Tag: Improvements

Manually selecting the current through the 1 W amber LED with a switch actually intended for LED flashlights:

The resistors on the low side of the LED use the MP1584 regulator for current control, with the orange wire feeding the resistor voltage into the error amplifier.

The 15 Ω unswitched resistor sets the LED current at 53 mA = 0.8 V / 15 Ω, with the LED dissipating about 100 mW. The resistor dissipates 43 mW.

Closing the switch puts the two parallel 4.7 Ω resistors in parallel with the 15 Ω resistor to produce 2.0 Ω, which sets the LED current to 390 mA and runs it at 950 mW. Each of the 4.7 Ω resistors dissipates 140 mW.

That much power raises the aluminum body to 50 °C = 120 °F: definitely uncomfortable but probably survivable for the LED inside.

Eyeballometrically, a decimal order of magnitude difference in the LED current produces an obvious brightness difference. My first try ran the LED at 500 mW (a binary order of magnitude less than 1 W) and wasn’t visually different. Given that the LED will run from the Bafang’s headlight output, saving power isn’t all that important.

If this is the first time you’ve encountered parallel resistors, this is why your calculator has a reciprocal button: the total resistance is the reciprocal of the sum of the reciprocals of all the resistances:

1/R = 1/R₁ + 1/R₂ + …

A real engineering calculator does not have a shifted reciprocal function.

Mostly because I wanted to verify that it really worked:

The Arduino Nano runs the default Blink program that all the knockoff manufacturers use as their final QC test.

The MP1584 specs say the Enable input can accept a logic signal up to 6 V, the Nano runs at 5 V regulated down from the 6.3 V from the bench supply, and the 1 W red LED now flashes 1 s ON / 1 s OFF.

The current feedback works as it did before, too, which is comforting.

The Nano adds 20 mA to the bench supply, so the whole affair runs at 220 mA = 1.4 W. Of course, it’s now at a 50% duty cycle, so that helps.

I doubt hand-hewing an astable multivibrator is the right way to add blinkiness, but it’d definitely be playing on hard mode.

A red 1 W LED works just as well as the amber LED from an MP1584 regulator hacked into current feedback mode:

I started with the same 1.65 Ω sense resistor and got the same 484 mA current, with the LED forward drop at a surprisingly high 3.3 V = 1.6 W. Ouch.

Adding a 1 Ω series resistor to get 2.65 Ω lowered the current to 300 mA with a forward drop of 2.45 V = 740 mW.

Running the numbers suggested a 2.3 Ω sense resistor made from a pair of parallel 4.7 Ω resistors, which produced 346 mA and an LED drop of 2.66 V = 920 mW. The resistor dissipates 280 mW.

The bench supply provides 6.3 V @ 200 mA = 1.26 W, so the overall efficiency is 94% and the LED burns 73% of the input.

I expected the red LED would have a lower forward drop than the amber LED, but it’s actually higher.

Word: Trust, but verify.

This actually worked out the way I expected:

The PCB is the generic MP1584 buck regulator, as seen before in its normal voltage feedback mode, rewired to get feedback based on the LED current, so that it adjusts the output voltage to maintain a constant LED current, regardless of LED forward drop variations.

Pin 4 normally sees the output voltage divided down to the 0.8 V error comparator reference voltage:

Yes, the MP1584 is “not recommended for new designs”, which surely accounts for the myriad cheap regulators built around it. Somebody picked up a great deal on a vast pile of obsolete ICs and is passing the savings along to us; there are exactly zero hits for MP2338 buck regulators.

Putting the ballast resistor on the low side of the LED turns it into a current sensor:

Pick R to drop 0.8 V at the desired LED current and It Just Works™.

The two 3.3 Ω resistors in the top photo produce a 1.65 Ω sense resistor to set the LED current at:

485 mA = 800 mV / 1.65 Ω

It actually works out to a bit higher than that, because I stuck a 100 Ω resistor in series with the feedback input. The PCB still has the 8.2 kΩ resistor from the original voltage divider, so the error amp sees only 99% of the sense voltage, but it’s close enough.

With 6.3 V and 0.28 A = 1.76 W from the bench supply over on the left, the regulator puts 490 mA through the LED. The LED drops 2.54 V = 1.24 W and the resistor drops 0.809 V (that 1% thing) = 0.4 W for a total of 3.35 V and 1.64 W. The regulator is 93% efficient, although the resistor burns a quarter of the energy.

One could use a Hall effect current sensor and an op amp circuit to deliver the proper feedback voltage without resistive loss, but I think burning half a watt is Good Enough for the purpose.

One could add parallel resistors with MOSFET switches to set the LED current. An unswitched resistor would set the lowest current, with switched parallel resistors lowering the resistance, raising the current, and brightening the LED.

The PCB leaves the Enable input floating with an internal pullup. Grounding the pin shuts off the LED as you’d expect, so I can blink the LED without any further hassle.

One could imagine simultaneously blinking and brightening the LED as needed.

That was surprisingly simple …

The PCB wrapping a buck regulator around an MP1584 chip uses a tiny trimpot to set the output voltage:

The 01D resistors use the EIA-96 identifier series and are 100 kΩ.

Based on simpleminded testing, a 1 W amber LED drops about 2.5 V at 430 mA. A 1 Ω ballast resistor drops another half volt and burns a quarter of a watt, sufficient to cover some LED forward drop variation.

The trimpot is entirely too twitchy, so I replaced it with an SMD resistor:

The trimpot read 26.5 kΩ after I extracted it, but I surely nudged it a smidgen in the process.

For the record (first column is SMD topmark, second is measured resistance):

• 3012 = 29.9 kΩ (!!) → 3.67 V into a 100 Ω resistor
• 2492 = 24.9 kΩ → 3.19 V : 2.63 V @ 550 mA = 1.45 W
• 2362 = 22.6 kΩ → 2.97 V : 2.52 V @ 450 mA = 1.13 W
• 223 = 22.0 kΩ → 2.91 V : 2.484 V @ 425 mA = 1.06 W

With 6.3 V @ 210 mA = 1.3 W from the bench regulator, the resistor now burns 180 mW at 425 mA and the LED burns 82% of the input power.

Letting it cook overnight settled out with the LED at 2.47 V and 440 mA = 1.09 W, with 6.3 V at 220 mA = 1.4 W from the bench supply. The LED dissipates 78% of the input power and the resistor burns 190 mW = 14%, so the regulator uses 120 mW = 8%.

I can come close to the final output voltage by plugging the new resistor value and the 8.2 kΩ resistor (on the PCB) into the MP1584 datasheet equations, but figuring the resistor to get a specific output voltage seems largely empirical.