The goal is to have no power consumption after the stepper motor has performed its action. I heard you can use MOSFET or relays for this. It'll be nice if there is a mechanical timer that can cut power via hardware, by literally opening the circuit from the 5V power source. I plan to power them all from a 5V source. I know with the Pro Mini you can go into deep sleep but the LCD and Darlington will still draw significant current I think. This is because the program don't need to be ran again for days. After the user start the program, there is a wait of 5 min, and the stepper motor performs an action, afterward I want to cut power to everything to save energy. Quote from: engineheat on February 12, 2019, 04:16:44 am I got an Arduino application that uses a 16X2 LCD and a ULN2003 darlington array for driving a stepping motor. The most common way to prevent this is to restrict physical acces to the button, for example to mount it in a slight recess. This may be an issue if the circuit is carried around in a rucksack or your back pocket or can get stuck in any other way. As long as the button is pushed, it is not possible to turn the power supply off. It turns the whole shebab on when it is first pushed, and when the circuit is already on it is used as a push button on an input of the AVR pin. In this circuit switch S1 has therefore 2 functions. If the button is released then pullup resistor R14 pulls pin PD7 high again. When the ATMEGA328 is on however, T2 pulls pin PD7 low whenever the button is pushed. When switch S1 is initially pushed the ATMEGA328 and T2 are at 0V and the transistor just acts as a diode. T1 is the same as the NPN transistor in Peabody's power circuit, but a led is added because it was a convenient place of a low current path. T3 has the same function as the Peabody's MOSfet. It also shows the voltage regulator (IC2) which might as well have been a LM7805. Power is coming in from a barrel jack or 9V battery. If you like schematics, here is the power supply circuit of the transistor tester. When the arduino is finished with it's job, it makes the I/O pin low, which turns off the NPN transistor, the voltage on the Gate is pulled high by the 100k resistor and the MOS fet turns off. The normal duration of a button push can be 500ms or lower, so that is all the time the arduino has. This opens the NPN transistor, which keeps the voltage on the gate of the Fet low after the button is released. When the arduino receives power, the first thing it should do is to make the I/O pin high. You can add some voltage regulator between the Drain of the Fet and the Arduino, or use the arduino power supply circuitry if it has any. and this effectively forms a short between Source and Drain. When the push button next to the transistor is pushed, the gate of the fet is connected to GND and the Fet turns on. When power is first applied the NPN transistor is off, nd therefore the 100k resistors pulls the gate to the source of the Fet and the Fet stays "off". I'll say the same, in different words, sometimes that helps. You are correct in the description of how the circuit works. What's the easiest way to do this? Thanks Then the user can walk away without having to remember to power off everything. Next time, the user can just power everything back up to restart the program. I got an Arduino application that uses a 16X2 LCD and a ULN2003 darlington array for driving a stepping motor.
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