Powering a project from a DC power brick

[BytesGuy]

Hi all,

New to the forums, but not a total electronics newbie (just a bit rusty due to my comp sci degree taking over my life for a few years)! Now it's over I can get tinkering again

I have a project I am working on that is powered by 5V DC and now I want to design a PCB for it. Is the best course of action to whack in a 7805 and just use something like a 9V DC power brick or is there a more efficient way of doing it? What kind of heatsink (if required) would you reccomend for the 7805? I imagine they get quite hot.

I hope I have posted this in the correct section.

Cheers 

[andtfoot]

Is there an issue with just running it from a 5VDC supply in the first place?

[BytesGuy]

Not really. I just have lots of 9V bricks laying around, so it would be nice to put those to use.

My only concern with a 5V supply was if I plugged in a 9V brick and blew the whole lot by mistake. It's just the sort of thing I would do

[mariush]

There are plenty of linear regulators like 7805 (ld/lm/nxp/exar etc 1117 for example) which can output 5v: http://www.digikey.com/product-search/en/integrated-circuits-ics/pmic-voltage-regulators-linear-ldo/2556290?k=1117

I would recommend a 1117 because it only needs about 1-1.2v above 5v to work, so your device would run for longer time from a 9v battery. 7805 needs around 2v if I remember correctly. A 9v battery can go as low as 5.5-6v before dying, so it would suck to replace a 9v battery at 7-7.2v because the regulator doesn't work properly at such voltage.

The 1117 would also allow you to use 7.5v DC adapters without worrying the voltage may drop on that unregulated adapter below 7v (due to poor adapter quality, voltage drop on long cable etc etc).

Depending on how much current you use, you maybe would like to add a diode and a small capacitor (100-220uF 16v for example) in case you accidentally put an AC adapter instead of a DC one. The diode would also add reverse voltage protection but the 0.3-0.5v drop on the diode would further decrease the voltage detected by the linear regulator (if you put it between battery and regulator for reverse voltage protection).

Of course, for reverse voltage protection you can also use mosfets which is also more power efficient .. no power wasted in diode.. Afrotechmods has a video on youtube about this.

[sleemanj]

is there a more efficient way of doing it?

Depends on your requirements
 current
 physical space
 clean-ness
 cost
 complexity
 power efficiency

A linear regulator (eg 78xx series) is clean, simple and cheap, but the current you can draw is necessarily limited by the power you can dissipate given the heatsink you can put on it.  It's also very wasteful, all the power you don't need [ (Vin-Vout)*Current=WastedPower ] is thrown away as heat.

A switching regulator (buck, boost, or both) is less clean, a bit more complex, a bit more expensive, but far more efficient at higher currents, and if you don't need a massive heatsink anymore then it could be more space efficient.

Both have their place.

[BytesGuy]

Thanks for the replies 

There are plenty of linear regulators like 7805 (ld/lm/nxp/exar etc 1117 for example) which can output 5v: http://www.digikey.com/product-search/en/integrated-circuits-ics/pmic-voltage-regulators-linear-ldo/2556290?k=1117

I would recommend a 1117 because it only needs about 1-1.2v above 5v to work, so your device would run for longer time from a 9v battery. 7805 needs around 2v if I remember correctly. A 9v battery can go as low as 5.5-6v before dying, so it would suck to replace a 9v battery at 7-7.2v because the regulator doesn't work properly at such voltage.

The 1117 would also allow you to use 7.5v DC adapters without worrying the voltage may drop on that unregulated adapter below 7v (due to poor adapter quality, voltage drop on long cable etc etc).

Depending on how much current you use, you maybe would like to add a diode and a small capacitor (100-220uF 16v for example) in case you accidentally put an AC adapter instead of a DC one. The diode would also add reverse voltage protection but the 0.3-0.5v drop on the diode would further decrease the voltage detected by the linear regulator (if you put it between battery and regulator for reverse voltage protection).

Of course, for reverse voltage protection you can also use mosfets which is also more power efficient .. no power wasted in diode.. Afrotechmods has a video on youtube about this.

I won't be powering from a battery, but thanks for the info on the low dropout regs, I'll bear that in mind for future projects. Will definetly take a look at the reverse voltage protection as that was something else I was going to ask as the jacks are sometimes center positive or center negative.

is there a more efficient way of doing it?

Depends on your requirements
 current
 physical space
 clean-ness
 cost
 complexity
 power efficiency

A linear regulator (eg 78xx series) is clean, simple and cheap, but the current you can draw is necessarily limited by the power you can dissipate given the heatsink you can put on it.  It's also very wasteful, all the power you don't need [ (Vin-Vout)*Current=WastedPower ] is thrown away as heat.

A switching regulator (buck, boost, or both) is less clean, a bit more complex, a bit more expensive, but far more efficient at higher currents, and if you don't need a massive heatsink anymore then it could be more space efficient.

