Amjad Yaghi

Aerodesign Payload ·

LED driver practice questions

Created a set of flashcards for some other members to learn about our LED driver.

What I’m trying to achieve

Over the last few days I have been working on creating a set of flashcards other members on Aerodesign can use to self-study the LED driver we are using to power our beacons.

Today I would like to publish them on meej.ca.

The circuit and its path to “stability”

+24 V LED1 LED2 to LED11 LED12 LED current 0.0 mA hoop around each of the landing gears drain 0.0 V Q1 IRFZ44 RS 6.8 Ω tone pin 0 V RG 1 kΩ gate 0.00 V ROFF 100 kΩ Q2 2N3904 0.00 mA

*The gate-side currents are about 50 times smaller, so their dashes are sped up 50 times.

0.00 µs

Before the pin goes high The pin is low, so Q1 is closed and nothing flows. Press play, or step through the six stages.

Stage 1 of 6 · 0.0 µs The pin goes high and starts charging Q1's gate through RG.

Stage 2 of 6 · 1.6 µs Q1 starts to open and pulls the drain down. While the drain falls, the pin's 2 mA flows on through Q1's gate-to-drain capacitance, leaving nothing to raise the gate, so it holds near 3 V. This flat step is called the Miller plateau.

Stage 3 of 6 · 5.0 µs The drain has fallen far enough for the LEDs to switch on, and their current climbs.

Stage 4 of 6 · 6.8 µs RS reaches about 0.6 V. Q2 turns on and holds the gate, so the current levels off near 98 mA.

Stage 5 of 6 · 13.0 µs The pin goes low, and the gate drains back out through RG, closing Q1.

Stage 6 of 6 · 14.2 µs Q1 has closed and the LEDs are dark. The gate keeps draining toward 0 V.

0 100 LED current (mA) 0 5 Q1 gate voltage (V) 0 20 Q1 drain voltage (V) 0 1 Q2 current (mA) 1 2 3 4 5 6 0510152025 time after the pin goes high (µs)

Replayed about 500,000 times slower than real life. Our fastest beacon (1340 Hz) holds the pin high for about 373 µs. Here we're only holding it for 13 µs just to show the shape better.

Above is a detailed simulation I ran in ngspice, using datasheet values for our MOSFET and BJT. I broke it down into six distinct stages.

Stage 2… the most interesting one

It’s hard to fit everything in the animation, so I’ll expand on what’s happening in stage two here. The gate capacitance stops charging right at the threshold voltage. This is because with the newly opened pathway from drain-to-source, a dropping voltage on the drain side forces the current incoming from the pin (through RGR_G) to land on the drain side (only briefly).

In the cards I mentioned that no steady current flows through the gate. Here we can see what the brief current is and why it exists. The gate and the drain form a small capacitor, and a capacitor passes current whenever the voltage across it is changing. During stage two the drain falls from about 25 V to about 10 V in 3.3 µs, so:

I=Cgd ΔVΔt≈400 pF×15 V3.3 μs≈1.8 mAI = C_{gd}\,\frac{\Delta V}{\Delta t} \approx 400\,\text{pF} \times \frac{15\,\text{V}}{3.3\,\mu\text{s}} \approx 1.8\,\text{mA}

That’s nearly all of the current coming from the pin:

Ipin=Vpin−VgateRG≈5 V−3 V1 kΩ=2 mAI_\text{pin} = \frac{V_\text{pin} - V_\text{gate}}{R_G} \approx \frac{5\,\text{V} - 3\,\text{V}}{1\,\text{k}\Omega} = 2\,\text{mA}

With nothing left over to raise the gate, it stays flat. This brief current stops once the drain stops falling.

Driving questions

  • To what extent should I explain the individual components?
  • How shall I handle the “what-if” type questions people naturally have about circuits?
    • “What if I flip the power?”
    • “What if I forget a resistor?”

Spent a lot of time getting the format right… I settled on a set of ~55 questions.

Here’s the link: meej.ca/learn/led-driver

Next

  • Send this off to the new recruits and hope it was useful!