Measuring current consumption of lunetta

[ianbax]

Hi all,

You may have seen on another thread me angsting about a noisy power supply. Now I think I have that problem solved with a power supply based on an LM7805 regulator.

Given the max current for an LM7805 is 1.5A and I have space for 12 modules in my system I'm curious to find out how close to 1.5A total current I am.

I think I'm probably OK. For example - my counter module based around the 4040 has 12 LEDs with 2.7k current limiting resistors on them, so each one is drawing ~2-3ma so what 36ma if they're all briefly lit at the same time? Add to that the chip consumption itself (?is this on the datasheets?)

12 modules consuming 100ma each could be taking it close.

[Grumble]

I bought this at Aliexpress. Very cheap, very usefull
It displays voltage ánd current, but be aware that this measuring device also consumes power!

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[PHOBoS]

doubtful you'll go over the limit.

The voltage drop across a LED is around 2~3V, but let's make it worse and say it's 1.5V.
That means that there is 5V - 1.5V = 3.5V across the resistor
3.5V / 2700 ohms = 1.3mA*
12LEDs so 12x 1.3 = 15.6mA if they are all on at the same time

The current draw of the chips themselves is practically negligible (for CMOS) and there is a larger current draw
from any pulldown resistors on inputs, but that's still not much. Say you use 10K (I usually use 100K myself) that's 5V/10K = 0.5mA per input.
5mA if you'd have 10 inputs with 10K pulldown resistors that are all being driven high at the same time.
add some other losses here an there and maybe you get to 25mA ?
at least under certain conditions, most of the time it will probably be much lower.


Having said that you do need some proper heatsinks to push a 7805 to its limit.
They do have thermal protection so shouldn't break but it might make your lunetta seem haunted (great for this time of the year!)
I did see you mentioned 12V. so 12V - 5V = 7V across the 7805
7V x 1.5A = 10.5W it would have to dissipate in heat.
That might not even seem like a big number but it can fry your fingers in a flash.


so no, I don't think the power consumption itself would be an issue, BUT keep heat dissipation in mind.


* if you check the datasheet for most CMOS chips you'll find that the max output current at 5V is actually lower
which will result in a lower output voltage and in turn lower current.

[ianbax]

Thanks Phobos - I forgot to take account of the the voltage drop on the LEDs. I'm learning...slowly!

I'd read that about the higher the input voltage to the 7805 the harder it has to work and I'd use another power supply but (another thread) weirdly this 12v power brick I had lying around is the quietest I've found. I've ordered some heat sinks - may as well do things properly. I actually - perhaps foolishly - did the finger test the other day but barely felt any heat (only 4 modules and a breadboard running, though). Had trouble with my multimeter which only wanted to work in 10A mode but I clocked just 25ma from those 4 modules (a 4093 oscillator, 4040 counter and transistor based envelope generator)

[PHOBoS]

The finger test is a well proven method, just make sure to wet it first. If it sizzles it's too hot
(of course never use it on some random gear where something might carry a high voltage)