The beacon LED driver

9 V.

The voltage drops around a loop add up to the supply: $$24\,\text{V} - 15\,\text{V} = 9\,\text{V}$$

Electrons move from negative to positive, and current is drawn from positive to negative.

Electrons are negative, so they leave the battery's negative end and travel around to its positive end. Current was defined as flowing from positive to negative before people knew electrons were a thing. This is called conventional current.

200 mA, from Ohm's law.

$$I = \frac{V}{R} = \frac{2\,\text{V}}{10\,\Omega} = 0.2\,\text{A}$$

0.4 J every second, which is 0.4 W.

$$P = VI = 2\,\text{V} \times 0.2\,\text{A} = 0.4\,\text{W}$$

50 mA.

A single loop carries the same current through every part in it.

9 V.

“The voltage at a point” means measured from GND, and the dot is 9 V above ground.

Its electrons are all locked into bonds.

Current is moving charge, and every outer electron is held in a bond between two atoms, so almost none can move.

It adds a free electron.

Four of phosphorus' electrons join bonds, and the fifth has no bond to join, so it wanders free. Swapping in atoms like this on purpose is called doping, and silicon doped with phosphorus is N-type: its free electrons are what let current flow through it.

It leaves a hole.

One of its bonds is short an electron, and a neighbouring electron can hop into that gap, which leaves a corresponding gap in its place. Silicon doped like this is P-type. Its holes are what let current flow, and as they move they act like positive charges.

Free electrons near the boundary cross over and fill holes on the P side.

That leaves a thin strip along the boundary, reaching a little way into each side, with no free charges at all. It's called the depletion region.

It blocks current.

With nothing free to move, it acts like an insulator across the path. Inside it, each boron atom now holds an extra electron and each phosphorus atom has lost its spare one, so the P side of the strip is negative and the N side positive. That charge pushes back on more electrons trying to cross, so the strip holds its width until a battery squeezes it thinner or pulls it wider.

A, with + on the P side.

It pushes holes and electrons toward the boundary and squeezes the depletion region thin, and past about 0.6 V for silicon, current flows. In B the battery pulls charges away from the boundary, the region widens, and almost nothing flows.

The battery keeps making new ones.

Its + end pulls electrons out of the P side, which leaves fresh holes behind, and its − end pushes fresh electrons into the N side. As long as the battery is connected, the supply never runs out. The 0.6 V mentioned before is what lets charges cross the depletion region and start this loop of pushing electrons in and pulling them out.

The left in both.

The triangle's flat back is the P side, called the anode (A), and the bar is the N side, the cathode (K). On our LEDs the anode is the longer leg, and the cathode sits by the flat edge of the rim. Current flows the way the triangle points, from anode to cathode.

No.

A diode only passes current in the direction its triangle points.

20 mA.

The LED takes 2 V, which leaves 3 V across the resistor: $$I = \frac{3\,\text{V}}{150\,\Omega} = 20\,\text{mA}$$

16 mA.

The LED still takes about 2 V, leaving 2.4 V across the resistor: $$I = \frac{2.4\,\text{V}}{150\,\Omega} = 16\,\text{mA}$$ With a plain resistor, the current follows the battery.

No.

The LED blocks current this way, so no current flows and the whole 5 V lands across the LED backwards. It gets dangerous once that passes the LED's backwards rating, which is 5 V here, so this battery sits right at the limit.

19.2 V.

The voltages of each LED add in series: $$12 \times 1.6\,\text{V} = 19.2\,\text{V}$$

Only if they leak the same.

A backwards LED still lets a tiny leakage current through, and the same current has to pass through both. The LED that leaks less needs more voltage to pass it, so it takes the bigger share. Leakage varies a lot from part to part, so the split can be very uneven.

A MOSFET.

It's a kind of transistor, a valve for current: the voltage on its gate sets how much flows from drain to source, and the gate itself draws no steady current.

No.

The path crosses from N to P and back to N. Whichever way you push, one of those crossings is backwards.

It makes a channel.

The strip of electrons under the gate acts as N-type silicon joining source to drain, so current can flow. Enough electrons gather only once the gate passes the threshold voltage, and the further past it the gate goes, the more gather and the more easily current flows.

Only for a moment.

The gate, the glass and the silicon under it form a capacitor. Current flows while it charges up to 5 V, then stops.

Yes.

The gate is 5 V above the source, past the 2 V threshold, so the MOSFET conducts from drain to source.

Off.

The gate's voltage relative to the source is below the 2 V threshold: $$5\,\text{V} - 4\,\text{V} = 1\,\text{V}$$

No.

The gate is insulated from the rest of the MOSFET, so no steady current flows into it.

The one at 10 V.

Just past the threshold a MOSFET only opens a little, and the further the gate goes above the source, the wider it opens.

