Electronic – Theory behind the “Joule Ringer” circuit

This one's a year old, but I'll necro it because I think I can say something useful.

Joule Thief circuits got a bad rap some time ago because some people in the over-unity, free-energy crowd went nuts over it, so lots of people won't really spend any time talking about it. Turns out, you still can't get something for nothing. So, moving on.

Pretend for a moment, the battery in the moment isn't connected. No current, no voltage anywhere else in the circuit. That's our starting point.

When the battery is first connected, the only path for current to really flow is into the base of the transistor. As the transistor gets biased, the current from the collector to the emitter will rapidly increase as a multiple of the current flowing into the base, depending on the exact transistor used. The increasing current will start to store energy in the secondary winding, just like any other inductor.

See those dots are the top and bottom of the transformer? A current flowing into the top of the coil on the left, will turn into a current flowing out from the bottom of the coil on the right. This current won't particularly be able to drive the LED, so its going through the transistor.

What happens next is a little bit hard to explain. The easiest way to explain it is to follow those dots. The current that is now flowing from top to bottom on both sides of the transformer generates opposite polarities of voltage from each other. And the current on the right side is higher, thanks to the transistors amplifier action. So the coil on the left gets a voltage boost from the coil on the right, and this boost opposes the tiny current that flows into the base of the transistor, shutting it down.

Well, the current in the right hand coil can't exactly just stop; it stored energy in the mutual magnetic field that has to go somewhere. As that field begins to collapse from lack of anything sustaining it, it starts to push at higher and higher voltages. Eventually, this voltage gets high enough to forward bias that LED, and the right hand coil completes its discharge cycle while the LED emits light.

The Joule Thief is not magic, it works the exact same way as any other boost converter. It just so happens to be a very clever use of mutual inductance to set up an oscillating switch to create the inductive kick, so that it can work from extremely low voltage sources.

So, there are only three real things to change - the transformer, the transistor, and the LED. Some LEDs are fairly dim by design, even when properly supplied. Assuming this isn't the problem, that leaves the transistor and the toroid.

Without doing the math, I'd say its safe to say you want a transistor with a fairly high beta value (the ratio of Collector current to base current) that can handle quite a bit of current.

The websites posted in the comments are pretty accurate. You need as few coils as possible around a reasonably sized toroid to store the most energy possible in a very short time. Don't forget that 1 volt across a very low resistance wire can still make for a significant amount of current, so don't use itty bitty magnet wire. The other feedback coil (left hand) can be relatively wimpy, in comparison - transistor base current through that resistor should be on the order of microamps.

LEDs would get dim in these circuits in the primary winding had way too much inductance, the transistor was relatively high in on-state resistance, or, quite possibly, from not having the coils wound in opposition to each other - the LED might find just enough juice from the battery to bias weakly, and the feedback path would simply hold the transistor in a hard-off reverse biased state.

Doh! So that's what that bit on Volt-second balance was I read last week!

Alright, well, the clever approximation I was looking for turns out to be well-known, of course, and essentially rests on a balance of energy argument.

So here's what the derivation of an expression for \$D'T_s\$ looks like.

The general principle used is what's called inductor volt-second balance, which basically derives from the fact that the energy discharged from an inductor each switching cycle equals the energy stored in it during that cycle. This relies on the converter being in steady state, which holds here. The converter operates in Boundary Conduction Mode (BCM) and the inductor current (and therefore energy) is zero at the beginning of the on-stroke and returns there at the end of the off-stroke.

Calculating the volt-seconds for the on-stroke is straightforward:

\begin{align} V_{Lon} \cdot DT_s & = (V_{in} - V_{CE}) \cdot DT_s \approx (1 - 0.1) \cdot 65\,\mathrm{\mu s} = 58.5\,\mathrm{V\mu s} \end{align}

For the off-stroke it's a little trickier, but helped by that linearity in the inductor voltage I was mentioned in the OP. The voltage curve across the diode (\$V_C\$) looks like this (yellow trace):

We can approximate the downramp in \$V_C\$ as a straight line from \$V_{Dmax}\$ to \$V_{Dth}\$, where \$V_{Dmax}\$ is the forward voltage drop across the LED at \$I_{Cmax}\$ and \$V_{Dth}\$ is the forward threshold voltage of the LED. These might be available on a datasheet, but might need to come from measurement or extrapolation. \$V_{Dth}\$ is the easier to determine of the two, I expect.

Having those figures we can quickly arrive at an average \$V_L\$ (inverting conventional polarity for clarity):

\begin{align} V_{Loff} & = \overline{V_C} - V_{in} = \frac{V_{Dmax} + V_{Dth}}{2} - V_{in} \end{align}

Equating on-stroke and off-stroke volt seconds gives us the balance relationship:

\begin{align} V_{Lon} \cdot DT_s & = V_{Loff} \cdot D'T_s \\ \\ (V_{in} - V_{CE}) \cdot DT_s & = \frac{V_{Dmax}+V_{Dth}-2V_{in}}{2} D'T_s \\ \end{align}

... and rearranging gives us an expression for \$D'T_s\$:

\begin{align} D'T_s & = DT_s \frac{V_{in}-V_{CE}}{\frac{V_{Dmax}+V_{Dth}-2V_{in}}{2}} \\ \\ & = 2DT_s\frac{V_{in}-V_{CE}}{V_{Dmax}+V_{Dth}-2V_{in}} \end{align}

Substituting values from this example gives:

\begin{align} D'T_s & = 2(65\,\mathrm{\mu s}) \frac{1-0.1}{5.55+3.2-2} = \frac{117}{7.75} = 15.1\,\mathrm{\mu s} \end{align}

Which is commensurate with, if somewhat less than, the value of \$16.7\,\mathrm{\mu s}\$ produced by the simulation.

Anyway, I'm pretty sure that's right and that gives me what I need to move forward with the derivations. T and f are an easy step away and I expect I'll be ready to move on to bigger and badder converters after that :)

Let me know if I've gotten anything wrong.