Electronic – How do transistors and MOSFETs generate pulsating signals?


I see many circuits in which transistors or MOSFETs are used as "automatic" on and off switching devices. Can somebody explain how a MOSFET manages to pulse its gate within a circuit up to many hertz without outside interference?

An example would be the circuit of a joule thief. How is it possible that you use a couple of resistors and MOSFETs and make a pulsating signal?

I want to add a circuit as an example so somebody could maybe explain how the MOSFETs in this inverter cycle work and how they manage to turn on and off. It transforms 12 V DC into alternating current.


simulate this circuit – Schematic created using CircuitLab

Best Answer

What you are asking about is oscillation. It's a very broad subject and spans everything from mechanical oscillators (like a "grandfather's clock's pendulum and escapement mechanisms tied to its gearing to the clock face) to crystal oscillators to simple relaxation oscillators (both flyback and astable), which also have a mechanical equivalent. A comprehensive view of the entire topic would occupy many books.

But we can pick on exactly the case you mention -- the so-called "Joule thief" circuit found in many different incarnations. The simplest form is something like this:


simulate this circuit – Schematic created using CircuitLab

The left side is closer to how you'd build it. You fold a wire in half and then thread it through a toroid core, building a "counter-wound auto-transformer" of sorts. It will have three contacts, which include both original ends of the wire plus a third contact where you folded the wire before making the transformer. Also, if you follow the usual instructions for making this transformer, the inductance of \$L_1\$ equals the inductance of \$L_2\$.

The right side is closer to a schematic representation designed to understand how the circuit works. Note that all I've done is some modest re-arrangement. It's still the exact same circuit as on the left. Nothing has changed. But it is easier to use the right side when explaining how it works.

Note the dots. This is important for understanding how it works.

When the battery is first attached, the currents all start out at zero. Since there is no current just yet, the voltage drop across \$R_1\$ is also zero. So initially, the battery voltage, less the \$V_\text{BE}\$ junction voltage, appears across \$L_2\$. But while \$L_2\$ does momentarily resist a too-rapid change in current, it does allow change to occur. Within a very, very short time the battery voltage, less the \$V_\text{BE}\$ junction voltage, appears across \$R_1\$ and this supplies some base current to \$Q_1\$, turning \$Q_1\$ on.

Once \$Q_1\$ is on, its collector pulls down hard on \$L_1\$, turning the LED off and causing the full battery voltage (less a small \$V_{_{\text{CE}_\text{SAT}}}\$ for \$Q_1\$) to appear across \$L_1\$. This battery voltage across \$L_1\$ causes the collector current (and the current in \$L_1\$) to rise rapidly but at a controlled rate. So the current ramps upward in \$L_1\$ and in the collector of \$Q_1\$.

If you ignored \$L_2\$, the base current will be something like \$I_{_\text{B}}=\frac{V_{_\text{BAT}}-V_{_\text{BE}}}{R_1}\$. But, because \$Q_1\$ has turned on, there is now almost the full battery voltage across \$L_1\$. The transformer behavior causes the same voltage to appear across \$L_2\$. And here, the dots become important. The more positive end of \$L_1\$ is where the dot is at. So the more positive end of \$L_2\$ will also be where its dot is at. So that point is more positive than the battery voltage. This is very important to its function for a variety of reasons: (1) it boosts the battery voltage providing still more base drive current; and, (2) it adds "positive feedback" that reinforces the on state of \$Q_1\$. So the actual current in \$R_1\$ will be more like \$I_{_\text{B}}=\frac{2\cdot V_{_\text{BAT}}-V_{_\text{BE}}-V_{_{\text{CE}_\text{SAT}}}}{R_1}\$. And that fact will keep \$Q_1\$ on for a somewhat longer time.

Eventually, one of two things happens. Either the transformer's toroid core saturates, leading to an extremely rapid change in \$L_1\$'s current and quickly exhausting the \$\beta\$ current gain of \$Q_1\$, or else the \$\beta\$ current gain of \$Q_1\$ is exhausted before the toroid core saturates. Either way, \$Q_1\$'s \$\beta\$ current gain is exhausted and \$Q_1\$ (even with its enhanced base current) can no longer support the ever-increasing current that \$L_1\$ "wants" when a fixed-voltage is applied across it.