Both have their place.

It's a project from well over a year ago that I have left on several broadboards, so I will need to fix up some of the loose wire and measure it, but I think the current will be no more than 500mA in total as a finished product. As it is right now it's probably 100mA max. I had planned it to be expanable though, so I could possibly be looking at up to 1A, but again I'm not quite sure at the moment. Space, clean and complexity, not too concerened, but I would like to keep the cost down. I'll definitely have a look at the switching regs, they sound quite interesting!

[Rick Law]

Hi all,

New to the forums, but not a total electronics newbie (just a bit rusty due to my comp sci degree taking over my life for a few years)! Now it's over I can get tinkering again

I have a project I am working on that is powered by 5V DC and now I want to design a PCB for it. Is the best course of action to whack in a 7805 and just use something like a 9V DC power brick or is there a more efficient way of doing it? What kind of heatsink (if required) would you reccomend for the 7805? I imagine they get quite hot.

I hope I have posted this in the correct section.

Cheers 

I suggest pay close attention to heat sink needs.

Little wall warts are my principal mean of power supply to my various projects.  Before I got careful with heat sink, I blew a 7805 and an LM317's.

Using your figures,you are drawing 100mA, and regulating from 9v to 5v - that is 400mW to dissipate.  Not too bad.  But if you are drawing 500mA as you said in your later post, that is a good 2W to dissipate.  The darn thing may get fried rather quick with that kind of dissipation.

Unregulated wallwarts tends to put out much higher voltage.  My "9V" puts out nearly 15V...  The unofficial "rule of thumb" is, if it is too hot to hold with bare hand, you need heat sink.  I blew at least one 7805 and at least one LM317 before I take it serious.  Now I default to using heat sink "unless proven otherwise."

[BytesGuy]

Hi all,

New to the forums, but not a total electronics newbie (just a bit rusty due to my comp sci degree taking over my life for a few years)! Now it's over I can get tinkering again

I have a project I am working on that is powered by 5V DC and now I want to design a PCB for it. Is the best course of action to whack in a 7805 and just use something like a 9V DC power brick or is there a more efficient way of doing it? What kind of heatsink (if required) would you reccomend for the 7805? I imagine they get quite hot.

I hope I have posted this in the correct section.

Cheers 

I suggest pay close attention to heat sink needs.

Little wall warts are my principal mean of power supply to my various projects.  Before I got careful with heat sink, I blew a 7805 and an LM317's.

Using your figures,you are drawing 100mA, and regulating from 9v to 5v - that is 400mW to dissipate.  Not too bad.  But if you are drawing 500mA as you said in your later post, that is a good 2W to dissipate.  The darn thing may get fried rather quick with that kind of dissipation.

Unregulated wallwarts tends to put out much higher voltage.  My "9V" puts out nearly 15V...  The unofficial "rule of thumb" is, if it is too hot to hold with bare hand, you need heat sink.  I blew at least one 7805 and at least one LM317 before I take it serious.  Now I default to using heat sink "unless proven otherwise."

Wow, I didn't realise unregulated supplies could be that far off! I will definitely take caution with choosing a heatsink, I have a few small ones and some really huge ones like this:



Possibly a bit overkill!

[fcb]

It's pretty rare to find a non-smps wall-wart these days - unless it's in your parts bin. 9v smps wall-warts tend to be 9v.

Depends what circuit your powering and how acurate your 5v needs to be - the 5v smps wall-warts *tend* to be fairly close to 5v - certainly close enough that if I was powering a simple microcontroller I wouldn't be bothered.

If you've got critical analog that's a different problem.

[BytesGuy]

It's pretty rare to find a non-smps wall-wart these days - unless it's in your parts bin. 9v smps wall-warts tend to be 9v.

Depends what circuit your powering and how acurate your 5v needs to be - the 5v smps wall-warts *tend* to be fairly close to 5v - certainly close enough that if I was powering a simple microcontroller I wouldn't be bothered.

If you've got critical analog that's a different problem.

No analog (though there might be in the future), but the voltage does need to stay fairly stable around 5V. I'll probably just go for a 7805 with a heatsink really.

[Jon Chandler]

A well-engineered cell phone charger can be found at most thrift stores for a buck.  Cheaper than you can build anything.

[Rick Law]

[quote author=Newbzors link=topic=24247.msg352255#msg352255
Wow, I didn't realise unregulated supplies could be that far off! I will definitely take caution with choosing a heatsink...

[/quote]

I was shock at that one too.  Now I don't take the nominal voltage on faith.  I use my DMM to do a quick check.

