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Power regulation options for 5V and 18V circuit
Posted by
dsegel
on 09 Jan, 2018 18:18
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I'm building a circuit that needs both 5V and 18V available. Current draw will be on the order of 200mA max for the 5V line, and 325mA for the 18V. I was planning on using a 20-24VDC input and running it through an LM7818 and LM7805, but I wanted to ask first because that doesn't seem like a very efficient solution and I'm worried about the 7805 overheating since it would be operating at the top of its range and burning off a lot of power as heat.
So are there better options? Should I go for a lower input voltage and use a boost converter for the 18V? Or vice-versa? Are there any common solutions to this problem?
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LM317 has a rating up to 1.5 amps and 35 volts. i use them al the time for timer ic's and mosfet drivers outputting up to 10 amp pulses . never had any heating issues even without heatsinks they stay cool. you will need 2 resistors to set the voltage tough, but these regulators are solid and also have shortcircuit protection. otherwise you could use a tiny switch mode power supply a buck or boost converter. there are small units that are cheap. but i'd stay with the lm317, that's the cheapest solution. the 18 volt rail won't waste much power when the input voltage is 20- 24 volts, since it only has to bring the voltage down by a little. And in my experience bringing 24 volts down to 5 volt at 200 ma will not cause heating. Steve ward uses lm7805 and sorts for his tesla coils logic and drivers and they do the job just fine, so they're rugged i suppose. also linear regulation causes the least noise whereas smps and sorts brings noise that you should compensate for with bigger electrolytics and sometimes even shielding (also more expensive).Also even these power supplies cause heat. you could also use zener diodes but they will surely heat the most i suppose. Also if you state the application people can help you better. is it for logic, for inductive (motor) or resitive loads (lights)
so what's the project and how big is the housing,budget...? these are also important factors for deciding. i'd say lm317 all the way for up to 800ma.
Hope this helped.
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#2 Reply
Posted by
dsegel
on 09 Jan, 2018 19:32
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Thanks, I'll look into the LM317.
The 5V is for logic-level controlling L293Ds and an Arduino Nano, the 18V is for some small electromagnet coils that are powered by the L293Ds for polarity purposes. This will all fit into a box about 2x2x6 inches, and I need components that are not much bigger than a few TO-220s and associated caps and resistors. External power (the 20-24V) will be coming from a wall wart.
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#3 Reply
Posted by
soubitos
on 09 Jan, 2018 19:48
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Since this is an arduino design, you can and probably should get the 5V from the arduino power supply side. Trying to drop from 20-24V to 5V with say 7805 even at 200-300mA will get hot sooner than later.
If i had a 5V power supply for the arduino part of my design, i would increase its A output say by 1A which is easier and use a small booster circuit for the 18V rail which @325mA draw is not at all impossible even from as law as 5V supply... it might not be as efficient at first glance but it is much better than trying to introduce a 20-24V supply in the mix which is a whole other power supply design on its own....
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#4 Reply
Posted by
dsegel
on 10 Jan, 2018 00:13
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That's essentially what I'm wondering about - should I drop the 18V to 5V to power the arduino, or just bring in 5V and boost it to 18V for the coils. I guess I'll look at boost circuits next.
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#5 Reply
Posted by
Zero999
on 10 Jan, 2018 09:47
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The 18V circuit will be fine with an LM7818 and modest sized heatsink 20oC/W.
The LM7805 could be connected to the output of the LM7818 but the heatsink will need to be larger, to account for the extra power dissipation.
Another option is to connect a series resistor before the LM7805 to drop the extra voltage. The LM7505 needs 8V to regulate properly and the minimum input voltage is 20V, so that's a voltage drop of 12V, at 200mA R= V/I = 12/0.2 = 60R, so use the next lower E12 value of 56R. The resistor will dissipate 56*0.22 = 2.24W, so use a 3W resistor.
As mentioned above, you might want to consider a switching regulator, especially for the 5V output, as with a linear regulator, the voltage drop all gets burned as heat.
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#6 Reply
Posted by
soubitos
on 10 Jan, 2018 10:54
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The 18V circuit will be fine with an LM7818 and modest sized heatsink 20oC/W.
The LM7805 could be connected to the output of the LM7818 but the heatsink will need to be larger, to account for the extra power dissipation.
Another option is to connect a series resistor before the LM7805 to drop the extra voltage. The LM7505 needs 8V to regulate properly and the minimum input voltage is 20V, so that's a voltage drop of 12V, at 200mA R= V/I = 12/0.2 = 60R, so use the next lower E12 value of 56R. The resistor will dissipate 56*0.22 = 2.24W, so use a 3W resistor.
