L7805 Voltage Regulator Circuit Question

DDJ2011

Member
Hi,

Daft question time. I want to set up a circuit using a 12V 1A wall wart through an L7805 circuit to provide 5v for my picaxe circuits. I have attached a circuit diagram from the 7805 data sheet that shows 2 caps, but when looking for the caps the only 0.33uF (330nF I think) I can find are these.

Have I found the right thing or am I missing something here?

Thanks for your help.

DDJ

L7805 circuit.png
 

moxhamj

New Member
http://www.ti.com/lit/ds/symlink/lm340-n.pdf

There is a little * saying that capacitor is only needed if the regulator is a long way from the power supply filter. But that would be true if you are using a wall wart.

The value is a bit unusual. I'd probably just use a 1uF cap instead as these are much easier to find.

Having said that though, my personal preference is to put a 25V 470uF electrolytic there instead. It is probably overkill as wallwarts usually have a similar capacitor inside, but at least this covers the possibility that the people making the wallwart didn't skimp and put a lower value capacitor in.

I tend to put more capacitors than are strictly needed, eg 0.1uF next to the 7805 and also one 0.1uF next to every chip on the board. Plus I also add a 22uF tantalum across the output of the 7805. No particular reason for doing this other than this was what was on all the computer boards I pulled to bits as a kid and I'm sure the engineers who designed those boards put the components in for a good reason.
 

SAborn

Senior Member
I agree with Dr_Acula on using a larger cap on the input side of the reg, i done some testing a while back and found any thing less than 330uf gave no filtering to ripple, and 470uf is a good choice.

Dont rely on the wall wart having good filtering as often they dont, add it to your own circuit to ensure you cover the filtering needs, because you might change wall warts and the next could have poor filtering and then find things dont work or worst you damage the circuit.

Personally i use at least 470uf on the reg input and 10uf on the reg output and 0.1uf decoupling caps through out the circuit across all chips, its just a standard practice for me and it always works without fault.

Problems come from under filtering and not from over filtering.
I also add a diode to the input of the reg for protection, in the case a wall wart is used with a backwards wired plug, it can save a lot of bad language and keep the magic smoke in, should a mistake be made.

Establish yourself a good universal regulator circuit and use the same circuit in all designs, then you have a piece of mind they will all work with any wall wart, without the need to review each circuit to suit.
 

Pauldesign

Senior Member
Hey DJ,

OP:D but just for your interest sake; for a good design, u might also need a heatsink since you're using a linear regulator to step down 12V to 5V (not efficient). The rest 7V will be dissipated as heat.
In future it might worth using but a switching regulator (e.g. L5973D) although that might also depends on your application, but it's more suitable esp if your input voltage is more than twice the regulator output voltage.
 

BeanieBots

Moderator
I'd go with Doc's suggestion. (Nice to see you back Doc, seems ages since you last posted)
A lot depends on the EXACT regulator you are using. Some won't work at all without 100nF or similar very close to their output, other will oscillate with more than about 10uF on their output. Go with whatever the datasheet for YOUR regulator suggests. More often than not 100nF each side will be fine.
 

MartinM57

Moderator
for a good design, u might also need a heatsink since you're using a linear regulator to step down 12V to 5V (not efficient). The rest 7V will be dissipated as heat.
Yes, you (please) might, but you might not. 7v is not "dissipated by heat" - about 7v x the current drawn (i.e. the power) is dissipated as heat in a linear regulator.

As OP has not defined the current (mA/Amp) draw of his "PICAXE circuits", then the jury is out.

But if you said that you were happy with say 0.5W of power dissipation (and you could go look in the datasheet to see what temp the regulator will get to with this dissipation and no heatsink - it's called degreesC/watt) then that implies you can draw 0.5/7 Amps from the regulator = 71mA with no undue heating effect
 

Pauldesign

Senior Member
Yes, you (please) might, but you might not. 7v is not "dissipated by heat" - about 7v x the current drawn (i.e. the power) is dissipated as heat in a linear regulator.
Yeah Martin, i meant to say 7W. typos once more!!! damn.

Anyway, all i was pointing was (which i didn't explained in details assuming DJ might know), there will be unused 7V from his 12V 1A PSU if he uses the L7805 linear reg, and at full load (which i assumed he's not only going to use Picaxe to blink LEDs) there will be 7W (7Vx1A) of power wasted as heat which is inefficient and will over heat the regulator.

