LM317LZ - how complicated does the circuit need to be?

I am beginning to think about how to supply current to my project.

The Arduino is powered by a 9v wall adaptor with a 1A supply capability.

I will eventually have six other circuits each needing to draw 3.3v and about 15ma.

I think I have these options:-

• Try to power each circuit from a digital pin with resistors to reduce the voltage from 5v to 3.3v
• Split the power cable before it reaches the Arduino and then reduce the voltage on the 'branch' with a voltage regulator
• Use a second 3.3v 1A wall adaptor to supply the 6 circuits with power

With options 2 and 3, would I need to tie the grounds of the Arduino and the six circuits together? Is the answer "It depends"?

Many thanks

yes grounds have to be "tied".

1 sounds bad, 2 would probably be the best. 3 is just cumbersome. IMHO

1will not work.
Just use a 3V3 regulator hung off the arduino's 5V pin.

Grumpy_Mike:
Just use a 3V3 regulator hung off the arduino's 5V pin.

Hi Mike

So, in a way, you're suggesting I use an alternate to 1? That all the circuits can be run directly from the Arduino, without splitting the power before the Arduino - but only if I take off from the 5v pin and only with the 3v3 regulator. I guess this is because the 5v pin has a much higher current draw facility than a digital pin?

THANKS

the 5v pin is directly off of the input power supply. it goes through a vreg first though so you could use a 9v supply for ex.

sirbow2:
the 5v pin is directly off of the input power supply. it doesnt got through a vreg.

aha - interesting. So how does it go from 9v to 5v? Does that mean I can run any current through it my PSU can supply without worrying about damaging the Arduino?

oops sorry, it is regulated from the input at 5v but your 3.3v vreg may have a large vdrop and 5v may not be a high enough voltage for the regulator to produce constant 3.3v. maybe jsut go off directly from the power supply would be best.

OK so if I need to supply 4.5v to five circuits each drawing 15ma, I could use a vreg hung off the 5v pin. Problem is, I can't find a vreg to turn 5v into 4.5v. Is it that big a difference? I'm guessing it is. I'm trying to replace 3 x AA batteries you see.

here: Mixed-signal and digital signal processing ICs | Analog Devices

BUT im guessing it will have a voltage drop to large for 5v. what i mean is that if you use a 4.5v regulator, you probably cant put 5v in, you have to put 6 or 7v to counter the internal voltage drop of the regulator.

you could also just stick in a 1N14001 diode in series on the 5v line going to the other circuits, as the 1N4001's voltage drop is .7v which would drop you down to 4.3v.

for 3.3 volts @15 mA I'd use one 1.2v battery + MCP1640 stepup dc/dc converter.

Also there are LM25xx for 3.3 volts which you could supply from the 9v adapter.

3.3v regulators usually work from 5 volts even if it is just on the margin.

I normally don't use linear regulators at all but definitively LDO 3.3v can work from 5 volts.

I would not worry too much about 6x 15mA using LDO from 5v.

4.5v from 5v -> use a 1n4001 or some schottky diode, you don't get 4.5 volts exactly but what is the difference between 4.5 volts and 5v? It makes a difference for LEDs since their gradient is very sharp, but usually not for ICs.

sirbow2:
here: Mixed-signal and digital signal processing ICs | Analog Devices

Looks good - but very expensive at £6 inc. VAT!

How does one adjust the output voltage? By varying what resistors are used after the vreg?

Thanks

you could use a LM317 which varies its output through a pot, except use a normal resistor to keep it regulated to 4.5v

sirbow2:
you could use a LM317 which varies its output through a pot, except use a normal resistor to keep it regulated to 4.5v

Now you're talking! £1 at Maplin.

So if I'm giving this vreg (the LM317) 7.45V via a power supply, I then use a resistor after it to get my 4.5V? Is there any limit to the current it can supply? Not sure how these things work...

Thank you!

Hi all!

I bought an LM317LZ from Maplin, and I bought the two resistors which, in combination will get me 4.5V

What I don't know, is how to wire it up? I had thought it was just a case of putting the LM317LZ on a bit of stripboard with the two resistors and away we go...

But now I am reading about heatsinks, capacitors etc? I feel a bit out of my depth...

Thanks for any and all help

hi all, would love some pointers.

I've spotted on an electronics forum :

"connect a 390? resistor to the adjust pin from the output pin. Connect a 150? resistor from the adjust pin to ground. Connect your input voltage to input. Connect your output to output." (this is an example for 4.5V)

Can it be done as simply as this? i.e. no capacitors or other funny business

Thanks

i.e. no capacitors

No you can't do this without capacitors and expect no trouble. Put a capacitor across the input and another across ( that means to ground) the output. If you don't have a data sheet that tells you the values to use then you won't go far wrong with a 4u7F for each.

You could try zener diodes. They're pretty cheap and not really all that complicated.

Mike, thanks for your help here!

I've made a model in Fritzing to show how I would wire the vreg circuit.

448×629 62.7 KB

Can someone please confirm this should do the trick?

Thanks

apart from the caps

Yes, You should turn polarity of C2, rather switch the connecting wires such that the capacitor + side remains up, which makes the picture clearer as well.

While topologically your circuit is correct:-
For best results the capacitors should be as physically close to the chip as possible and the capacitor leads should be as short as possible.
I would put the regulator chip in three adjacent holes so the capacitors can go straight across the chips and then wire the other stuff from that.