is turned off, a voltage spike could occur across it. Solution? Put a reverse-biased “snubber” diode in parallel with the bulb (1N4005 works well).
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An incandescent light bulb has very low resistance when it’s cold. Because of this, there will be a current spike through the transistor when you first turn it on. Solution? Here are a few to consider: 1) Use a FET or bipolar transistor that’s designed for high-current switching. 2) Stick a small resistor in series with the bulb (conservative but wasteful). 3) Stick a small inductor in series with the bulb (make sure you put a diode across it).4) “Slowly” turn on the transistor using an RC filter. But this is wasteful for a FET (the FET will consume a lot more power).
In response to Crafter Man’s analysis, my own thoughts from a couple of minutes of looking at it :
The '317 isn’t going to put out 1.2V unless it can get constant current; C2 prevents that and is probably an important part of the oscillator.
I even took a look at the datasheet (National’s pretty good about schematics, especially with those parts), and it looks like (p.12) it’ll work until C2 charges up, then shut off, etc.
Though again that’s only a few minutes of observation. That’s not the way I’d want to use that part. The 555 circuit is a much better suggestion - in fact, I’ve used it for the exact same application (blinking lights).
You have a good point there. The weird thing about the circuit is that there’s no resistor between the LM317’s OUT and ADJ terminals; instead we have C2. I’m not sure how the LM317 behaves under this condition. If we assume C2 has a very high “resistance” at DC, then the equation seems to predict that the output voltage would be 1.2 V. But then again, it may not work correctly with such a high resistance (not enough bias current?).
We have a bunch of LM317’s lying around here. They’re more plentiful than jellybeans. I may do some experimenting and get back to ya.