This is a JTAG connector used for development and testing. Here's a picture of an old Pi B+ featuring the signal names printed on it:
AFAIK it's useless unless you have VideoCore documentation and tools from Broadcom.
There is more than one type of motor driver board using the L298N.
The typical board has two voltage inputs and a common ground.
One voltage input is to drive the motors. The other voltage input is to provide logic power to the module.
Typically the board has a jumper which can be fitted to supply logic power from the motor supply. If that is fitted DO NOT ...
6V motors usually work fine with 5V (other than the fact that they run at 80%..85% of it max speed). However, powering a motor from the Pi is only possible for very small motors, which have stall current that the Pi can provide without a significant voltage drop. Even toy motors are often rated for 2A stall current or more, which can easily reboot the Pi ...
Since there are only output from the Raspberry Pi that should drive the 4026 7-Segment Counter there are only need for a driver circuit.
This can be done with transistors, MOSFETs or a driver IC.
A simpler way is to change the voltage for the 4026 from 9 volt to 3,3volt its within the IC's working parameter (3-15v).
And then replace the current limiting ...
I don't see any reason for what you describe to have killed the Pi. I regularly use UBECs to power the Pi and that is pretty much the same (apart from the output voltage being preset).
As to what happened in your case. Either the wrong polarity of the input voltage was quite a lot more than 5V.
The "Coin acceptor" GND (or minus) needs to be connected to the Raspberry Pi's GND.
And replace the voltage regulator with a voltage divider (https://en.wikipedia.org/wiki/Voltage_divider).
Where Vin is the voltage from the ""Coin acceptor", I assume its 12 volt and the 3,3 volt that the Raspberry Pi accept.
So the formula is Z2 = 3....
There might be other things wrong with your design, but one that jumps out of the figure in your question is the Voltage Regulator step down to 3 V.
You don't need a "voltage regulator" per se - you simply need an interface to deliver either 0V or ~3.3V to the GPIO from the coin line. There are several ways to do this, the two most obvious are:
Any GPIO which is not connected to a voltage will float between high and low.
You need to give the GPIO a fixed voltage by supplying a pull to 3V3 or a pull to ground.
The weak internal pulls (about 50kohm) can easily be overcome by noise.
Try adding an external pull of around 5kohm or lower until the GPIO is stable in your environment.
What sort of frequencies are you talking about?
pigpio can do this with waves (hardware timing).
My new lg library can do this using software timed PWM.
E.g. Python http://abyz.me.uk/lg/py_lgpio.html#tx_pwm
pigpio example using pigs (command line)
In my younger years (long before microprocessors) I spent some time working on automotive electronics. This is a HOSTILE environment - motorcycles are an order of magnitude worse.
It is possible BUT is somewhat a specialist field. You NEED an isolated supply, extensive shielding, all inputs should ideally be galvanically isolated and any connections need ...
It's not clear from your question that you need to measure current. As a practical matter, measuring voltage is often easier.
@Dougie comment suggesting an [opto-coupler (a.k.a. opto-isolator)] is a good one as it provides galvanic isolation - an important safety consideration working with mains voltages. As you've seen in the comments, there are various ...
AWG = "American Wire Gauge" is a standard for the thickness of wire (check Wikipedia). AWG 22 is equivalent to a wire surface of about 0.3 mm^2. That's fine to be used in breadboards. If you don't need to transmit a lot of power, that should be fine. If you use it also for the power lines, you might need thicker wire or take several wires for ...
If I understand your objectives correctly, it seems that your idea is generally sound. But, the devil is in the details as they say.
Reviewing your diagram:
It appears that the purpose of the Low Battery Cutoff Switch and the N.O. 12 V Relay is to signal the RPi (via GPIO input) that it should execute a shutdown soon. However, 12.3V seems too high for a low-...
Is it possible to replace it to save this rpi?
Maybe... But it is literally impossible for anyone here (remote, no hands-on) to tell you if your board is repairable or not. I don't know if the odds improve significantly if we were "hands-on".
All you can do is try replacing the component that you blew off the board, and see what happens. The RPi ...
After spending ~ 20 min searching for an actual data sheet on the SH5461AS - and not finding one - I'm of the opinion this is JAPCJ (just another piece of Chinese junk). My best advice is chuck it in the bin & find a part by a reputable manufacturer. There are lots of manufacturers that publish spec sheets (even on parts as old as this one) - here's an ...
What you are missing is a specification: "How much current does your LED Tinsel require to operate?" Give us that info, and we can give you a definitive answer - without it, we have to guess - and that's a waste of everybody's time, no?
The GPIO pin can not supply enough power to drive your LED Tinsel. You can do some research on that. ...
When the GPIO pin is set to OUT mode, it can be in two states, LOW or HIGH. Have you put your pin into HIGH state? If yes, then it's possible that the GPIO doesn't provide enough current to power a LED tinsel.
The converter you have found is tiny and there is no way it could provide 3A. From the looks it could continuously deliver 0.3..0.5A, not more.
For comparison, this is how an actual 4.5A buck converter looks like:
When you connected it to the Pi, the converter got overloaded, and, having no overload protection (also not surprising for $1/piece module) it ...
Note: I'm not a 100% sure so please correct me in the comments.
I think you could use 4 USB hubs/splitters that would connect to 4 10A bricks. You would then get 20 Micro USB cables to connect to the individual Pis from the Hubs. Because you have a 10A Power supply, it would supply 2A each to the Pis.
Another solution, if you're willing to buy a power supply,...
Here's one way to do it: combine two (or more) batteries in a "wired-OR" configuration. This is fairly generic as you provided no details on your battery voltages. If you want to edit your question to add those details, we'll try to be more specific.
simulate this circuit – Schematic created using CircuitLab