Hello all, quick (and probably dumb) electronics question. Suppose Calculon had 3 7805 voltage regulators in parallel, like such:
Would such an arrangment share the load evenly?
Hi:) what if i use lm2940ct-05 insted of 7805. I know its stil gonna use a lot power but its much better then 7805. So can i get aproximately 3am if use 3 2940 and how efficient ist going to be? Thanx:)
It is still a linear voltage regulator, so all the same points apply.
DC-DC converters are not so much expensive than linear regulators. It's the future! When you have a battery, you don't want to waste 60% of energy...
Sometimes I plug my Spider only to get 5Volt from my drill battery :p
Thumbs up for the Red back spider controller; I have one. Very happy with it :)
Or you can just buy the regulator that the Red back spider controller uses, it's a LM2676-5.0. Can deliver 3A @ 5V.
At first one regulator might deliver more load than the others but as it heats up it should start delivering slightly less voltage (within the normal opperating range) allowing the others to catch up.
Linear regulators such as the 7805 are not a good choice with such a large change in voltage. If for example your circuit above is supplying 2A @5V (10W power) then it is drawing 2A @ 12V (24W) from the battery. 14W is being wasted as heat.
Your circuit above at full load will need a heatsink the size of a brick. If you add a fan then you can make your own "Heaterizer".
Do yourself a favor and get a DC-DC converter. There are many available for automotive use that will accept anything from 12V-24V on the input and give you a 5V output.
Or buy a Red Back Spider controller, it has a 3A DC-DC converter built in.
I'm guessing you're trying to get more current. With this in mind you might consider a DC - DC converter. More efficient than 7805 regulators and very low cost on ebay.