As mentioned above, you might want to consider a switching regulator, especially for the 5V output, as with a linear regulator, the voltage drop all gets burned as heat.
OP says he wants to get away with it with a few small components, 78xx regulators plus heatsinks dont account for that spec LOL.... since its an arduino thing, he probably has a 5V rail already, no need to create another one and add a 20-24V rail on top of that... 325mA @ 18V is not hard to create with modest cost and small size booster circuit and the choices there are endless... at least that would be my approach to solve this... 1 beefier 5V power supply which can be off the self for cheap or can make one also cheap and a booster and you are done... having a 5V rail for the arduino side, then a 20-24V power supply from which to create 5+18V is a total waste of resources and adds both in complexity and cost
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#7 Reply
Posted by
Zero999
on 10 Jan, 2018 11:36
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The 18V circuit will be fine with an LM7818 and modest sized heatsink 20oC/W.
The LM7805 could be connected to the output of the LM7818 but the heatsink will need to be larger, to account for the extra power dissipation.
Another option is to connect a series resistor before the LM7805 to drop the extra voltage. The LM7505 needs 8V to regulate properly and the minimum input voltage is 20V, so that's a voltage drop of 12V, at 200mA R= V/I = 12/0.2 = 60R, so use the next lower E12 value of 56R. The resistor will dissipate 56*0.22 = 2.24W, so use a 3W resistor.
As mentioned above, you might want to consider a switching regulator, especially for the 5V output, as with a linear regulator, the voltage drop all gets burned as heat.
OP says he wants to get away with it with a few small components, 78xx regulators plus heatsinks dont account for that spec LOL.... since its an arduino thing, he probably has a 5V rail already, no need to create another one and add a 20-24V rail on top of that... 325mA @ 18V is not hard to create with modest cost and small size booster circuit and the choices there are endless... at least that would be my approach to solve this... 1 beefier 5V power supply which can be off the self for cheap or can make one also cheap and a booster and you are done... having a 5V rail for the arduino side, then a 20-24V power supply from which to create 5+18V is a total waste of resources and adds both in complexity and cost
You seem to have misread the first post. He has 20V to 24V on the input, which needs to be converted to 5V and 18V. I agree, using a 5V supply and a boost converter would be more efficient, but if he only has 20V to 24V available, it's not possible. Building a boost converter, is also much more complex, than an LM78xx regulator, although using a cheap pre-built module would solve that problem.
I've just realised I missed that using the LM7818 with 20V in would be marginal, because it ideally needs 3V of headroom, although at only 325mA, you'll probably get away with it, especially if it isn't critical.
Using buck converters to generate 18V and 5V is another option, but again it's more complex than a linear design, although pre-assembled modules would overcome that.
I'm building a circuit that needs both 5V and 18V available. Current draw will be on the order of 200mA max for the 5V line, and 325mA for the 18V. I was planning on using a 20-24VDC input and running it through an LM7818 and LM7805, but I wanted to ask first because that doesn't seem like a very efficient solution and I'm worried about the 7805 overheating since it would be operating at the top of its range and burning off a lot of power as heat.
So are there better options? Should I go for a lower input voltage and use a boost converter for the 18V? Or vice-versa? Are there any common solutions to this problem?
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#8 Reply
Posted by
soubitos
on 10 Jan, 2018 11:46
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I stand corrected.... i missed the part where the 5V rail was for the arduino, i thought it was there to power other part of the build.... I agree with your thinking yet, it is dirt cheap and easy to find a 5V power supply, it can be an old 1A capable mobile phone charger ffs..... and like you said, an off the shelf booster module would work just fine for the 18V rail required.... if OP has the possibility to get both (5V PSU and Booster module) he would be fine
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#9 Reply
Posted by
ogden
on 10 Jan, 2018 12:11
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It is no brainer to convert digital 5V rail from power 18V rail, not opposite. So use 18V buck and 5V linear 7805 reg powered from 18V. It's doubtful that Nano will draw more than 50mA, so you can get away with linear reg for 5V.
Boost converters are are inefficient compared to buck because of higher currents, thus higher losses in switch and inductor. Also for 18V@0.33A (6W!) boost you need high power 5V rail, around 2A. Such a current needs much bigger capacitors on 5V rail than in buck+linear topology.
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#10 Reply
Posted by
soubitos
on 10 Jan, 2018 13:42
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So you suggest having a 20-24V power supply stepped down to 18V via a linear regulator and then a 5V linear regulator... OP asks for 18V/325mA + 5V/200mA that is a total of 6.85W available power plus the power converted to heat from the regulators....
