If the relay module needs pulling high to turn it off, then you need a pull-up resistor to 3.3V, not a pull-down resistor to GND. I'm not sure what the 1K resistor is for.
On most microprocessors, a GPIO pin can be in one of three states:
Input, high impedance
(Because it's a GPIO pin, not a GPO pin).
On powering up, the safest ...
As an aside: you are using a relais module, with optocouplers on its inputs, so your schematics are a bit misleading.
Anyway: the Pi could be doing all sorts of things with its IO pins on start-up, I don't know the Pi well enough to tell you exactly what happens, but high-impedance is the most likely.
You need a pull-up resistor instead of a pull-down ...
how to access all 256 pins to read and write from/to them simultaneously?
Not at all. Simultaneity requires the same amount of inputs, and you have less.
Normally, you'd just use a parallel-to-serial shift register as input, or a serial-to-parallel shift register as output here, if you need to interact with all pins regularly, instead of just selected ones.
I want to improve the precision of the draw and I don't know if the problem is a sequential software problem
I suspect you are currently limited by the absolute resolution of one step (1.8 degree). Parallel programming can't fix this. It will however be required later on when you need to calculate a path in advanced from a defined source to follow at a ...
Wire lengths + breadboard!
If you look at the datasheet, chapter 8.2.1, they show that the input and output caps and the diode should be connected at a single point close to the GND pin of the IC.
Looking at your picture, you are using a GND rail, as well as "long" (non negligible inductance) GND wires everywhere.
I strongly suggest that you try to ...
The common 9 volt rectangular batteries cannot supply 3 Amps - they are designed for low current applications - I think even 100 mA is a bit much for them.
The LM7805 is a linear regulator and will waste over a third of the power you draw from the battery as heat.
Four AA cells and a DC-DC converter (switching power supply) may work, but 3 Amps is really too ...
It is a CMOS chip with TTL compatible IO levels. It will understand 2V as logic high. Make sure you don't accidentally set the chip to read mode, so that it does not output the 5V supply to the 3.3V IO pins of RasPi.
The normal logic inputs are recognized as a logic 1 if the voltage is above Vih (2.0 volts).
However, CMOS devices often take current from the supply if the input level is not close to the supply rail.
The level at which they specify the standby current when using CMOS level inputs is 100uA(Isb2), but to guarantee that the CE needs to be at Vdd-0.3v.
If you ...
I just need to spin the motor, one direction, for about a second. It is for an automatic food dispenser, so just a burst for a second or so
Unidirectional control of a DC motor is typically accomplished with an N-FET switching the motor's negative power lead, while the positive remains permanently connected to the supply.
Many of the best modern FETs for ...
@mguima has provided a useful link in his answer. BUT do note Chris Stratton's comments also.
Important considerations include:
Any motor tends to make electrical "noise" which a microcontroller is very sensitive to.
Usually the power needs of a motor exceed that which what can sensibly be fed from a Pi header directly.
The servo voltage needs ...
1.) whith positive voltage current will flow through Q2A and as it is a current mirror, it will also open Q2B. Together with the parasitic diode of the mosfet, it will rise the voltage at R24. This will open the mosfet -> normal operation.
In case of under/reverse voltage, the circuit will close the mosfet (current mirror has a high gain, so it will ...
I use a 9 volt battery snap connector which connects the battery and barrel jack of the arduino uno(I am assuming that the arduino uno schematic is same as original one) as shown here. Raspberry Pi 3b(original schematic) has a USB type power connector and you cannot use a 9 volt snap connector like arduino. It is better to use a power bank or a wall ...
While breadboarding is not ideal, I'm not so sure that's the problem.
Look at page 5 of that datasheet : notice the conditions for the output voltage specification:
0.5 A ≤ ILOAD ≤ 3 A
It's fairly common for switching supplies to require some minimum load before they can regulate properly. Otherwise the output voltage drifts up out of specification.
I'm not ...
The second solution with lead acid would be the easiest solution, bare in mind that lead acid need to be refreshed after deep discharge, and maintenance charged.
Li-ion need extra care when charging and discharing, at least least dubble protection for OVP and over current i recommended preferable with a hard fuse aswell. Besides the charging circuit. ...
The RPi4 standard psu is 5V 3A. As mentioned linear regulators are not going to be very efficient in voltage conversion so you will be losing a lot of power and therefore duration the battery could have provided. You want to look at a switching based converter and a higher capacity battery technology. Take a look at https://www.adafruit.com/product/1385 ...