Powering multiple devices with one 9 V DC power

Hi all!

I have an 8mm film scanner that has

1) Arduino UNO 2) 4.5V seven LED light source 3) 3.3V Canon IXUS 95 camera on DC adapter

I also have an old ADSL power source rated at 9V / 1500 mA. I am now using it for powering the Arduino alone, but it occurred to me that what if I could use it for all three separate power needs.

What would the wiring be like? I could 3D print the housing for the wires and connectors, but I lack the electronics knowhow.

See http://www.sabulo.com/sb/category/8mm-film/ for the project itself.

Many thanks in advance!

what if I could use it for all three separate power needs.

Yes you can.

What would the wiring be like?

It depends on how much current you draw from the 3V3 line. If it is less that 100mA you can connect it to a Uno’s 3V3 output, otherwise you need a 3V3 regulator circuit.

4.5V seven LED light source

Should run off 5V, but you need a current limiting resistor on each segment. If you haven’t got them already then it is a crap design anyway.

The LEDs may be burned by 5V but Arduino will work at 4.5V (unless you connect it to computer via USB).

The LEDs may be burned by 5V

No.

Unless the OP is confusing the forward volts drop of an LED with the working voltage, which is why I said what I did about the resistors.

Thanks for the answers already.

I was thinking of having 9V enter at one end of a box, with three outlets on the side, one for Arduino at 9V, one for lamp at 4.5 V and one at 3.3 for the IXUS. I gather that to be possible, with suitable resistors?

Hi. I reckon your best bet is a couple of DC-DC converters. These so-called 'buck' converters can take a wide range of DC input voltages and drop them safely down to a lower voltage, usually adjustable but also available with fixed-voltage outputs. They are ridiculously cheap from China these days. Any self-respecting electronics tinkerer/hacker should keep a few around! The more expensive modules have screw terminals so there's not even any soldering required. A quick search on ebay or Amazon will find lots of them, although expect a few week wait if ordering from the Far East. Here's a typical example from ebay I found straightaway:

http://www.ebay.co.uk/itm/Buck-DC-DC-Step-2016-Converter-Module-Car-Voltage-Down-5V-9-12V-4-38V-to-3-3V-/172417320515?hash=item2824df8643:g:bigAAOSwB09YM9wg

Hope this helps. Jim.

I gather that to be possible, with suitable resistors?

No you can not use resistors to reduce the voltage because the current draw for each device is not constant and therefore the voltage output will not be constant. The whole point of a power supply is that the voltage is constant no matter what the current.

You need a voltage regulator, the DC/DC one cited above is best but you can use cheaper linear ones.

Please provide a link to those 4.5V LEDs because I am sure you are mistaken about those.

To know "best" voltage conversion you should know how much current (typical and maximim) each device needs. For low currents you can use dropout regulator, for higher currents it is better to use switching one.

Thanks a bunch folks! The light source is a LED flashlight from Clas Ohlson, with 3 AAA batteries.

I'll look into the links provided.

The light source is a LED flashlight from Clas Ohlson, with 3 AAA batteries.

Oh dear. Do not simply replace the batteries with power supply. Those sort of things only work because of the impedance of the battery. You replace that with a power supply with it's low impedance and the LEDs will fry.

Grumpy_Mike: Oh dear. Do not simply replace the batteries with power supply. Those sort of things only work because of the impedance of the battery. You replace that with a power supply with it's low impedance and the LEDs will fry.

Okay, that's a good point. Do you mean I should measure the impedance of the circuit with the LEDS and the batteries, and attach a suitable resistor between the power supply and the lamps? Or should i measure just the battery pack?

Sorry for asking all the simplest questions, but I am a newbie here, and your help is invaluable.

usenetfan: Okay, that's a good point. Do you mean I should measure the impedance of the circuit with the LEDS and the batteries, and attach a suitable resistor between the power supply and the lamps? Or should i measure just the battery pack?

Sorry for asking all the simplest questions, but I am a newbie here, and your help is invaluable.

Measure current the LEDs drain from batteries (and voltage for reference) and supply them with similar CURRENT.

Do you mean I should measure the impedance of the circuit with the LEDS and the batteries, and attach a suitable resistor between the power supply and the lamps?

Well in essence yes, but how are you going to do that? You can't use a resistance meter.

So measure the current the LEDs take from the battery call it I. Then disconnect the current meter and measure the voltage on the battery when the LEDs are lit call it V. Then calculate the resistance value when you are using 5V to power the LEDs using:- R = ( 5 - V ) / I If the result is less than 30R then you should use a higher voltage.