I hope somebody can explain why the current is much higher when a step-up converter is used, see detailed info below.
Setup-1, the LiDAR is powered by the Vcc port (3.3V) which is boosted to 5V using a DC-DC step-up converter [1].
Setup-2, the LiDAR is powered by an external 5V power supply.
In setup-1 the measured LiDAR current is ~240mA, in setup-2 the current is around 110mA. According to the datasheet [2] the LiDAR peak current is 140mA.
Why is the current in setup-1 so high and exceeding the LiDAR peak current? It is caused by the step-up converter but why?
The lidar wants I=140mA under U=5V that means it requires P = U.I = 140x5 = 700mW of power
so your step up converter needs to provide those 700mW but it's only fed with 3.3V so the intensity needed is higher 700 / 3.3 = 212mA
the delta you see between 212mA and 240mA is because the converter is not totally efficient and may be only 90% of the power is converted so it draws 212 / ~90% = ~240mA
makes sense?
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side note : in your second drawing you need to join the GND of he lidar with the GND of the arduino so that they agree on what LOW and HIGH are. This also assumes the Lidar's Tx sends information at 3.3V and not at 5V otherwise you need a voltage adapter to not destroy your arduino's Rx pin.
One question about your side note: I dont understand why the lidar GND needs to be connected to the arduino GND. It's an external power source which has nothing to do with the arduino right? LiDAR Tx is at 3.3V so that is fine
remember that a voltage is a potential difference, that is a value of the potential in one point minus the potential in another accepted point we called usually GND.
with two different power supplies, the potential of their respective GND might be different and when you communicate through Rx and Tx, you do so with HIGH or LOW voltage. if the Arduino does not have the same GND reference as the LIDAR, then a HIGH for the LIDAR might not be seen as a HIGH by the Arduino.
@samaritaan11 if you want to use the step up regulator, you should power it directly from the battery, perfect, not from the 3.3 volt output of the Arduino board.
I use small lipo cells (or batteries of them) and step up (or down) for 5 or 3.3 volts as needed.
You don’t have a lot of current available from that pin and having strong current go through your arduino is not great in general. If the load on the other side is inductive or noisy you can also get that noise back to your arduino…
Cheap regulator dissipate the extra power as heat and so this is not very efficient either (and too much heat is not good for your arduino).
So tread carefully when you use your arduino as a power supply… remember its primary function is being a MCU…
Thanks Jackson for all the great tips! In order to save power i want to turn on/off the lidar. I thought a mosfet would be a good idea to achieve this. Got any feedback on this new electric circuit?
I am searching for a mosfet at the moment, and its quite difficult to find a suitable thru-hole mosfet.
IRLZ44 is a popular choice.
Alternatively, get some small AO3400's or something else in SMD form factor, one a couple of those little PCB's with an SMD footprint and the contacts broken out to regular 2.54mm-spaced pins on the edges of the PCB.
Boost converters have a relatively high quiescent current, so you should want to cut the power to the converter, not the device it is powering. Some boost converters have an Enable input, which would mean you don't need an external MOSFET, so maybe try to find one like that.
Learning new things by the minute, thanks! I am trying to figure out the electric circuit with the mosfet infront of the boost converter, but i can't figure it out.
