I found a car-cigarette-lighter USB charger on the street with a missing contact spring and some broken plastic. It’s labeled as having one 1-amp USB port and one 2.1-amp USB port, although in fact these are wired strictly in parallel. It’s fairly transparently a simple buck converter on a single-sided PCB (silkscreened “HT-668”), built around a buck-converter through-hole 8-pin chip labeled “LC51”, followed by some kind of fraction-like sigil. There is some residual flux from sloppy hand soldering on the bottom of the board.
“LC51” is not a useful marking for finding the correct datasheet on search engines; it turns up many datasheets for chips used in this application but they all have the wrong pinout.
C1 is a 22-μF 25-V electrolytic across the nominally-12-volt input, with its - terminal on pin 4 of the IC and its + terminal on pin 6. L1 is an unmarked ferrite-core inductor across pins 2 and 3, and "D1" (with the “1” silkscreened upside down) is what looks like an ⅛-watt rectifier (labeled, I think, “N5 MI”) across pin 4 (ground) and pin 2 (the stripey end of the diode package). “CS” is a 100-μF 10-V electrolytic on the output, which is connected to pins 4 (on its - terminal) and pin 3. There is a blue LED connected across pins 4 and 5. Pins 1 and 7 are shorted to pin 6. The USB connector shield is tied to pin 4.
Amusingly, there are no resistors.
So I infer the following pinout for the chip:
Except for the indicator LED, this is a textbook buck converter circuit. It seems strange to me that there are 3 pins for Vin and only 1 pin for Vout, but I guess most of the output current (7/12 at nominal 12V input) really flows through the diode and not the chip. Still, one would hope that the amount of current flowing to ground through the chip’s internal circuitry is at most a few milliamps, not hundreds of them, so you’d want the same number of input and output pins.
None of the components have any visible damage, but on connecting it to a 12-volt power supply sourcing an amp or two to a LED illumination panel connected in parallel, the onboard blue LED lights up, but there is no 5-volt output. None of the components heat up noticeably.
After unplugging, some residual voltage is visible on both capacitors, tens to hundreds of mV, so they are not shorted.
The diode reads as a 170-mV voltage drop with this multimeter a friend lent me, but I’m not sure how correct that is. That seems too low to be a working diode. But in the other direction it looks like an open circuit, so I suspect the meter. However, unless I’m reading this wrong, the meter is telling me not only that the diode has an implausibly low forward voltage drop but also that it is installed backwards, i.e., passing current in the +5V-to-ground direction. Maybe the diode is blown and the voltage I’m seeing is actually from reverse-biasing the output electrolytic, or some kind of input protection circuit in the SMPS chip itself?
If we assume that the ⅛-W rectifier diode is a 300-mV Schottky, then it can carry an average of 417 mA, which means that the whole board can only supply 12/7 of that — 714 mA — without overheating the diode. So the “2.1 A” output rating molded into the plastic is probably unreasonably optimistic. There is also nothing connected to the data pins (two of them may be shorted together, but this looks like an accident) so no smart current negotiation can be going on here, but even without current negotiation, you could plug two 500-mA devices in at once, which would mildly overheat the diode.
It probably isn’t really okay to use a 25-volt electrolytic on the input of this charger, either, because car electrical systems are reputed to have inductive spikes that go higher than that. You’d probably be okay if you never left it plugged into the cigarette lighter when the car was turning on and off.
So the design seems a bit dubious.
Most of the chips that come up in the “LC51” datasheet search use fixed-frequency oscillations in the tens to hundreds of kHz. Maybe if I hook this up to a voltage source I could observe its oscillation frequency.
I don’t know the inductance of the inductor, so I don’t know how fast its voltage would ramp up and down, but at 1000 mA difference between input and output you would drain the output 100-μF capacitor from 5 V down to 0 in 500 μs, and from 5 V down to 4.5 V (which I am inferring is probably about the limit of the acceptable voltage deviation, although I don’t remember from the USB spec) in 50 μs. So it seems like a fair bet that the chip’s designed oscillation frequency is at least a few tens of kHz to keep the thing from falling over without a larger output cap.