I'm not going to elaborate more on power dissipation as there is a lot to take into considerations which in his case might be irrelevant. Well DJ, depending on your application; in case y can't touch your L7805 reg continuously for more than 10sec (assuming your circuit connections are correct and regulator is OK), think no further.

I hope it now make sense.

Anyway thanks, at least, i'm keeping you guys busy as moderators. No offence :D
 

Pauldesign

Senior Member
Yes, you (please) might, but you might not. 7v is not "dissipated by heat" - about 7v x the current drawn (i.e. the power) is dissipated as heat in a linear regulator.
Yeah Martin, i meant to say 7W. typos once more!!! damn.

Anyway, all i was pointing was (which i didn't explained in details assuming DJ might know), there will be unused 7V from his 12V 1A PSU if he uses the L7805 linear reg, and at full load (which i assumed he's not only going to use Picaxe to blink LEDs) there will be 7W (7Vx1A) of power wasted as heat which is inefficient and will over heat the regulator.

I'm not going to elaborate more on power dissipation as there is a lot to take into considerations which in his case might be irrelevant. Well DJ, depending on your application; in case y can't touch your L7805 reg continuously for more than 10sec (assuming your circuit connections are correct and regulator is OK), think no further.

I hope it now make sense.

Anyway thanks, at least, i'm keeping you guys busy as moderators. No offence :D
 

SAborn

Senior Member
A word of warning when using switch mode 5v wall warts, i have had them kill picaxe chips on 2 occasions, when the DC was plugged into the picaxe board and then power applied to the wall wart.

Some wall warts (SM) appear to take a little time to stabilize in output voltage, resulting in higher than 5 volts to the circuit for a short period.

I simply dont use the 5v switch mode ones now, and if i need to will power the wall wart first then plug it into the picaxe board.

I do think i had a couple of dodgie wall warts (albeit they were new) but they do exist and some caution should be taken when using them.

Never had a problem with the old transformer wall warts, or boards that have voltage regs on the boards just the 5v switch mode ones plugged direct into the board.
 

DDJ2011

Member
Follow Up Questions...

Thanks all for your help - I will post up a proposed wall-wart/voltage regulator circuit later on.

One thing does strike me though - MartinM57 said this:

As OP has not defined the current (mA/Amp) draw of his "PICAXE circuits", then the jury is out.
How do I estimate (or better measure) the current across my whole circuit.

In addition, is there a link anywhere that defines the max current I can put through a picaxe chip? I'm guessing 1A straight in will burn it out so I would need a resistor in series to the V+ to drop my 1A down to something suitable.

If I remember my senior school physics right, V = IR which means R = V/I

If I need (say) 50ma at 5v to run the picaxe safely then R = 5/0.005 = 1,000 ohms or 1K.
 

nick12ab

Senior Member
How do I estimate (or better measure) the current across my whole circuit.
Use a multimeter, or if you haven't got any sort of ammeter make one out of a low value resistor, op-amp and a PICAXE with analogue input.

In addition, is there a link anywhere that defines the max current I can put through a picaxe chip?
Search for the part number of the PIC (e.g. PIC18F25K22).

I'm guessing 1A straight in will burn it out so I would need a resistor in series to the V+ to drop my 1A down to something suitable.
Only if that is what's coming out of it too! I like to imagine voltage as being forced and current as being taken - just because 1A is available doesn't mean that the load will always consume 1A, it could consume however little it likes - 1A is just the upper limit.

You only need the current limiting resistor for something that would otherwise blow up at the given voltage - if you power an LED with a forward voltage of 3.3V with a 5V supply then it'll overheat and die quickly without a resistor but it will be just fine on a 3.3V supply.
 

DDJ2011

Member
Hi Nick,

Thanks for the answer. I guess I measure from the output of the voltage regulator to the ground back to the power supply to measure the current across the whole circuit?

Search for the part number of the PIC (e.g. PIC18F25K22).
Is there a link (or doc ref) that tells me which picaxe corresponds to which PIC chip? I don't remember seeing it in the manuals anywhere.

Thanks,

DDJ
 

nick12ab

Senior Member
Thanks for the answer. I guess I measure from the output of the voltage regulator to the ground back to the power supply to measure the current across the whole circuit?
Just wire the ammeter in series with the whole circuit.

The thing about making your own with a resistor + amplifier + ADC input is possible but not recommended for newbies and you'd need a multimeter to find the exact value of the resistor anyway otherwise your reading would be very inaccurate.

