Hi.
After months of studying to learn the basics of electronics and then kidcad, I made this 5V power supply scheme that uses an lm7805. I wanted an opinion from the most experienced. If it is possible.
Here is my schema.
Many thanks.
I would replace the 0.22uF with 1uF-10uF electrolyte, check datasheet for your part for recommended value.
R1 seems a bit low for ~15v you have over that led. Either move it after the regulator or use 1.5-2k resistor. Unless of course you have a 200mW+ diode and you want it to be very bright.
Also power flag on the ac line doesn’t do anything but you can keep it as visual indication of power input.
Important thing to consider is power dissipation on your regulator. If your load is X amps then the regulator will dissipate about (15-5)*X Watts or 10X. If you have a standard TO220 package with no heatsink I wouldn’t load it more than 40-50 mA which is not that useful. I recommend attaching a heatsink, with that you should be able to go to 100-150 mA. Any more than that and I would use a switching regulator instead of linear one.
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I’m commenting more on style issues than actual circuit function. I recognize that style is highly personal (or organization if making schematics for an organization), so feel free to ignore me. 
In addition to the above note about the PWR_FLAG symbols, your other two PWR_FLAG symbols seem to be randomly placed. I tend to use them to indicate the source of power. I mean, if you need to place the PWR_FLAG symbols to make ERC happy, may as well also use them for a visual indication on the schematic. Thus I’d move the PWR_FLAG near the input of your 7805 to the + output of the bridge rectifier subcircuit. Also, I’d move the PWR_FLAG on the GND line to the - output of the bridge rectifier subcircuit.
Looking at your annotations around your bridge rectifier subcircuit it looks like you are using local labels. (The tool tip calls them “net labels” but I call them “local labels” to indicate scope. On your single page schematic scope isn’t important, but it’s a habit that I have so when talking about multi-page schematics the scope is very important.) Neither of the “A” labels, nor the “+” label are connected to anything (that is why there is a small square). Good thing that the “A” labels aren’t connected to lines, that would short out your AC input. I would change those local labels to regular text (the tool bar button third from the bottom that is a capital “T”). That way you can’t accidentally connect the AC sides of your bridge rectifier subcircuit.
Is your AC input polarized? I know AC doesn’t have a plus and minus side, but it does have a source and a return. If your AC input isn’t polarized and the source is on your J1 pin 1 (odd you have pin 1 on the bottom, but that really shouldn’t matter) and then the fuse blows, your circuit is still energized and any accidental connection to earth would still get current flow. Yeah, 12V is low voltage and the circuit probably can’t supply too much current, anywhere, so this probably isn’t a life-or-death risk. But it’s good to be in the habit of recognizing these issues and if mitigating them only costs some thought, why not mitigate?
I would rather use a switching DC DC converter than a Linear Voltage regulator due to its effectiveness. For very low power consumption and cost worries, this could be acceptable at some places, but thinking that 1A fuse, that means you’re ready to consume 12W from the source. If everything would be ideal, you’ll lost 7W as heat in this design, and that means a good heat sink would be required. So cost effectiveness would be gone. Switching DC DC converters are very cheap and neat these days. Try to adapt them. They need a little bit more experience looking from the EMC/EMI stand point but since you are educating yourself, this should be the path you’d like to follow.
Other point also mentioned by @qu1ck, 220 Ohms for that poor LED would be very low and probably its life will be very short. You should increase it at some KOhms, like 5.6K? Just refer to LED datasheet and try to calculate necessary resistor based on it.
For the power input, if you use a barrel jack, you can get rid of the diode bridge and could use a stock wall adapter which is very easy to find. 5.5/2.1mm models are most common types.
You should filter your input if you’d like to have a good power supply. But never use an RC filter on a power path. Add an LC filter calculating with the cost.
You should also worry about ESD. Use a TVS diode at the input to eliminate unwanted spikes from the input.
Since this is a DIY supply, shorts on the output would be very likely. Therefore think about changing the fuse with a PTC or eFuse. This would give you rough ideas on how to design circuits which will self protect without any intervention.
Good morning MuratUrsavas, Quick and Sembazuru and others.
I thank everyone for the quick answers.
