I have setup a grid pcb that further distributes power and receive signal from sensors, I know low voltage can detroit if not damage sensors , so I am thinking to put BD-LLC to distribute power output to 6 sensors connected to PCB shown below. [ignore 24 volt power that feeds solenoid it is separate and comes from power adapter not from mega though the wire has same red colour] the 5 volt wire connecting to main connector from other connector is not visible. rear view

front view And I have Arduino Mega from which I am drawing power 5 volt (using digital multimeter I found it's 4.98 volt). And all sensors are connected to this 5volt parallely. So the question is should I put this BD-LLC on above PCB to distribute correct power output to sensors ?

  • A logic level converter has nothing to do with power conversion. To answer your real question though there are two more pieces of information that are needed: the current draw of your board, and the gauge of your 4m long power and ground wires.
    – Majenko
    Dec 25 '16 at 10:15
  • 4m long wire adds 2.79 to 3ohm resistance, the board is on 6.98 voltage input from VIN pin of step down from 24 volt Dec 25 '16 at 10:29
  • The voltage at the board end will depend on the current draw of the board, and if that varies then the voltage will vary. You may be better off using a 24V -> 5V step-down at the board and providing 24V to the board instead (known as "point of load regulation") to ensure the board runs at 5V.
    – Majenko
    Dec 25 '16 at 11:35
  • In that case I would have to send back 5 volt to VIN of Arduino mega Dec 25 '16 at 11:37
  • Why would you want to? The Arduino is already 5V.
    – Majenko
    Dec 25 '16 at 11:37

Transmitting power over longer distances requires careful consideration. The actual voltage you will get at the remote end depends on two major factors:

  • The resistance of the wire, and
  • The current draw of the remote device

If you regulate your 24V incoming power down to 7V for the Arduino and then send the Arduino's 5V over 4m of wire to power your remote board the actual voltage the board will receive depends on the amount of current the board draws.

If this current is tiny then the voltage drop for the wire will be small and can largely be ignored. For instance, you quote about 3Ω resistance for 4m of wire, which to me seems incredibly high (32AWG which is like hair thin is only ~0.5Ω/m), but anyway. If your board draws 100mA, over your 4m wire, which is actually 8m when you take into account the length of the ground wire ("There and back again - an electron's tale.") you would drop (2*3Ω)*0.1 = 6*0.1 = 0.6V, so your 5V would turn to 4.4V by the time your board sees it. Not only that but the ground wire would be offset at the board end by 0.2V, which may impact LOW logic level detection.

So ideally you need to both minimise and mitigate that voltage drop. And both can be done using a buck regulator at the board and powering it from the 24V supply instead of 5V.

By using a buck regulator you massively reduce the current over the wire (for 24V to 5V you could be looking at around a quarter of the current or even less, depending on the efficiency of the regulator) which reduces the voltage drop on both the power and ground wires. It also guarantees that the board runs at 5V regardless of the current draw of the board.

Here's a diagram:

enter image description here

For digital sensors this loss of voltage could cause damage if any incoming logic signals are above the absolute maximum for the sensor (which is usually a function of the supply voltage, say Vcc + 0.3V). For analog sensors it will affect the values you are reading, since the voltage emitted from the sensor is often a function of the supply voltage, and if that is not what you expect it to be your calculations from the ADC values will be incorrect.

To answer your original question "can I use a LLC to distribute power" the answer is "no". The LLC can allow you to interface logic signals between two different voltage levels, but not affect power at all. And further more it requires that you have both voltages availabe for the LLC to operate, and you don't have both voltages, you only have the 5V minus the wire's voltage drop available, so you can't use it to mitigate any offset in the logic levels either.

  • How do I figure out current draw at remote device(pcb sensor grid) in my case? Do you mean sum up power of all sensors connected to grid? Dec 25 '16 at 12:18
  • Sum up the current requirements for all the sensors. You can usually find that kind of information in the datasheet for the sensor.
    – Majenko
    Dec 25 '16 at 12:19
  • I do have to step down converter but it does really bothers me Dec 25 '16 at 15:47
  • What bothers you about it? The colour?
    – Majenko
    Dec 25 '16 at 15:49
  • Baaah u still there, I meant having to put another step down converter Dec 25 '16 at 15:50

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