Having a multimeter is always recommended.
Is there a link (or doc ref) that tells me which picaxe corresponds to which PIC chip? I don't remember seeing it in the manuals anywhere.
In the Options dialog in Programming Editor, just select the PICAXE you're using and it'll tell you what the part number is below the chip selection combobox (e.g. "A PICAXE-28X2 chip is marked PIC18F25K22")[hr][/hr]After reading jtcurneal's post, I just want to note that my idea (or if you haven't got any sort of ammeter make one out of a low value resistor, op-amp and a PICAXE with analogue input) did basically the same as what he says but by measuring the drop with another PICAXE's ADC (the amplifier just amplified the voltage for better resolution).
 
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jtcurneal

Senior Member
You can also measure the current by placing a low value resistor in between your power supply and your circuit
and measure the voltage drop across the resistor and use ohm's law to compute the current.

1 ohm works very well
0.1 volt (100 millivolts) drop across the resistor = 0.1 Amp (100 miliamps)
0.2 volts (200 milivolts) drop = 0.2 Amps (200 milliamps).

the voltage drop across a 1 ohm resistor should not cause a problem with your circuit
and lets you check the current draw as you add or take away components.

Joel
 

Reloadron

Senior Member
I built a similar little supply once using a 12 volt 1 amp wall wart. It was the traditional transformer flavor of wall wart and that little science experiment can be seen here in 4 images. The wall wart was one of a bucket of the creatures I have and before using it I did look at the DC Output on a scope and it was clean. It was also in the neighborhood of 14 or 15 VDC out. The LM7805 and associated capacitors went on a small piece of circuit board from Radio Shack, hey, I had it lying around. The input side of the 7805 consist of a .1 uF and 22 uF cap as does the output side. While the regulator never seems to run even warm to the touch as can be seen in the images a simple little clip on heat sink can be slipped over a TO220 case LM78XX series regulator very easily. My choice in filtering was more a matter of what I had lying around than a science, though I would suggest the .1 uF on in and out as a minimum, pretty much follow the LM78XX data sheet.

As to the current draw. While I have not collected a mountain of data I can tell you a lowly little PICAXE 08M while doing nothing draws about 740 uA which amounts to nothing. Obviously a larger chip will draw more current even at idle but the chips themselves will draw current as a function of what they are doing. Much like a sound system on low volume draws little current but turn it up, the furniture is moving around and it draws high current. Makes it hard to really define what the current will be. However, I think it is safe to say if the regulator is only to be used to power a PICAXE the power draw will never heat the LM7805 to any degree. That assumes you only power the chip and not several relays and lord knows what like motors or whatever.

The little power supply pictured has worked just fine for several years. I would suggest in time you invest in a good basic DMM. You don't need a $400 USD DMM but I do recommend something above a $9.95 USD Home Depot special. A reasonably good DMM is a good investment and will be your new best friend for many years to come. :)

Hope That Helps
Ron
 

Goeytex

Senior Member
I have 5 Picaxe Chips on a breadboard all powered up and doing normal tasks. The breadboard includes a 5V LDO regulator.
The total current drawn for all 5 plus the regulator is just at 100 ma.

PS: You cannot do serious work with Picaxe projects without a decent DMM.
 

DDJ2011

Member
Hi,

Would appreciate thoughts on the attached schematic. Taking the advice given above on capacitor values and using a diode I have drawn a schematic of how I think the power supply should look.

However, not certain where I would put the diode. I have added two to the schematic - can you advise which one (or both?) should be there and whether the model I have selected looks sensible?

Is there anything else I have missed for this particular module?

Many thanks,

DDJ

12v Power Supply.jpg
 

MartinM57

Moderator
1N4106 are 12v zener diodes and are the wrong choice. You just need standard diodes - 1n400x (x=1, 2, 3 or 4 - whatever's easiest to get) for both

D1 should be on on your 12v in, pointy end to C1, to protect the circuit should you connect the 12v the wrong way round.

D2 across the regulator is OK, but not needed in most circumstances. You can have it in your circuit if you wish....
 

srnet

Senior Member
And myself, I would have a 0.1uf or 1uF ceramic capacitor from the regulator input to ground and output to ground, as close to the regulator that you can get them.
 

DDJ2011

Member
Updated schematic attached - is this where diode D1 should be?