At this point, \$Q_1\$ goes out of saturation and goes into active mode. It does this by relaxing its grip on its collector, allowing the collector to float. \$L_1\$, however, won't have any of this. It was quite happy before increasing its current rapidly and it already now has a high current in it which it demands will continue. Just the same, \$Q_1\$ is done with this and allows the voltage at its collector to rise back upwards. That drops the voltage across \$L_1\$ a little, but even with a smaller voltage across \$L_1\$ it only means a smaller increase in \$L_1\$'s current. But increase it still means. But \$Q_1\$ can't increase. It just can't. So the collector voltage goes still higher and higher, trying to stop the increase. But \$L_1\$ doesn't care. The only way the current in \$L_1\$ can decline is if the voltage across \$L_1\$ flips over and changes sign. Which is exactly what happens. The voltage at the collector of \$Q_1\$ rapidly flips and becomes higher than the battery voltage, so that the sign of the voltage across \$L_1\$ can change.

Now, \$L_1\$ still has all that current in it which has to go somewhere. Guess what? There's that handy LED over there! That looks like a good place to dump that current. So the voltage rises at the collector of \$Q_1\$ until the LED turns on. Now, this is a white LED and it probably needs something like \$3.5\:\text{V}\$ to operate. Well, \$L_1\$ has no trouble helping out there. It immediately modifies the voltage at the collector such that the LED can in fact turn on and accept the inductor's current.

But this also means that the voltage across \$L_2\$ flips over, as well! Remember, this is a transformer. \$L_2\$ was, previously, adding voltage to the battery voltage to help increase the base current. But now, because \$L_1\$ reacted so quickly to reverse its voltage in order to dump current into the LED, it also reverses the voltage across \$L_2\$, too. (It can't help doing that.) Now, this means that \$L_2\$ subtracts from the battery voltage and basically turns \$Q_1\$ completely off.

There's a moment we missed, here. That's just at the place where the collector voltage is rising up, but the voltage across \$L_1\$ hasn't quite reversed itself, just yet. As the collector "lets up" and floats upward, there is a diminishing voltage across \$L_1\$. This diminished voltage across \$L_1\$ yields a similarly diminished voltage across \$L_2\$ (transformer action.) That leads to a lower base drive current in \$Q_1\$. Which means that \$Q_1\$, which was able to handle more collector current beforehand, can handle just that much less collector current. Which means the collector has to rise still further as \$Q_1\$ approaches being turned off. \$L_1\$ is very unhappy with change in \$Q_1\$, too, and reacts. If the current in \$L_1\$ can't increase, and can't even stay the same, there is only one response possible -- the magnetic field must start to collapse. The moment this takes place, the voltage across \$L_1\$ reverses itself, the collector voltage rises above the battery voltage, the voltage in \$L_2\$ also reverses itself and greatly reduces the base current towards zero, and this whole process rapidly feeds on itself. Very quickly \$Q_1\$ is turned completely off.

Now that \$L_1\$'s magnetic field is collapsing, it's current can decline as it drives current into the LED. Eventually, the magnetic field energy has completely collapsed to zero and no more current is possible. At this point the voltage across \$L_1\$ returns to zero, the voltage across \$L_2\$ also returns to zero, and now \$R_1\$ can supply a starting base current needed to turn \$Q_1\$ back on, which then places a voltage across \$L_1\$, leading to a supporting voltage across \$L_2\$ that increases the base current, again, and the cycle repeats itself another time.

This whole process takes time as it stores increasing energy in \$L_1\$. However, eventually, the BJT cannot continue to support those increases in the magnetic field and then the magnetic field must collapse. This collapse is used to turn the BJT off and drive current into the LED. When the stored energy in the magnetic field is exhausted, the process repeats.

So one of the keys is the temporary storage of energy "somewhere." This can be done by temporarily storing energy in magnetic fields (inductors), temporarily storing energy in electric fields (capacitors), or both. You can slosh the energy back and forth between magnetic and electric fields, too (tank circuit.) But you need a place to temporarily store energy. That's one of the keys. With that key, plus a way of providing sufficient positive feedback to keep things from finding a "quiescent point" in some halfway-place, gives you an oscillator. The trick, as always, is working out good ways to achieve both in a simple circuit.