My 4 from the part bin that is in constant service:
(1) 6v unregulated with low-load/no-load 12v
(2) 6v unregulated with low-load/no-load 9.8v
(3) 9v unregulated with low-load/no-load 12.3v
(4) 9v unregulated with low-load/no-load 15.7v

The other 3 in constant service are regulated ones:
12V = 12.1V
5V = 5.2V
5V = 5.6V

As you can see, the unregulated ones are all over the map.  When under higher load, voltage drop brings it closer to to the nominal voltage, but when powering something low-load, there can be unpleasant surprises. (I was using my 9V/15.7V to power a small utility light).  Now these four all making a stop at either a 7809 or an LM317 before powering anything.

[mariush]

That's good Rick, but never forget that those linear regulators have a large voltage drop, 1-2 volts. 

Some of those 6v unregulated power supplies that are 9.8-12v with no load, can get pretty close to 6v with just 50-100mA. If your project needs 5v and uses more than that, then your LM317 will have too little input voltage.

There's a huge selection of LDO chips (low drop linear regulators) which are also cheap.

For example (surface mount, but through hole to-220 and similar versions available)

up to 10v input, 300mA output, 0.2v voltage drop at 100mA : http://uk.farnell.com/microchip/mcp1802t-5002i-ot/ic-ldo-5v-300ma-sot-23-5/dp/1578372
up to 16v input, 300mA, 0.3v voltage drop at 300mA : http://uk.farnell.com/microchip/mcp1755st-5002e-db/ldo-0-3vdo-0-3a-5v-2-3sot223/dp/2306611

So you could basically even take 5v from a usb charger as input and get 4.8v+ at the output, which is close enough to 5v for a microcontroller and various 5v stuff to work without problems.But, you'd protect the devices from usb chargers that output 5.6v or thereabouts at low loads.

Really, these regulators are cheap in volume and just as easy to use as the classics.

[Rick Law]

That's good Rick, but never forget that those linear regulators have a large voltage drop, 1-2 volts. 

Some of those 6v unregulated power supplies that are 9.8-12v with no load, can get pretty close to 6v with just 50-100mA. If your project needs 5v and uses more than that, then your LM317 will have too little input voltage.

There's a huge selection of LDO chips (low drop linear regulators) which are also cheap.

For example (surface mount, but through hole to-220 and similar versions available)

up to 10v input, 300mA output, 0.2v voltage drop at 100mA : http://uk.farnell.com/microchip/mcp1802t-5002i-ot/ic-ldo-5v-300ma-sot-23-5/dp/1578372
up to 16v input, 300mA, 0.3v voltage drop at 300mA : http://uk.farnell.com/microchip/mcp1755st-5002e-db/ldo-0-3vdo-0-3a-5v-2-3sot223/dp/2306611

So you could basically even take 5v from a usb charger as input and get 4.8v+ at the output, which is close enough to 5v for a microcontroller and various 5v stuff to work without problems.But, you'd protect the devices from usb chargers that output 5.6v or thereabouts at low loads.

Really, these regulators are cheap in volume and just as easy to use as the classics.

Yeah, those volt variation is annoying.  I have to match the brick to the tasks.

Thanks for that reminder (LDO).  I have to write that down.  I actually have not experimented with the LDO types.  That is on my menu to learn, but last time I ordered parts, I forgot to include them.  I think I am going to try a couple of them next.


Rick

[Zero999]

Little wall warts are my principal mean of power supply to my various projects.  Before I got careful with heat sink, I blew a 7805 and an LM317's.

Using your figures,you are drawing 100mA, and regulating from 9v to 5v - that is 400mW to dissipate.  Not too bad.  But if you are drawing 500mA as you said in your later post, that is a good 2W to dissipate.  The darn thing may get fried rather quick with that kind of dissipation.

Were you using proper branded parts?

The LM7805 and LM317 have built-in protection which turns off the output when the chip gets too hot. The usual causes for destruction of the IC are: over voltage, reverse voltage or back feeding.

If the chip gets too hot to handle and the output voltage falls to zero, try disconnecting the power, leaving it to cool and it should resume normal operation, unless it overheats again.

[Rick Law]

Little wall warts are my principal mean of power supply to my various projects.  Before I got careful with heat sink, I blew a 7805 and an LM317's.

Using your figures,you are drawing 100mA, and regulating from 9v to 5v - that is 400mW to dissipate.  Not too bad.  But if you are drawing 500mA as you said in your later post, that is a good 2W to dissipate.  The darn thing may get fried rather quick with that kind of dissipation.

Were you using proper branded parts?

The LM7805 and LM317 have built-in protection which turns off the output when the chip gets too hot. The usual causes for destruction of the IC are: over voltage, reverse voltage or back feeding.

If the chip gets too hot to handle and the output voltage falls to zero, try disconnecting the power, leaving it to cool and it should resume normal operation, unless it overheats again.

Yes, it was fairchild (I think).  Either way, since I want it to operate comfortably with any of my strange power bricks, I add a cheap heat sink.  With that, I can get 5V out of any of my bricks from 7V to 19V.  For extended runs, I match the bricks a lot more closely so I am not doing a house-warming party with wall warts.