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Need opinions on regulation.

sndscientist

Jul 10, 2013
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To avoid going into great detail on the project at hand. I am stuck at an impasse. My working voltage is 24VDC. from the 24 i need about 5. Sounds simple enough so far, pop in a 7805 voila i get my 5.x volts. Well here is my issue, I need only around 10-15ma. The 7805 tends to act a tad odd at the lower end of the current spectrum. Which brings me to start looking into a 5.1v zener. I do have a few one watt zeners here but this is an efficiency question as well as a longevity one. this circuit will be powered 24/7 the 24 volt supply is a regulated low ripple linear supply. all i'm asking is which of the two do you think i should use?
 

BobK

Jan 5, 2010
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A 7805 will not be much more efficient than a zener. In both cases, you need to drop 19V across either across a resistor or the regulator, which is 0.36W, which should not be a problem. A zener with a 1K resistor 1/2W resistor should work fine.

Bob
 

sndscientist

Jul 10, 2013
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i was leaning toward that myself, i was just looking for opinions in case i missed something. the 10-15ma was a worst case scenario. The actual current is right around 1ma with no indicators. thanks for your answer
 

swagguy8

Dec 10, 2014
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Don't use a LM7805, those aren't really good, instead use the Reference voltage version of the LM317( it think its tl317) that produces around 50ma i think.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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The 317 is a variable reg. I would use a 7805. The dissipation is a little high to use the smaller 78L05.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Oh I think swagguy8 was talking about a TL431. You could use one of these instead of a zener however the component count would be significantly higher for no good reason.
 

sndscientist

Jul 10, 2013
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Actually the TL317 does ~100ma but thats also more money into this project that i don't need. I actually built a teeny zener circuit. as mentioned above the 5.1v zener and the 1k 1/2 watt resistor. the sensor this is running draws about 60ua.and it's feeding an IRF630 so there is no current draw.

i uploaded a picture of one of the sensors. it's a standard adjustable PIR. it has a working voltage of 4.5 - 20v. the chip attached behind is the IRF630. the PIR board outputs a high of the VCC when activated, it rails it low when it shuts off. the ideal situation is to fit this entire circuit into a single gang box, which is why smaller is better. you can see my zener circuit on the right hanging in front of the battery. the FET is there to control a set of LED fixtures i built with a constant current regulator set at 630ma. the CCR is installed in one of the fixtures heat sunk to the steel case.i need the 24 volt supply because each fixture has a drop of 9.6 volts or 19.2 total. so as you can see the smaller the circuit, and the fewer components the better, was gonna post this in projects but i figured it's still in build.

**edit** Yellow LED is driven by HEXFET. not the PIR board directly
 

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BGB

Nov 30, 2014
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FWIW: I have been using 7805's and have been having a different issue (going from around 12-14v):
they get hot during use, and then go into thermal shutdown when they get too hot.

I have basically made them work with big gobs of metal-dust mixed with clay gobbed on as a heatsink and blowing on them with a fan (a PC cooling fan), but yeah, kind of lame (they seem to only keep working if the fan is blowing on them).

it isn't even really all that much power either, generally around 750mA or so (mostly running a Raspberry Pi and a few other things).

had tested using two 7805's in parallel, but didn't really see much of a difference (in terms of heat generated).

not sure if there is really a better option for this.
 

BobK

Jan 5, 2010
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A 7805 drawing 750mA from a 14V supply will dissipate (14 - 5) * 0.75 = 6.75W. Of course it will get hot. You need a major heat sink for that. The T0220 case can only dissipate about 1W without a heat sink.

The alternative is a DC to DC converter, which will save power and not get hot.

Bob
 

BGB

Nov 30, 2014
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A 7805 drawing 750mA from a 14V supply will dissipate (14 - 5) * 0.75 = 6.75W. Of course it will get hot. You need a major heat sink for that. The T0220 case can only dissipate about 1W without a heat sink.

yeah. just 750mA seemed like not a lot of power, and the things are rated for 1.5A, ...

the first one I tried using was a straight clay heatsink, but as it got a little bigger I noticed that it didn't get hot uniformly (some parts got hot, others were still cool, and generally was a pretty big mass of clay).

with the metal-dust and clay heatsink, the heatsink seems to at least get hot all over, and presumably cools better (at least, it is a bit more effective with a lot less material, but still not effective enough to work without a fan).


have noted that it seems to not get as hot when running off of a 9v back up battery (currently consisting of 6 alkaline AAs wired in series), but this was mostly intended to deal with momentary disruptions to the 12-14v source (I am using a lead-acid battery as the main battery, but power may be momentarily disrupted depending on motor activities, and the 470uF caps I have available don't seem to be sufficient to deal with this on their own, as brief losses to 5v power are enough to cause the RPi to crash or reboot).

note that input diodes are used, so that the 12v and 9v inputs are isolated.


The alternative is a DC to DC converter, which will save power and not get hot.

Bob

I had looked, most options here seemed to be either building my own, or buying something in PCB form. couldn't really seem to find anything quite as convenient as the 7805, but I may not have looked enough.
 

BobK

Jan 5, 2010
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There are actually DC to DC converters with 3 pins that will replace a 7805, but they are expensive.

The reason it gets less hot when reducing 9V to 5V as opposed to 14V to 5V is that regulators of this type must dissipate all of the power of the lost voltage at the current drawn. So for 9V to 5V, the power dissipated is 4 * I, and for 14V it is 9 * I, more than twice as much.

Bob
 
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davenn

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Sep 5, 2009
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The reason it gets hotter when reducing 9V to 5V as opposed to 14V to 5V

you sure you got that around the right way ? ;)

it contradicts what you went on to say, which was correct

So for 9V to 5V, the power dissipated is 4 * I, and for 14V it is 9 * I, more than twice as much


cheers
Dave
 

BGB

Nov 30, 2014
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you sure you got that around the right way ? ;)

it contradicts what you went on to say, which was correct

yeah, I think a typing error...


for once, I agree with you Colin :)

D

in my case, the power efficiency isn't really a big issue, at least vs the annoyance of needing a cooling fan taking up space and being in the way (also sort of needed for the H-bridge transistors though).

this is because, at around 750mA, it is a pretty small piece of the power-use pie (apart from leaving it running for many hours, there wont be much effect on the battery, driving the thing around uses a lot more power a lot more quickly).
 
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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While I was away, we seem to have gone from 20mA to 750mA.

I too would agree on the use of a small SMPS. It will end up smaller, and cooler than a linear reg. The only issue is the noise on the output, but I can't see why this should be significant from anything I've heard about this project.
 
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