12v Power Supply v2.jpg


And myself, I would have a 0.1uf or 1uF ceramic capacitor from the regulator input to ground and output to ground, as close to the regulator that you can get them.
@stnet - I have two electrolytics in there already - do you mean ceramics as well or instead of, as the earlier consensus was to put in the ones I am showing now?
 

SAborn

Senior Member
No the diode is wrong on the cap (D3), it should be where you have the 12v label.

0.1uf caps would be ceramic or polyester etc. and would go in parallel with the electro's should you choose to add them.
 

srnet

Senior Member
In wired capacitors I would stick to 0.1uF, the are only about 1p each. The wired 1uF can be fairly pricey by comparison.

These days, for PCB designs, I would use the SMT 1206 ceramic chip capacitors, the 1uF in particular can be a lot cheaper than the wired part.
 

premelec

Senior Member
Considering all the PS questions I wonder if we shouldn't have a whole separate forum for power supplies.... but no forum for resistors... :)
 

DDJ2011

Member
So, I'm not sure why I need the 2 caps on either side, but this is my guess...am I right?

The larger cap caters for larger power fluctuations and the smaller (quicker to recharge) deals with smaller ones.

Apologies - I have read all of the manuals at least once but had forgotten about that circuit. My bad.
 

eclectic

Moderator
So, I'm not sure why I need the 2 caps on either side, but this is my guess...am I right?

The larger cap caters for larger power fluctuations and the smaller (quicker to recharge) deals with smaller ones.

Apologies - I have read all of the manuals at least once but had forgotten about that circuit. My bad.
Here's a starter:
http://www.picaxeforum.co.uk/showthread.php?19066-Transients-decoupling-bypassing-capactiors-on-PICs-and-other-things&highlight=capacitor%2A

e
 

Paix

Senior Member
The larger value capacitors take care of relatively slow fluctuations, but because of their construction they have small elements of inherent series resistance or inductance which make them deviate from being 'perfect capacitors'. These defects make them less able to bypass some of the high frequency transients. These are instead handled by the low value decouplers which are more efficient at passing high frequency noise and spikes.

I'm sure that if my layman explanation is significantly flawed, that someone will please correct me :)

I think that I'm at least playing in the right ball park.
 

DDJ2011

Member
Stripboard Design

After a long absence, during which time I haven't progressed this at all, could you review my stripboard design?

This is the agreed circuit from earlier in this thread:

12v Power Supply v3.jpg

This should be the same circuit represented as a stripboard using PEBBLE:

12v to 5v Power Supply stripboard.png

This is the PEBBLE save file:

Code:
Diode|0|166|67|4|1N4001|CR?||2|1|IC||diode_124
Diode|0|193|257|2|1N4001|CR?||2|1|IC||diode_122
Capacitor|470 uF|92|162|3|Capacitor|C?||2|1|||cap_123
Capacitor|1 uF|92|189|1|Capacitor|C?||2|1|||cap_121
Capacitor|1 uF|119|379|1|Capacitor|C?||2|1|||cap_121
Capacitor|10 uF|119|406|1|Capacitor|C?||2|1|||cap_121
Miscell||79|27|1||XX?||1|32|Ground||misc_32
Transistor|BC559|118|281|1|L7805|Q?|||1|IC||transistor_11
Miscell||160|27|1||XX?||1|33|+12v||misc_33
Wire||91|226|11||11|#000000|1|11||10|
Wire||145|308|11||11|#FF0000|2|11||10|
Miscell||187|434|1||XX?||1|33|Miscell||misc_33
Miscell||79|434|1||XX?||1|32|Ground||misc_32
Wire||172|226|11||11|#FF0000|1|11||10|
Miscell||192|275|1||XX?||1|36|||misc_36
BREADBOARDSTYLE=BB49
SHOWTHETOPAREA=false
All help gratefully received!
 

Dippy

Moderator
1. I thought Pin2 was ground and 3 was O/P? (based on the supplied schematic and "the same")
You have Input, Output, Ground.
Are you using some other regulator?
If so, please say so.

2. Left hand diode does nothing unless you've cut the track under it (you don't say so).

3. Have you left enough space for fat 470uF to sit down?

4. You have a wire under the right-hand diode.
Is it going under the board?
Why not just move it 2 blobs to the right?

Note: Don't forget when assembling that the electro caps are polarised.
Ques: Are the 1uF caps Ceramics? Seems overkill/overbudget to me. I'd have them as 100 nanoF (0.1F , u1F)
 
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