New answers tagged power
0
You have to note that if you go in "power down" mode, you will not be able to get the time as millis() will return the wrong time.
If you just want 15 minutes, you can just put it inside a for loop:
for(int j=0;j<112;j++) LowPower.powerDown(SLEEP_8S,ADC_OFF,BOD_OFF);
So the next instruction will be executed approximately after 15 minutes (15 * 60 / 8 = ...
0
In your function lowPower(), instead of repeating the LowPower.powerDown(), loop over that call enough time to make up the total sleep-time you need. That's the general case. For your specific case, where you need 15-minute intervals and 2-hour intervals, lowPower() should loop 113 times (15:04, since 15 *60 doesn't divide evenly by 8). In your loop() ...
0
Here's a sketch that uses the LowPower library. You'll have to "tweak" the batteryCounter and sensorCounter variables to get the times a little closer to your specifications.
#include "LowPower.h"
unsigned int sleepCounter = 0;
unsigned int sleepCounter2 = 0;
unsigned int batteryCounter = 110; // 8 sec X 110 = 15 min approx.
unsigned int sensorCounter = 900;...
0
Try to use an external timer. I have the same problem approximatly.
0
The ground of both computers are connected also via their power lines. You get a ground loop that might result in malfunctions and/or defects.
Think about using a galvanically separated connection, for example via an optocouple.
0
If my supply increase from 12V 1.5A to 4A. My arduino will draw around 800mA @5V (571mA @7V) plus solenoid 1.2A @12V, is it a good idea to get 12V 4A (48W) power supply? Since Filip Franik told me that it is a good practice to get 2 times the current I need just in case.
I know there is the efficiency of the converter is calculated by dividing the output ...
2
If the sensor does provide two defined voltage levels, you do not need a pullup or pulldown resistor, but if it's only a switch: either (closed) connected to Vcc or GND, or (open) not connected, you need a pullup or pulldown resistor to get a defined voltage level in case there's "nothing else" (except the wire to the switch catching noise out of the air) ...
2
I realize that power is flowing the entire time the switch is open, which may (or may not) drain the battery.
No, it doesn't. You get a small spike as the switch is opened or closed, as the gate capacitance is charged or discharged, but other than that there is negligible current draw by a pin, which remains static regardless of the direction of the biasing ...
2
The input current does not depend on the pin level. Have a look at this discussion from avrfreaks. The input impedance of a digital pin is extremely high. The leakage current of the Atmega328P (here as example chip) is mentioned to be 1uA for both pin states. So this current is both unavoidable and neglectable from a normal view.
But the input impedance of ...
3
So you want to use the 5V voltage regulator on the Arduino serve as the supply for your 5V components, and feed the 5V regulator from the 12V supply that powers your solenoid?
The problem with that plan is heat. The voltage regulator on the Arduino is a poorly heat-sinked (heat-sunk?) "linear" regulator. A linear voltage regulator is basically a solid state ...
0
An Arduino UNO has a max current of 20 mA per IO Pin and a max current of 200 mA of all Pins together. If you intent to drive the senders with the pins then this is impossible.
If you power the Arduino with USB, the VCC current is limited to 500 mA (thermo fuses). If you power it externally, it depends on the power supply. Using the voltage from the Vcc Pin ...
1
You can use the adapter because:
the adapter uses an AMS1117 to generate 3.3V to supply the SD card and the level shifter
the level shifter is a 74LVC125 which uses 3.3V and works with 3.3V (and accepts up to 5V as input level)
The adapter works for 3.3V and 5V systems as long as 5V power is available.
1
Does this mean that even with the external power supply the Arduino will be connected to the UGND, and in effect the protective earth?
Yes.
Would it be ok to remove the jumper connection between UGND and GND?
No. I don't know quite why they provided that jumper. I can see no scenario in the datasheet where it suggests the two can be separate.
Or ...
-1
This is not an Arduino question BUT definitely do NOT try this!
The Foundation Hats Master has recommendations.
"It is possible to power the Pi by supplying 5V through the GPIO (sic) header pins 2,4 and GND. The acceptable input voltage range is 5V ±5%. ⋯ Implement a duplicate power safety diode ⋯ supply 5V at a minimum of 1.3A ⋯ Under no circumstances ...
3
The TIP120 is possibly the worst transistor you could choose for this job. It is a Darlington Pair transistor, and those have a (relatively) massive forward drop of between 2 and 4 volts. It works with the LED because an LED needs a tiny amount of current to operate compared to the motor and has a fixed forward voltage of its own. The motor doesn't, and is ...
0
There is a maximum current you can use from the 5V pin(s), so it is good to split the current going into the Arduino.
You can use a breakout board for a USB, connect the USB to the breakout board, connect it both to the Arduino, and in parallel to the sensors; meaning the current used for the sensors is not going through the Arduino. You might need 2 break ...
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