2 picaxe 1 supply

alex126

Member
Hello. I have a question. There is a circuit with 2 picaxe 20m2 which go to light several leds by 2 ULN2003. Totally, it'll be 14 leds. If, basically, there is only 1 power supply (let's say 24v) and it will be:
Power supply / LM7805 / picaxe 20m2 / ULN 2003 / and the same power supply to ULN2003. All this twice.
Is there any precautions to have for the pic's or ULN2003 when powering from the same source.
Thank you
 

premelec

Senior Member
when you say "same power supply" do you mean 24v or 5v ? If the 5v only goes to the 2 PICAXEs shouldn't be a problem - however the 24 LEDs current can add up and cause heating if running through the linear regulator with its voltage drop...
 

alex126

Member
I mean, there will be a voltage regulator (LM7805, for example) before every picaxe 20, to have 5 v for each from the same 24v supply.
 

Hemi345

Senior Member
That shouldn't be an issue. Just make sure you use suggested input and output capacitors on each regulator and decoupling caps next to each 20M2.
 

hippy

Technical Support
Staff member
24V down to 5V is quite a drop, which usually means a lot of heat generated in the regulator at higher currents, but it should be fine for the few tens of milliamps the PICAXE will draw.

More important might be the amount of current you are sinking through the ULN's.
 

Hemi345

Senior Member
So 7 leds x 20mA on each LM7805 is easily doable. Even all 14 leds on one LM7805 is okay but make sure you provide adequate heatsinking for it.
Another option is to use one of these Ebay modules with a switching regulator to step down 24V to something like 7V then use a linear regulator to step that down further for cleaner 5V or 3.3V power. It's the route I've taken on a few projects with high input voltages.
 

westaust55

Moderator
With the linear LM7805 type voltage regulators, much over 0.8 watts In losses/heat and you need a heat sink. Then the higher the power lost as heat (watts) The more heatsinking you need.

if each PICAXE can have all 7 LEDs on at once then the total current is
7 * 20 + say 10 = 150 mA or 0.15 amps.
The heat generated by the 7805 bases on 24 V input is
(24 - 5) * 0.15 = 2.85 Watts
So a larger heat sink is essential.

mad others have mentioned, reducing the voltage to the 7805 down to something like 9 volts would be ideal.
Then the heat losses would be
(9-5) * 0.15 = 0.6 Watts
And no heat sink is necessary - the 7805 will get slightly warm.
 

premelec

Senior Member
I've had good results using inexpensive step down converters... no heat sink needed... and don't forget that all V- points need to be joined together...
 

depeet

New Member
AS Hippy already mentioned, from 24VDC to 5VDC is a serious voltagedrop. The LM7805 would will get heated up considerably. Its maximum current is around 1.5 Ampere. The maximum input voltage is 25VDC. In a previous project my regulator was heating up the enclosure is was mounted in. I replaced it with a buck-converter. If you still want to use the LM7805 and want more current then this circuit could help you. But keep in mind that you'll need heatsinks and a good ventilation to keep the temperature of the components low. https://www.electroschematics.com/lm7805-voltage-regulator/
 

Technoman

Senior Member
As an other factor, your don't even need a low noise power supply as required in an hi-fi analogue amplifier. In order to get the highest energy efficiency (little heat), the use of step down converters is your best choice. By now, you can get these devices at a low price ; it was not obvious years ago.
 
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