Motercycle powered circuit advice and review

I’m creating an Arduino powered device to display a scooters speed and other information on a small OLED screen.
The device will be powered from the scooters 12v battery and will use a 5v voltage regulator to regulate the voltage down for the Arduino and the OLED.
The Arduino is an ATMega328p I’ll be soldering on a bread board.

I’ve read about the voltage spikes that may occur starting up a veichle and how zener diodes can help handle such spikes and this is the circuit I’ve come up with so far

  • Please review the circuit I’ve designed.
  • Is the circuit realistic, will it work or will it get fried by 12v battery and its spikes.
  • I’m fully aware of the inefficacy of Linear voltage regulators.

My math
The Arduino consumes 230ma and the OLED consumes 30ma (worst case)

At a steady 12v, the Linear Regulator (U1) should be disappointing 1.61 W.
(12v - 5v) * (200mA + 30mA) = 1.61W

At a voltage spike of 80v (will a scooter even go this high), the power dissipation goes really high, but since its for such a short period of time I don’t think it will be a problem (will it???). Maybe increasing R1 could help.

R1 Voltage 55v (80v - 25v)
R1 Current 550mA (55v/100 Ω)
Diode Power Dissipation 13.7w (550ma * 25v)
Linear Regulator Power 4.6w (20v *230ma)

Thanks.

Hi,
Welcome to the forum.

Sorry, I don’t have time to check your calcs but your approach looks good
.
The 100R sereis input resistor will drop 230mA * 100 = 23V ? ? ?

I needs to be much lower like 5R 5 * 230mA = 1.15V
Use a 1W resistor.

Check your zener specs, I think they provide an current impulse parameter, a 1W zener would be nice and safe.

Not sure if you have seen this project for info?

Tom… :slight_smile:

That 100 Ohm resistor is going to mess you up... Calculate the voltage drop across 100 Ohms at 260mA. :wink:
[EDIT] Tom beat me to it! ...You're not going to get 5V out of the regulator or 260mA.

Short-duration power dissipation shouldn't be a problem, but a short-duration voltage spike could still damage the voltage regulator (if you didn't have the Zener). We really don't know what the Zener will be exposed to, but if it fries it will probably short, protecting your Arduino. The series resistor may then burn-up and open (depending on it's power dissipation and rating), or you can add a fuse.

(U1) should be disappointing 1.61 W.

Dissipating. :wink: You'll need a heatsink.

...I've used an Arduino in a vehicle just using the Arduino's on-board regulator with no problems. I'd be inclined to just try the 7805 and see how it goes... If it burns-up go "back to the drawing board". Try it on the scooter with just some LEDs or something to see if it gets killed by a voltage spike.

Thanks TomGeorge and DVDDouge for the feedback.
Given the cheap cost of electronic components, it would be be best for me to perform tests on the scooter itself to see how things go.
As for my Math, it seems I forgot to account for the 100R resistors affect on the input of the Voltage regulator, though I don't know how to do that at the moment.

The mentioned Scootputer project directly connects the 12v battery to the Arduino Board but I wont be using an Arduino Board, just the chip.

CptMichles700:
\As for my Math, it seems I forgot to account for the 100R resistors affect on the input of the Voltage regulator, though I don’t know how to do that at the moment.

V = I * R

I = 260 mA = 0.26 A
R = 100 Ohm

V = 0.26 * 100 = 26V

That’s your voltage drop over that resistor. Obviously at a 12V input you’re never going to get 260 mA running through the resistor.

At 5 Ohm:
V = 0.26 * 5 = 1.3V

No problem there, enough voltage left for your regulator. Even if you draw quite a bit more current you’re fine.

But mind the power dissipated by the resistor! It goes cubed with the current.

P = I2 * R or P = V * I

At 5 Ohm and 0.26A:
P = 0.34W (so a 1W resistor would do - it’ll get warm).

Do make sure you have ventilation or your enclosure will get really warm, with the heat from that resistor and the 7805 regulator.

But if you add a bit more and go to, say, 500 mA:
Voltage drop V = 0.5 * 5 = 2.5V. No problem here, 9.5V left for the regulator, which then will be dissipating 0.5 * 4.5 = 2.25W.
Your resistor however is now dissipating 2.5 * 0.5 = 1.25W already! That’s too much for a 1W rated resistor, even if well exposed to the air.