In any case, pending your reply, I would suggest the following circuit for reliable operation. This will require a P-channel MOSFET, which is different from the two MOSFETs you tried earlier, which are all N-channel. This will also use two resistors. I am making an assumption that your speaker module simply requires two wires at feed it 4 volts, and does not care whether we add a switching circuit to either wire, the positive or negative wire.

This type of circuit would be described as an inverting, low-side MOSFET switching circuit. The inverting part means that when the MOSFET is fed a lower voltage, that causes the transistor to become active, whereas a non-inverting circuit would require feeding the MOSFET with a higher voltage to make the transistor become active.

Low-side switching refers to the fact that the load (ie the speaker module) is permanently attached to the higher voltage (the high-side) and we are manipulating the low-side. Not all electronic loads can be used with low-side switching, but this is the easiest mode to implement using a single MOSFET transistor. As a general rule, to do low-side switching always requires a P-channel MOSFET. WARNING: a P-channel MOSFET has its drain/source reversed from how an N-channel MOSFET is usually wired up. Observe the schematic very carefully.

As for why we cannot do high-side switching (which would use an N-channel MOSFET), it is because a typical N-channel MOSFET requires that the gate be a few volts higher than the source. But consider that when the transistor turns on, the drain and source become almost-similar voltages. So if the drain is attached to 4 volts, and as the transistor becomes active, the source rises to something like 3.95 volts, then what gate do we use to keep the transistor active? If we give 4v to the drain, then the gate-to-source voltage is a mere 0.05 volts, which is insufficient to keep the transistor on. We would need an external source to provide more gate voltage, relative to the source pin. If we tried such a high-side switching circuit anyway, it would quickly oscillate: the transistor tries to turn on, then turns itself off, then back on, and so forth. Or it would sit comfortably at some half-way gate voltage, where the transistor is barely-on, barely-off. This is not useful as a switching circuit.

The way that my suggested circuit works is as follows: when the tripwire (marked as SW3) is in place, then R4 and R2 will form a voltage divider. Given that the battery supplies 4v, we can show that the voltage at the MOSFET’s gate will be 91% of 4v, or 3.64 volts. This should be just enough to prevent the P-channel MOSFET from becoming active. Note: a P-channel MOSFET becomes active when there is a low gate-to-drain voltage, with 0v causing the transistor to become active. In this way, with the trip-wire, the transistor will not allow current to pass through the speaker.

When the tripwire is pulled out, this breaks the connection to R4. That leaves the gate connected to only R2, which is connected to the negative side of the battery. Thus, any charge in the gate will seep away through R2, meaning that the voltage across R2 will equalize at 0v. This means the gate-to-drain voltage will be 0v, which means the MOSFET will activate. And that allows current to power the speaker module.

Note: one end of the tripwire (labeled #1 in the diagram) will still have 4v on it. If the tripwire is cleanly detached from the whole circuit, using your loop-of-wire and nails idea, then there is no problem. But if the tripwire is still hanging onto the 4v side of the circuit, then be careful that the tripwire doesn’t make contact with another part of this circuit. The R4 resistor will still be there, so there won’t be a short circuit or anything bad like that. But if that tripwire reconnects to the gate, then the transistor will deactivate again, stopping the music.

I wish you good luck in this endeavor!

I’m going to try to answer your situation, but although time appears to be of the essence, I need to first understand exactly what you’ve already tried. So bear with me for a moment.

The examples I found were very simple, involving an NPN transistor (2n2222), 10KΩ resistor, battery, and DC Piezo speaker.

With my initial attempt, I wired +4V from the switch to the transistor’s collector and then separated the collector from the base with a resistor. I connected the emitter to the pin that, when the switch is engaged, would send 4V through and power the module.

Does this diagram correctly describe what you tried as a first attempt?

Someone suggested that what I actually needed was a MOSFET …

I have the resistor connected between Gate and Drain, +4V going to drain, and the load from the module on Source.

With an RFP30N06LE, I get about 2V output to Source. With an IRF840N, I’m only getting 0.9V.

Do these diagrams match your circuits with each MOSFET?

What I am not able to understand, in your last photo with the MOSFET, is where the blue wire is going.

I see. Given those constraints then, I don’t see any option besides a new heater. Ideally, the new heater would be built with less circuitry, so there would be fewer things to break.

Looking at the Adax Clea product description, it seems overly complicated for a radiator, IMO. I’m not sure I’d want triac switching for something like a heating appliance. Resistive heating doesn’t strictly require silicon switches, when a relay should work. But I suspect an equally-svelt radiator that’s also simple may be hard to find.

Would it be possible to rewire the supply wires so that it provides 230v line and neutral? That should make it easier (and hopefully cheaper) to select a heater, although the heater would not be as powerful.

My experience is mostly with repairing lower voltage devices (eg 12v to 54v PoE). In your case, a phase to phase short has made quite the mark on that PCB, and being a much higher energy event than low-voltage DC, its possible that some delamination has occurred, with downstream effects on expected trace resistance, capacitance, and leakage/creepage.

