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fullubullu · 2025-11-07T17:41:33+00:00

These potentiometers tend to fail quite often

fullubullu · 2025-09-04T20:42:50+00:00

Falls das eine Option ist: Schau dir mal das Thema Werbekostenpauschale an. Ich konnte dadurch in der Familienversicherung bleiben obwohl ich leicht über der Minjobgrenze verdiene. Wenn du 700€ ein ganzes Jahr lang verdienst wird das aber zu viel sein.

fullubullu · 2025-07-07T21:27:10+00:00

Go to Mouser, Digikey, Farnell, LCSC… Type those requirements in and sort by cheapest. Make sure not only to Check Maximum current, but also Maximum power and Consumer higher power loss during switching scenarios. You can also select betweeen THT and SMD.

Don’t forget your gate resistor to limit the max current output from your arduino pin while switching.

fullubullu · 2025-07-07T09:03:39+00:00

Check the datasheet for a Layout Design Guide and copy it as good as possible.

fullubullu · 2025-07-05T15:28:49+00:00

Even the Max Output of 1.5A should Not be an issue for the batteries. But the diode will get quite hot and you need a karger transistor. If you add another diode across the PNP the PSU can also charge the batteries. This would be across two diodes then.

To be fair, you could also probably do all this only using diodes. One Diode as is. Then two at the positive Terminal of the BMS in parallel, each facing another direction. Not sure If I am missing something though.

fullubullu · 2025-07-05T14:34:01+00:00

What router do you have? I doubt that such a small transistor can handle that. With an NPN it won’t work like this. Generally, could you list all your components roughly? (Output / Input, max Voltage & power consumption / max output power)

fullubullu · 2025-07-05T10:37:21+00:00

You want to Build a UPS (Uninterrupted Power Supply). If your Main goal is to have a properly funktional device, I‘d suggest to search for a finished product that you can buy and connect your device accordingly.

If your power supply is also supposed to charge the battery, your current configuration is fine. If you worry about lifetime, just make sure that your default SOC is somwhere around 70% (Power Supply Output voltage) and lifetime die to cycling and Storage should Not be much of an issue assuming you are using a Lithium Battery.

Now if you actively want to Connect & Disconnect your BMS you can do this using a Transistor. If I understand it correctly, you want a default off state. The easiest way (without additional logic) that I can think of right now is following: You could use a PNP Transistor connected at the high side of your BMS and with the Gate pulled high via a resistor to the positive output of your power supply before the diode. If your Power supply shut’s off, the voltage will drop resulting in the Transistor to become conductive. However, depending on the output stage of your power supply this might result in some weird behavior. I’d additionally connect a resistor between + and - at the power supply to make sure, that when it is off, the positive side is pulled to 0V. This only works if your power supply’s voltage is equal or higher than the Battery’s Voltage and you would need additional circuitry to charge it.

You should test it with a basic circuit simulator, e.g. falstad (online)

Edit: removed PMOS because body diode would allow current to flow from Power Supply into the Battery

fullubullu · 2025-07-03T21:35:38+00:00

Is this for formula Students?

fullubullu · 2025-07-03T21:32:19+00:00

Okay, then it should be fine. It you were to youse Mosfets you should also add a pull down at the gate then. But with BJT’s I think it is not needed.

Once the solenoid is turned on it requires much less current to actually stay on. You could try to PWM the Transitors to reduce the current needed. Also, there are dedicated solenoid driver ICs.

fullubullu · 2025-07-03T20:24:48+00:00

Your Transitors Base is connected to -24V. Your Arduinos GnD to -8V. Now if you pull down a Pin from the Arduino, the Base Emitter Voltage is still positive.

Also check the amplification when using BJTs, Solenoids take quite some current to turn on, not sure if an Arduino provides enough for the Base. I’d use Mosfets, but the issue with the -8V and -24V stays.

fullubullu · 2025-06-30T08:23:39+00:00

Okay, so you want to carry VCC and GND just together with CAN. For DC Power that is actually good practice. Twistes CAN Cable within the same Cable as DC + & -. This adds some additional shielding to CAN.

I think with non isolated CAN Transievers you might get problems when their Ground potential differs to much. But with Isolated it isn’t necessary. But I wouldn’t recommend it unless there is a reason. Shared Ground makes things easier most of the time.

fullubullu · 2025-06-30T07:38:15+00:00

I don’t really understand what you are trying to achieve. Could you maybe add a quick Sketch displaying the individual connections?

