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flippityflop

heavy duty, realiable dc power supply

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been scouring the web for a good diy dc power supply that can power a lot of my experiments.
so this is the what i settled on:

http://www.eleccircuit.com/high-power-supply-regulater-0-30v-20a-by-lm338/

so, does anyone else know of a better design??
i really don't have the knowledge to analyze if it's reliable or have a lot of noise... that and i don't want to spend hours, scouring the web, you see...

if there's even a remote chance of this being overloaded, i'd like to add (if it isn't already there) a current limiter.... can it be easily integrated to this design?

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The circuit you found is poorly designed.

Its max output voltage is 20V, not 30V. The 18VAC transformer would produce a rectified and filtered +22VDC to feed the LM338 ICs if the filter capacitor is about 15 times larger.
If the output is shorted or is set to 4V with a 20A load then each LM338 will overheat and shutdown. The thermal stress might break one or more.

The voltage across each LM338 wll be 17.5V and with a current of 5A each they will dissipate 87.5W each when the output is 4V and much more heat when the output is shorted.
But one LM338 cannot dissipate more than about 50W if it is on a huge heatsink and has a high speed fan.

You cannot use a circuit breaker because the heating is the problem, not the current.
You cannot sense the heat inside each LM338 IC so a heat sensor also will not work.

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With all the series diodes feeding emitter-followers and the fuse then there is no voltage regulation.
It has the LED shown backwards which shows how much that site knows about electronic circuits.

The rectified and filtered 30VAC produces +40V. If the output is shorted or the voltage is set low and the load current is 20A then each of the 4 2N3055 transistors must dissipate 40V x 5A= 200W which is absolutely impossible.

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With all the series diodes feeding emitter-followers and the fuse then there is no voltage regulation.
It has the LED shown backwards which shows how much that site knows about electronic circuits.

The rectified and filtered 30VAC produces +40V. If the output is shorted or the voltage is set low and the load current is 20A then each of the 4 2N3055 transistors must dissipate 40V x 5A= 200W which is absolutely impossible.


crap. bunch of liars, they are.

i can just go with this:
http://electronics-lab.com/projects/power/001/index.html

but 3A MIGHT NOT BE ENOUGH....

can we tweak it a bit or make improvements in the design so that it can reach around 6A?? bit too much to ask.. but is it even possible?

if heat dissipation really is an issue, i can level all the ICs that needs cooling and just mount a single large copper plate or something and attach a dc fan....

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i see...

is that project based on switching-mode?

No, it is a linear power supply with no output ripple but its driver and output transistors get very hot so they need pretty big finned heatsinks.

in general, would power supplies based on switching-mode deliver higher current and even better voltage range?

A switched-mode power supply has high frequency ripple and its regulation is not as good as a linear regulator.
But it does not heat much.

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i see...

right now, power efficiency is just a plus, it's not like i'm gonna leave it for hours and hours on end. if heat dissipation is really an issue, a better solution (than having a big-ass common heat sink) probably would be to add more redundant components to distribute the load.... or substitute more expensive components....

guru, do you think that would work??

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A cheap linear power supply dissipates exactly as much heat as an expensive linear power supply. Redundant components do nothing.

A big heatsink with many fins distributes the heat to the surrounding air.



yeah, but what about using more expensive components??....

see i'm using this to play around with electrolysis (when i get too impatient, i remove the resistor on the electrode, and since it's a very ionic solution, the current definitely goes beyond 2A). so, i've broken 3 scavenged wall warts so far, so now i'm determined to have a heavy duty power supply with current limiting capabilities; and i'm sure it's gonna be indispensable in the future.

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Of course a high current, big and heavy power supply costs more than a low current, little and lightweight cheap little power supply.

You are supposed to calculate or measure the current then look at the datasheet for the power supply then select current-limiting parts so that you do not overload the power supply.

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