My approach is a common 5v-2A power supply delivering the 5V to begin with and a booster module to create an 18V rail
Good quality 5v-2A wall converter like this
https://tinyurl.com/y88ysqt3 cost 5.44$ and a booster module like this
https://tinyurl.com/yd9zceu7 for a total of 6$ and an extremely small footprint too which as i get it is also required. No heat issues to solve, no heatsinks to take space etc....
The parts required to build the booster with MT3608 are as little as 1 ic 1 diode 1 inductor a couple resistors and capacitors... total cost for parts alone can be as little as 0,20$ !!!
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#11 Reply
Posted by
dsegel
on 10 Jan, 2018 13:59
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I said I needed 20Vin because I was only thinking of reducing it, not boosting it. I could do 5V in and boost it for the 18V line, so I'm going to look at that now. This is all a learning experience for me; as long as it works in the end there's no right or wrong answer.
Thanks for all your help and advice.
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#12 Reply
Posted by
soubitos
on 10 Jan, 2018 14:24
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actually... for learning's shake... try both approaches and let us know which worked better for you!
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the lm317 is in a to22 package and will be the cheapest option. with that current draw they're fine without cooling. i used one in a variable power supply for a slayer exciter circuit that could draw up to 1 amp and it stayed cool without heatsink. after half an hour it was lukewarm. so try it they're simple, cheap, small and rugged.
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#14 Reply
Posted by
dsegel
on 10 Jan, 2018 15:53
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Is there any advantage to using an LM317 over fixed regulators like an L7818 or 7805?
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lm317 has a 1.5 amp rating and can regulate up to 35 volts. don't know about the fixed ones. in my country the lm317 is the only one available so never looked into the fixed ones datasheet. if they are rated the same i'd say use the lm7805 and 7818. if steve ward can use them in all his tesla drivers i'd say they're rugged. he uses a heatsink however to cool the one for the mosfet drivers, but that's when driving big IGBT's rated at hunderds of amps. just try them for you purpose they'll be fine, trust me.
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#16 Reply
Posted by
ogden
on 10 Jan, 2018 16:40
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lm317 has a 1.5 amp rating and can regulate up to 35 volts.
7805 in TO220 also goes up-to 1.5A. You shall consider lm317 only if you need to optimize your stock or nonstandard/trimmable voltages are needed.
This circuit is one why you could possibly want to use LM317, thou lot of new LDO's offer even better performance/precision:
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#17 Reply
Posted by
Zero999
on 10 Jan, 2018 18:51
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lm317 has a 1.5 amp rating and can regulate up to 35 volts.
7805 in TO220 also goes up-to 1.5A. You shall consider lm317 only if you need to optimize your stock or nonstandard/trimmable voltages are needed.
This circuit is one why you could possibly want to use LM317, thou lot of new LDO's offer even better performance/precision:
Just one question: why use the LM317 as well as the TL431 when the LM317 will do on its own?
Is there any advantage to using an LM317 over fixed regulators like an L7818 or 7805?
Yes, it actually has better characteristics. Compare the data sheets. It also allows you to do things like bypass R2 with a capacitor, to increase the ripple rejection.
So you suggest having a 20-24V power supply stepped down to 18V via a linear regulator and then a 5V linear regulator... OP asks for 18V/325mA + 5V/200mA that is a total of 6.85W available power plus the power converted to heat from the regulators....
My approach is a common 5v-2A power supply delivering the 5V to begin with and a booster module to create an 18V rail
Good quality 5v-2A wall converter like this https://tinyurl.com/y88ysqt3 cost 5.44$ and a booster module like this https://tinyurl.com/yd9zceu7 for a total of 6$ and an extremely small footprint too which as i get it is also required. No heat issues to solve, no heatsinks to take space etc....
The parts required to build the booster with MT3608 are as little as 1 ic 1 diode 1 inductor a couple resistors and capacitors... total cost for parts alone can be as little as 0,20$ !!!
There's no right or wrong way. Both ways are valid. 5V to 18V might be more convenient, as it's probably easier to get hold of a 5V 10W PSU, than an 18V PSU which isn't overkill, but it's not necessarily better.
Using a higher voltage, before the long piece of cable, then stepping it down at the other end, is better than doing the reverse. For the same thickness and length of cable, transmitting 10W of power will be more efficient at 18V than 5V. Work it out. Think about the current needed, the resistance and power loss, in the cable, at different voltages.