You all have understood that I am a beginner. Your answers give me many points for reflection.
I will have to study them.
I will need some time.
I hope you will be available for any questions I may have.
Thanks again
There is one other issue to be considered. Although many folks ignore this.
On powerup capacitor C1 is at zero volts. If the 12V source is very strong, the current of the 1st cycle will be high because until the capacitor starts to charge it is a short to ground.
You’ve not suggested the current required. If not too high you could add a resistor in series before the capacitor. Maybe 10 ohms, perhaps a little more.
A quick method to estimate the droop in C1 when the AC line is heading to zero is:
I = C dv/dt
Where I = the current C1 is expected to supply when the AC line is lower that the LM7805 needs.
C = capacitor in farads.
dv = the change (drop) in voltage of the capacitor caused by the load current
dt = the time of the above drop.
You would be estimating the drop in the rectified filtered voltage (represented by blue line)
@JohnRob IMHO putting a resistor is not a good idea, even though there will be very little power consumption.
As you can see in my previous post I had suggested to implement an LC filter, which would deal with the in rush current, too.
What are those two “A” texts?
They look like wire labels that are not connected.
It also looks like you have labels with “-” and with “+”.
Labels are used to give names to a net. But to do that you have to put the little square in the lower left corner of the label on a green wire, which makes it disappear . You can rotate and mirror the labels, which puts the attachment point in a different orientation.
I also prefer to draw bridge rectifiers like:
The difference is small and just visual, but it is more compact.
I also agree with putting some kind of filter with an inductor or ferrite core on the input. In the past I’ve had lots of troubles with microntrollers that got reset by interference from turning nearby FL or Halogen lights on or off, which conducts spikes through the power supply, and those spikes go right through the voltage regulator because it’s too slow to react to them.
I agree both with JohnRop of putting in a series resistor, and with MuratUravas of not putting in a series resistor.
I assume you get the 12VAC from a regular iron core transformer, which results in 16V to 20V DC after the bridge rectifier, which will cause a lot of dissipation in your voltage regulator. (A transformer with a lower output voltage is a much better choice here). Adding a Ceramic power resistor to dissipate a part of the extra power may be cheaper then a big heatsink for the voltage regulator.
Adding a small (0.5 to 10 Ohms) series resistor after an inductor lowers it’s Q and dampens its tendency to oscillate.
JohnRob mentioned the peak current at turnon. And this is indeed an issue, but the copper resistance of the transformer probably already adds enough resistance to keep everything in safe limits. This mostly becomes a problem with big transformers and big reservoir capacitors. For example with high powered audio amplifiers. Such amplifiers often have a series resistor in the primary winding which gets short circuited after a few seconds by a relay, when the electrolytic capacitors have charged.
It also becomes a big issue if you have a DC power supply on the left with a big reservoir capacitor, and then plug in your circuit via some connector. In that case you can have very high charge currents for the bulk capacitor in your circuit, and sparks from the connector.
Neither is the case here though. You probably do not need a series resistor for this in this case.
A disadvantage of a series resistor is that it always causes an extra voltage drop when it conducts current, so you have to be aware of this. But in this case you have lot’s of voltage available and this is not a big issue.
Sorry I’m late. I couldn’t play with my hobby.
I thank everyone for the advice which was very helpful.
One thing missing is a small capacitor on the input. A quick look at a datasheet suggests a 0.33 uF.
Make sure when you lay out the PCB that the two small (ie .33 / .1 uF) caps are very close to the actual regulator, otherwise there is the risk of oscillation.
Most thing already have been said, but I’d add a ceramic input capacitor, like the 0.22nF on the output also to the input. The LED resistor is probably too small, I’d use 5.6K or even 10K if the purpose is just to show that there’s power and you’re not actually intending to light something. But of course you could experiment with that to your liking (and it obviously depends on the LED type and color).
I’d replace C3 with a 10uF ceramic (or tantal) capacitor, you want to have a small one in parallel to a large one, not two relatively small ones. An additional electrolytic (maybe 100u-330u) on the output probably doesn’t hurt, either.
A ferrite bead (or two) on the input doesn’t hurt, either, if your load is sensitive.
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