Were this a low-voltage board, I personally wouldn’t be worried about those downstream effects. But for AC line voltage, I’d rather buy myself the peace of mind. Do keep parts from the dead board that are salvageable, but IMO, a thermal event on the AC side of a 400vac board would disqualify it from continued service.

P.S. does that circuit not have an onboard fuse? I’m not seeing one and I’m kinda surprised. Presumably an upstream circuit breaker or fuse was what tripped to stop this turning into a fire?

I’m taking a guess that perhaps the fridge makes similar assumptions that automobiles make for their lamps. Some cars that were designed when incandescent bulbs were the only option will use the characteristics resistance as an integral part of the circuit. For example, turn signals will often blink faster when either the front or left corner bulb is not working, and this happens to be useful as an indicator to the motorist that a bulb has gone bust.

For other lamps, such as the interior lamp, the car might do a “soft start” thing where upon opening the car door, the lamp ramps up slowly to full brightness. If an LED bulb is installed here, the issues are manifold: some LEDs don’t support dimming, but all incandescent bulbs do. And the circuit may require the exact resistance of an incandescent bulb to control the rate of ramping up to fill brightness. An LED bulb here may malfunction or damage the car circuitry.

Automobile light bulbs are almost always supplied with 12 volts, so an aftermarket LED replacement bulb is designed to also expect 12 volts, then internally convert down to the native voltage of the LEDs. However, in the non-trivial circuits described above, the voltage to the bulb is intentionally varying. But the converter in the LED still tries to produce the native LED voltage, and so draws more current to compensate. This non-linear behavior does not follow Ohm’s Law, whereas all incandescent bulbs do.

So my guess is that your fridge could possibly be expecting certain resistance values from the bulb but the LED you installed is not meeting those assumptions. This could be harmless, or maybe either the fridge or the LED bulb have been damaged. Best way to test would be installing a new, like-for-like OEM incandescent bulb and seeing if that will work in your fridge.

To start, the idea of charging in parallel while discharging in series is indeed valid. And for multicell battery packs such as for electric automobiles and ebikes, it’s the only practical result. That said, the idea can sometimes vary, with some solutions providing the bulk of charging current through the series connection and then having per-cell leads to balance each cell.

In your case, you would have a substantial number of cells in series, to the point that series charging would require high voltage DC, beyond the normal 50-60 VDC that constitutes low-voltage.

But depending on if charging and discharge are mutually exclusive operations, one option would be to electrically break the pack into smaller groups, so that existing charge controllers can charge each group through normal means (ie balancing wires). Supposing that you used 12s charger ICs, that would reduce the number of ICs to about 9 for a pack with a nominal series voltage ~400vdc. You would have to make sure these ICs are isolated once the groups are reconstituted into the full series arrangement.

Alternatively, you could float all the charging ICs, by having 9 rails of DC voltage to supply each of the charging ICs. And this would allow continuous charging and battery monitoring during discharge. Even with the associated circuitry to provide these floating rails, the part count is still lower than having each cell managed by individual chargers and MOSFETs.

It’s not clear from your post what capacity or current you intend for this overall pack, but even in small packs, I cannot possibly advise using anything but a proper li-ion charge controller for managing battery cells. The idea of charging a capacitor to 4.2v and then blindly dumping voltage into a cell is fraught with issues, such as lacking actual cell temperature monitoring or even just charging the cell in a healthy manner. Charge IC are designed specifically designed for the task, and are just plain easier to build into a pack while being safer.

I don’t think there’s a good way to adapt this circuit to provide current limiting on the 18v rail. Supposing that it was possible, what behavior do you want to happen when reaching the current limit? Should the motor reduce its output torque when at the limit? Should the 18v rail completely shut down? Should the microcontroller be notified of the current limit so that software can deal with it? Would a simple fuse be sufficient?

All of these are possible options, but with various tradeoffs. But depending on your application, I would think the easiest design is to build sufficient capacity on the 18v rail so that the motor and 5v converter inherently never draw more current than can be provided.

I suppose the first question is whether you had the baud rate set correctly. The photo of the “cleaned up signals” (not entirely sure what you did, compared to the prior photo) seems to show a baud rate of 38400, given that each bit seems to take about 25 microseconds.

As for the voltage levels, the same photo seems to show 5v TTL. So it doesn’t seem like you would need a level converter from 15v RS-232 levels. This is one of the few times where the distinction between a “serial port” and an RS-233 port makes a difference, but a lot of data center switches will deal using 5v TTL, because the signals aren’t having to travel more than maybe 5 meters

I’ve never had a need to use a Yubikey, but is that the button which spews gibberish into DMs? I’ve seen that a lot at work on Teams lol

It looks a lot like a Yubikey, which is used to securely authenticate to company resources like a VPN. Fortunately, unlike losing a hard drive, a Yubikey can be deauthorized by the company and thus the device becomes useless for malicious use.

So if you want to use that USB port for something else, it shouldn’t be a problem to remove a Yubikey for the prior user’s employer.

I’ve changed the setting to prevent the behavior, but the prompt is still missing.