Given the choice, assuming equal cost and availability of parts, I'd choose an 18V 1A mains power supply and a 5V buck converter, over a 5V PSU and 18V boost converter. Another thing is regulation: the 5V rail probably needs to be better regulated, than the 18V rail and having only a short PCB trace between the 5V buck and device will result in better voltage regulation
You'll find a similar scheme used in laptops. An 18V to 20V PSU used to charge the batteries and a some step-down converters to provide the various lower internal voltages to power the device.
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#18 Reply
Posted by
james_s
on 10 Jan, 2018 19:29
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No matter what linear regulator you use, they'll produce the same amount of heat.
Going from 18V down to 5V at a couple hundred mA I consider a switching regulator a no-brainer. Small buck converters are very easy to make now or buy as complete modules. They are far more efficient and run cooler, the only reason to use a linear regulator is if you need really clean power.
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less noise, cheaper and smaller so use the linear option, trust me on this. if it's going to be powered 24/7 i'd go with the buck/ boost but if it's for occasional use go with lm's . This is the last time i will say this or you guys will think i'm an old wining bat addicted in using lm317. i also use smps, buck and boost converters but not when i occasionally use the application and the power consumption is so low then it's a waste of money imo. i pay 0.50 euro's for an lm317 haven't come across quality buck/ boost converters for that price.
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#20 Reply
Posted by
ogden
on 10 Jan, 2018 20:22
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Just one question: why use the LM317 as well as the TL431 when the LM317 will do on its own?
Already mentioned: performance/precision. Reference of LM317 is specified as +/- 4% with +/- 0.7% additional temp drift for 0..125oC range. TL431 is 2%, B grade is 0.5%. In circuit TL431 does not dissipate heat, so circuit is much more thermally stable compared to LM317 alone. FYI circuit from TL431 datasheet.
Yes, it actually has better characteristics. Compare the data sheets.
Well, yes. Agreed. On the other hand 7805 shall be considered as close to obsolete today (more expensive than 1117), so we better compare LM317 to 5V 1117
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#21 Reply
Posted by
Gyro
on 10 Jan, 2018 20:26
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the lm317 is in a to22 package and will be the cheapest option. with that current draw they're fine without cooling. i used one in a variable power supply for a slayer exciter circuit that could draw up to 1 amp and it stayed cool without heatsink. after half an hour it was lukewarm. so try it they're simple, cheap, small and rugged.
Hmm, your circuit wasn't actually pulling anywhere near an amp then. Voltage dropped across the regulator times load current equals dissipated heat. Even at its minimum dropout voltage (2V typ. at 1A) it's going to be dissipating 2W.That's going to be pretty noticeable on a bare TO220 and with a short time constant. There's no getting around the laws of physics.
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#22 Reply
Posted by
ogden
on 10 Jan, 2018 20:34
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i pay 0.50 euro's for an lm317 haven't come across quality buck/ boost converters for that price.
If heat is not an issue (0.32A * (24-18)V = 2W peak), then indeed you can't beat price of linear regulator. Not always you can afford to build 2W heater out of power supply.
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#23 Reply
Posted by
Gyro
on 10 Jan, 2018 20:41
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Do factor in a heatsink though. Junction to ambient for a bare TO220 is 50'C/W, that will take you too close to shutdown at normal ambient.
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#24 Reply
Posted by
mariush
on 10 Jan, 2018 21:03
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If you have 20v or higher available and you're not afraid of making a small circuit board for a DC-DC converter, you could try using a chip like TPS54383 :
https://www.digikey.com/product-detail/en/texas-instruments/TPS54383PWPR/296-23087-1-ND/1790154It's two switching regulators in one, and works at high enough frequency that you can use surface mount inductor and ceramic capacitors and resistors to have a small board ... and it should be efficient enough that you wouldn't need to worry about heat and heatsinks and moving the heat away.
Chip is expensive in small quantities, but then again if you add two linear regulators and heatsinks for both and the footprint those two heatsinks will use on your board ... maybe it's not so expensive.
In other notes, you may want to consider designing your project to work from 7.5v to 12v, if you think your product it's going to be powered from a wallwart adapter. 7.5v to 12v adapters are very common and therefore cheap.
May want to think if you're gonna make a version to run from 4 AA batteries ... in that case you'll have a voltage range from around 4.8v (probably close enough to 5v that your circuit won't care) and up to around 7v - you could use a LDO with as much as 0.1v voltage drop to get ~5v from 4 AA batteries and use a boost regulator to create your 18v (though 18v @ ~200mA would be quite a load on AA batteries).