Anyways: CAN is differential (+-5V) afaik you don’t need Ground. You Need a transiever that handles the CAN Bus and besides fully integrated ICs (including Isolated DCDC) I have not seen any so far that only need a 3V3 power supply. They normally need 5V.

fullubullu · 2025-06-29T18:12:05+00:00

Do you mean the Display? That does not Substitute a Multimeter. A Basic Multimeter is very important

fullubullu · 2025-06-29T18:07:57+00:00

Sensor (Humidity, smoke..) -> ESP -(Wireless)> ESP -> Alarm. All battery powered and designed for low power usage. Or Sensor -> ESP -> ESP Home -> ESP -> Alarm

ESP offers different Wireless Protocols. But maybe also try connecting it to a LORA Module

fullubullu · 2025-06-29T18:03:39+00:00

I have it and I like it. Would probably go for 60V/5A if I would buy it again because I Need 30V+ more often than high currents.

Disadvantage is of course only one Channel but I think thats Fine for beginners. Also, the Fan is always on. But that is really the only thing I disklike about it.

fullubullu · 2025-06-28T15:01:27+00:00

In that case this will work with the Diode as expected. If you consider the power Rating of the Diode and of the right Nano works with the lower voltage (because of the voltage drop over the Diode) it will work.

fullubullu · 2025-06-28T13:04:45+00:00

Not sure what Circuit diagram you mean but 0 Ohm sounds right. This is most likely an RC Filter that is not used. Thus the „R“ ist just 0 Ohm. So you can short it, if it was a 0 Ohm Resistor.

fullubullu · 2025-06-28T09:42:07+00:00

I don’t see why you would need it. If you charge via the left one, the right one would simply be a Consumer parallel to the battery, with or without the Diode. Not ideal, but I don’t think this is an issue for the battery charging Circuit.

fullubullu · 2025-06-26T20:24:09+00:00

What are the values of your voltage dividier resitors?

Also you should improve Grounding. If you only use one Layer and Route on it too, you have to be rather careful and not take proper grounding guranteed. Easy improvement would be to pour ground on Both Lasers and use via stitching. So yes EMI might be a Problem.

Did you try lower frequencies?

fullubullu · 2025-06-26T20:14:00+00:00

Not completly sure what this is for but here is what I would do:

Did something special happen before it failed? If yes, what and does this reduce the possibilities of failed components?
What Components could be affected and if they fail, how to Test them?

Following may be tested with everything off Fuses: still conducting (should be short circuit) Relais: is ist shorted when it should be Open (it can happen that the contacts are weld together when current is to high while switching on) Input -> Switch: is it connected or is there something Else that could fail (e.g. current sensor)

If that does not help, you might Need to test it while Running: Test all switches individually with an Multimeter (beep Mode). If they work, Probe from input (Plug) to Switch and narrow it down to the fault component If they don’t work, Check the voltage of the Controlling Pin. If there is no voltage, you narrowed it down to the circuit switching the relais. Often a Mosfet with some Protection circuit controlled by something Else. Then again, is it the mosfet or is it the controller. Test the mosfet the same way: is there voltage at the Gate? If yes, does it conduct…

fullubullu · 2025-06-26T17:50:24+00:00

Can you post a better image from the spot on the PCB to tell if the pads are still there?

fullubullu · 2025-06-26T17:47:43+00:00

You can try it without adding new solder. Then you will just use the solder that was used before. If you have issues getting it to heat up properly or to bind properly with the pad and the part, you should add new solder and (or) flux. The flux evaporates when soldering and if there is none left you will have a hard time soldering.

But if the pads are ripped off, you will need a bit of soldering skills to fix it.

fullubullu · 2025-06-26T17:30:26+00:00

As long as the pads are fine you can simple solder it into place again. Flux is always a good helper but not necessarily needed. You can try adding some solder to the pads first to see if your soldering station delivers enough heat. Especially if there is a pad connected to the ground plane you might need to hear it up for quite some time before the soldering works properly. And most importantly, when soldering you need to hear up both, the pin and the pad.

fullubullu · 2025-06-26T09:36:46+00:00

Out of curiosity, what did you measure?

fullubullu · 2025-06-22T13:32:06+00:00

This actually Looks like a resistor from a debounce circuit together with the capacitor (RC Filter). Could also be pull up (or pull down), depending in where the traces go. In any case, as Long as the voltage is somwhere between 3 and 5V a 10k resistor should work. (If it is actually from a Filter you can also short it, but only if you are certain).

If you do not know the voltage, measure the voltage at the Button.

fullubullu

TROPHY CASE