You’ve disabled the automatic switching based on HDMI CEC, and yet the TV still automatically switches and without a notification/option in advance? This just sounds like a firmware update for the TV introduced a bug.

I’m in the same camp with the other commenter who suggested never attaching a so-called smart TV to the Internet, for then it can never perform an unwanted update. Because for whatever neat features an update may bring, it rarely can be reversed if proven to be undesirable. I’m staunchly in the “own your hardware” camp, so automatic-and-non-undoable updates are antithetical to any notion of right-to-repair principles, and will inevitably lead to more disposable and throwaway electronics.

[gets off soapbox]

Your best bet might to try attempting a manual software downgrade using a USB stick.

Based solely on this drawing – since I don’t have a datasheet for the PWM controller depicted – it looks like the potentiometer is there to provide a DC bias for the input Aux signal. I draw that conclusion based on the fact that the potentiometer has its extents connected to Vref and GND, meaning that turning the wiper would be selecting a voltage somewhere in-between those two voltage levels.

As for how this controls the duty cycle of the PWM, it would depend on the operating theory of the PWM controller. I can’t quite imagine how the controller might produce a PWM output, but I can imagine a PDM output, which tends to be sufficient for approximating coarse audio.

But the DC bias may also be necessary since the Aux signal might otherwise try to go below GND voltage. The DC bias would raise the Aux signal so that even its lowest valley would remain above GND.

So I think that’s two reasons for why the potentiometer cannot be removed: 1) the DC bias is needed for the frequency control, and 2) to prevent the Aux signal from sinking below GND.

If you did want to replace the potentiometer with something else, you could find a pair of fixed resistors that would still provide the DC bias. I don’t think you could directly connect the Aux directly into the controller.

are not audio drivers but PWM drivers

They can be both! A Class D audio amplifier can be constructed by rendering an audio signal into a PWM or PDM output signal, then passed through an RC filter to remove the switching noise, yielding only the intended audio.

That said, in this case, using the unfiltered PWM output would work for greeting cards, where audio fidelity is not exactly a high priority, but minimal parts count is.

This made me wonder if normal PWM controllers could be used to drive more power full LEDs.

What exactly did you have in mind as a “normal PWM controller”? There’s a great variety of drivers that produce a PWM signal, some in the single watt category and some in the tens of kilowatts.

Whether they can drive “more powerful” LEDs is predominantly a function of the voltage and current requirements to fully illuminate the LEDs, plus what switching frequency range the LEDs can tolerate. Some LED modules that have built-in capacitors cannot be driven effectively using PWM, as well as anything which accepts AC rather than DC power. You’d need a triac to dim AC LED modules, and yet still, some designs simply won’t dim properly.

My idea was to just remove the potentiometer and feed in music from Aux at that point.

You’ll have to provide a schematic, as I’m not entirely sure where this potentiometer is. But be aware that the output current needed to drive a small speaker is probably insufficient to light up a sizable LED, nevermind the possibility of not even having enough voltage to meet the required forward voltage drop of the LED.

Is there a chance of this working?

It might, but only if everything just happens to line up. But otherwise, it’s likely that it won’t work as-is, due to insufficient drive current.

BTW, you might consider posting about your finished results at !micromobility@lemmy.world . We enjoy all things bicycle-related there, especially when it’s solving a unique problem or solving it in unique ways.

(sorry for the long delay)

From your description, I’m wondering if the internal pull-up from the bike computer might actually be an active output, and that the open-drain buffer is causing the bike computer to give up sourcing that pull-up voltage. That is to say, if a larger-than-expected current is drawn from the bike computer, it might trigger a protection mechanism to avoid damage to its output circuitry.

To that end, I would imagine that either: 1) an inline resistor to limit drain current, 2) a push-pull buffer, or 3) both, would help rectify the issue.

My suspicion is based purely on the fact that getting stuck low for an open-drain device could be an issue “upstream”. If it were stuck high, I wouldn’t normally suspect this path.

If you still have the original configuration, measurement of the drain current would be valuable info, as well as the current when the buffer is omitted (when the motor and bike computer are directly attached, a la factory configuration). That would indicate if perhaps the currents are too mismatched.

For your edit #2, can you post a schematic of the relevant part of the circuit? It’s a bit hard to imagine how things are arranged, especially where your pull up resistor at the output of the buffer is.

If I had to guess, perhaps the buffer circuit is going onto latch-up due to ESD spikes, which is then locking the open drain to conduct, which is why you’re seeing a LOW output.

When in doubt, I suppose you can tack on more decoupling capacitors nearby the buffer’s Vcc.

When the buffer gets into this glitch scenario, is the output stuck at high or low?

Switching noise is naturally the first place to look, when an IRQ is firing rapidly and unexpectedly. But have you verified that your IRQ handler is completely handling each interrupt event? And that another interrupt event while handling the prior one will not lead to unusual behavior?

It could very well be a rare, spurious interior firing due to noise, but then exacerbated by an IRA handler that doesn’t clear properly, leading to high speed sprioois events.

What are the approximate sizes for the internal and external pull-up resistors you’ve attached? And what is the impedance for the actual interrupt source, when it actually fires?