The inductor presumably needs to be able to ramp its current up and down between 0 and 1000 mA over a similar time period — otherwise it wouldn’t be able to respond fast enough to things being plugged and unplugged. So a sensible inductor size would be in the neighborhood of 50–100 μH. When the chip’s switched output pin is floating, the inductor has to maintain some significant fraction of its ≈1000 mA against ≈4.7 volts (the 5 volts of the output capacitor minus the presumably 300 mV of the rectifier) for ≈50 μs, which would require at least 250 μH or so. When the chip’s switched output pin is high, the inductor’s current would have to be able to ramp up from 0 to ≈1000 mA when driven by ≈7 V (12 V - 5 V) in ≈50 μs, which would require no more than 350 μH. In practice the frequency is probably considerably higher, so the time period is probably more like 5–10 μs, with a correspondingly smaller inductance.
(But there’s no guarantee that they did in fact use a sensible inductor size.)
If the inductor were 100 μH, the LC time constant √(LC) would be 100 μs. I think the resonant frequency is 1/(2π√(LC)) ≈ 1.6 kHz, so it is surely much lower than the frequency being used.
Presumably its feedback pin is divided down internally and compared to a temperature-compensated internal bandgap reference; this feedback (plus rectifier and capacitor leakage currents) is the only open-circuit load on the power supply, and so it might be designed to be a heavy enough load to maintain regulation when nothing is plugged into it. The chip, though it’s a through-hole DIP, is a super tiny through-hole DIP, and although its package seems like ceramic rather than epoxy, it’s probably not capable of dissipating more than ¼ W, which implies a maximum current of 50 mA into the feedback pin, and probably more like 5 mA.
These are probably obvious, and maybe a bit goofy given that the thing is apparently broken.
Presumably the duty cycle of the chip’s switched output is variable from under 10% to at least 50% (probably 100%), and this is controlled by the voltage detected on the feedback pin. By cutting a trace on the board and hooking up the feedback pin to the center tap of a pot between the output and ground, it should be straightforward to get a more or less regulated, temperature-compensated output from 5 V up to at least 50% and probably 95+% of whatever the input voltage is; I’ve inferred that the current through the internal feedback divider is probably several milliamps in order to satisfy the minimum-load requirement of the internal circuitry, so the current through the pot would probably need to be tens of milliamps if you want it to be very precise. But that would imply a pot < 1kΩ, which would be hard to find.
Lower output voltages would require pulling the feedback output up toward some higher voltage. The simplest approach might be to totally fake the feedback — hook up a variable voltage divider across the input that lets you generate a voltage somewhere around 5 V to feed to the feedback pin, and use an external resistor to fulfill whatever minimum-load requirements the switcher has. That might work, but depending on the feedback scheme, it might always rail at Vout = 0 or Vout = Vin. Including some of the real output voltage in the summing junction should make it possible to stabilize it in such cases. Any of these schemes depend on a regulated input voltage, since the divided input voltage is going to be compared against its internal bandgap.
You might be able to get a regulated output current rather than voltage by the usual trick of floating the circuit’s ground at a distance below its filtered output determined by a current-sensing resistor in series with that filtered output. For example, if you want to set the output to 100 mA, you’d use a 50-Ω resistor. I’m not sure the circuit will start up properly in this situation, though, and because there’s 5 volts across the resistor, it’s potentially going to burn up a lot of power.
Soldering in a bigger diode would probably enable higher output powers, although without current negotiation, compliant USB devices will not really take advantage of that. Amplifying the output with a large output transistor to get tens of watts of output power might be a fun thing to try, too, though at the cost of losing the overcurrent protection presumably included in the chip.
If the output frequency is fixed, a resonant LC notch filter or two across the output should be able to attenuate the EM noise from the switcher by 20 dB or so.
Presumably you should be able to use the chip as a decent class-D audio amplifier by adding the audio input signal into its feedback pin in order to modulate its output voltage, which, as I said above, needs to be able to respond across the whole audible spectrum in order to avoid output overvoltage when you unplug USB devices. It should be able to output 5 watts or more of output at efficiencies of around 90% into impedances in the usual 4–8 Ω range, but you will need an ac-coupling capacitor on the output to avoid a humongous 5-V dc bias on the speaker output.
Depending on what the output frequency actually is, and whether it’s fixed or variable, you might also be able to use it as a micropower AM or FM radio transmitter if you bias the audio input into the right range (and add in the appropriate fraction of the filtered output voltage, if necessary) on the feedback pin. Even if the baseband output frequency isn’t in the right band, one of its harmonics may be, and by finely adjusting the duty cycle you can finely adjust the harmonics’ amplitudes. (See Processing halftoning for some notes on this in the spatial domain.)