Go below its source by more than its threshold.

A gate below the source pulls holes up under the glass, making a P-type channel between the two P islands. That's why P-channel parts usually sit on the high side, with the source at the supply.

A BJT, short for bipolar junction transistor.

The pin's 20 mA flows into its base, which allows a much larger current to flow from its collector to its emitter (2 A in this case).

B.

A BJT is drawn with a circle, a bar for the base and an arrow on the emitter, and its pins are the base, collector and emitter. A MOSFET's gate is drawn as a separate plate because it's insulated. Its arrow sits on the middle line, which is wired to the source.

The base is very thin.

Most electrons cross it before they meet a hole, and the collector's positive voltage sweeps them in. About 1 in 100 does meet a hole and leaves through the base wire, and that small flow is the base current.

The base-emitter voltage.

The base and emitter meet at a PN junction, whose depletion region pushes back on electrons trying to cross. Raising the voltage weakens that push. Electrons in the emitter move around with random energies, and only the most energetic make it across, so a small weakening lets many more through: each extra 60 mV lets about ten times as many cross.

Yes.

A BJT's base draws current. It flows in the base and out the emitter, and that small current lets a much bigger one flow from collector to emitter.

10 mA.

Collector current is gain × base current: $$100 \times 100\,\mu\text{A} = 10\,\text{mA}$$ as long as the supply can push that much through the load.

Almost none.

The base has to sit about 0.6 V above the emitter before current starts to flow. Remember, those PN junctions need a little push before significant current can flow through them.

About 0.6 V below its emitter, near 4.4 V.

That pushes holes from the emitter into the thin base, and most of them cross into the collector. Conventional current flows in the emitter and out the collector.

By the arrow on the emitter.

It points the way current flows: out of an NPN and into a PNP. One way to remember it: NPN, Not Pointing iN.

The MOSFET is likely ruined.

The gate sits on very thin glass, and the IRFZ44 (the particular MOSFET we're using) is only rated for 20 V between gate and source. Past that, electrons get forced through the glass and melt a tiny hole in it, which shorts the gate for good.

No.

The MOSFET is just off harder lol. Its gate is safe anywhere within 20 V of the source, either way.

The MOSFET can end up half on and overheat.

With nothing connected, stray charge can leave the gate at any voltage, including just past the threshold, where the MOSFET only opens a little. Say it lets 1 A through: the 6 Ω load drops 6 V, and the other 6 V sits across the MOSFET, so it turns $6\,\text{V} \times 1\,\text{A} = 6\,\text{W}$ into heat, enough to cook it. A resistor from gate to ground drains the stray charge and holds it off.

The pin or the transistor can burn out.

The base-emitter junction is a PN junction, so a little more voltage means a lot more current, and with nothing to limit it the base takes all the current the pin can give. A resistor in series with the base sets a safe current.

No.

It's saturated, which means fully on. The base asks for more current than the load lets through, so the collector falls until the load takes almost all 12 V. Once the collector is below the base, the base-collector junction gets pushed forward too and electrons flow back out of the collector. At about 0.2 V that backflow holds the current to what the load allows, and any extra base current is wasted.

Probably.

The 2N3904 (the particular BJT we're using) is only rated for about 6 V backwards across its base-emitter junction, and past that the junction breaks down and conducts. Current forced through it that way wears the transistor out and lowers its gain.

It regulates the current through the LEDs.

Whenever the tone pin is high it holds them near 98 mA, which keeps them from getting cooked (they're rated for 100 mA) and keeps them consistently bright while the supply sags, as long as it stays above about 21 V.

Yes.

This is the LED current, about 98 mA.

No.

Q1's gate is insulated, so no steady current flows in.

Yes, a steady milliamp or so.

The pin pulls Q1's gate up through $R_G$ the whole time the hoop is lit, so Q2 has to keep pulling it down just as hard to hold 98 mA. The two pulls balance, and the gate sits at one steady voltage.

It's an N-channel MOSFET that lets more current flow from drain to source the higher the voltage on its gate.

Here it's the valve the LED current passes through.

It decides the LED current.

While the hoop is lit, the circuit settles with about 0.67 V across $R_S$, so $$I = \frac{0.67\,\text{V}}{6.8\,\Omega} \approx 98\,\text{mA}$$

It reacts to the voltage across $R_S$, which sits across its base and emitter.

Around 0.6 V it starts to conduct and pulls Q1's gate down.

1. The voltage across $R_S$ rises.
2. Q2 conducts more.
3. Q2 pulls Q1's gate lower.
4. Q1 closes a little.
5. The LED current falls back toward 98 mA.

B.

This circuit was simulated, and the current was shown to climb and level off within about 7 µs without overshooting or oscillating. This means the system is overdamped. You can learn more about the way the circuit works step by step in this logbook entry.