Creating a stable 5V DC supply out of 6 V (RMS) AC

[Wrichik Basu]

Wrichik, you didn't mention which, if any, Ardiuno sensors you'll be using

I will be using only the TSOP 1838 infrared receiver. It's datasheet shows a circuit that can be used to suppress power supply disturbances, if there are any.

Oh ya, when you're doing this make sure you only connect the ADC to nodes after the rectifier that are referenced to the same ground as the Arduino. You can get into some hazardous situations for you and the equipment if you don't.

https://www.newark.com/pdfs/techarticles/tektronix/FFM.pdf

Yes, of course, it will always be after the rectifier. I know that microcontrollers cannot withstand AC.

 

[anorlunda]

Surely you have a spare USB phone charger around?

If you don't here's a tip I learned decades ago. Go to the nearest car rental counter. Tell them you left your charger in a rental car. They will show you to a bin full of chargers, cords, phones, and other devices that people left behind and they'll say, "Help yourself."

 

[dlgoff]

The best treatment I've seen is from this 1980 National Semi handbook (back in the days when tech companies would actually teach circuit design to their customers!).

I've got that National handbook and several other National handbooks (e.g. Audo applications handbook with tons of op-anp circuits Yes, back in the days there was an annual Mid-American Electronics Conference in Kansas, City that I would always attend and the chip vendors would hand out their free data books. I also have the full set of Fairchild Semiconductor handbooks and a Signetics's IC databook and their their Digital, linear, mos applications book. I don't use them much these days however.

edit: I just noticed that I also have a Hewlett-Packard Optoelectronics Designer's handbook

 

[AlexCaledin]

With 6V AC, the rectifier must be "doubling" (or how do you call it in English), and then the regulator ought to be switching, LM2575T−5G seems best (very reliable).

The thing is, the regulator input must have some "spare" volts, to give time for the microprocessor to finish the process in case of sudden AC supply disappearance. Even if there's no AC sensor to warn the processor, the "excessive" voltage (and capacitance) helps to withstand possible transient AC interruptions. The LM2575 allows up to 40V at its input!

 

[Wrichik Basu]

LM2575T−5G seems best

I will make a note of this, but unfortunately won't be able to buy it because the website from where I buy my components does not have this.

 

[Baluncore]

I will make a note of this, but unfortunately won't be able to buy it because the website from where I buy my components does not have this.

Why not just use a traditional 7805?
The circuit model shows it should work. All linear regulators will be equally efficient in that application.

 

[Baluncore]

- well, if you have a heat sink (especially at wintertime), you choose linear - whereas, having a fast diode + inductor, you choose a switching regulator)

There is no need for switching or low dropout regulators unless energy management is critical, such as when operating on batteries. The cost of the mains power will not be high. Sometimes it is better to keep the project simple.
The 5V regulator input voltage will average about 9 V, so the voltage dropped across the linear regulator will be 4 volts. At the maximum specified current of 1 amp, that will be 4 watt, so the heat sink will not need to be big. At the same time, the load will be dissipating a maximum of 5 watt.

 

[Wrichik Basu]

Why not just use a traditional 7805?

I am using the 7805 for this project. I just said that I will "note" it down, and might explore it in my future projects.

This is the final circuit for this project, with the corrections that you suggested:

 

[AlexCaledin]

- Here, I've disassembled a very old "linear" 5V supply unit for a microprocessor. The rectified voltage to feed the 5V linear regulator is 15V at no load. The secondary wire resistance of the transformer is few Ohms. So, a professional developer ought to think of possible voltage instability - because of the drop in the transformer at max. current and the primary AC instability and choose some higher regulator input voltage ( - which is why a modern "cool" switching regulator may be much better nowadays, requiring smaller transformer, too).
____________________________________________________________________
(yes, I was of the same opinion as Averagesupernova below - before being surprized by the good performance and reliability of that LM2575 switcher in a mass-produced device)

 

[Averagesupernova]

I would never persuade someone to use a switcher who has lacks experience as the OP in this thread does. Please @Wrichik Basu don't think I am belittling you in that comment. Nothing wrong with being inexperienced. After all, we were all inexperienced at one time (some of us still are in areas) and you are here asking questions and making headway.

 

[Baluncore]

This is the final circuit for this project, with the corrections that you suggested:

Different regulators need different input and output capacitors to maintain stability. A ceramic capacitor across the input will hide the connection inductance and resistance of the electrolytic reservoir capacitor from the regulator. A ceramic capacitor across the output hides the output wire inductance, so stability is not dependent on changes in the load.

I therefore suggest that you place a 100nF ceramic capacitor close to the 7805 input, and another close to the output.

 

[Wrichik Basu]

I therefore suggest that you place a 100nF ceramic capacitor close to the 7805 input, and another close to the output.

Along with the electrolytic capacitor?

 

[Baluncore]

Along with the electrolytic capacitor?

Yes. One ceramic capacitor is electrically in parallel with the electrolytic, but it is placed physically close to the regulator.

 

[Wrichik Basu]

Yes. One ceramic capacitor is electrically in parallel with the electrolytic, but it is placed physically close to the regulator.

This is the new final diagram, then:

 

[Rive]

This is the new final diagram, then:

If you want to max it out then a fuse on the mains side. Not really against overcurrent - more against transformer failure. So it depends on the type of the transformer, some does not require additional fuse.

 

[DaveE]

I would never persuade someone to use a switcher who has lacks experience as the OP in this thread does. Please @Wrichik Basu don't think I am belittling you in that comment. Nothing wrong with being inexperienced. After all, we were all inexperienced at one time (some of us still are in areas) and you are here asking questions and making headway.

Yes, I agree. Especially when the SMPS benefits aren't really necessary. However, National's Simple Switchers are really easy to use. They are highly integrated ICs and they've done the design for you for common applications, including layout and BOM.

 

[Wrichik Basu]

If you want to max it out then a fuse on the mains side. Not really against overcurrent - more against transformer failure. So it depends on the type of the transformer, some does not require additional fuse.

Safety precautions on the mains side are already present. I bought a custom-built extension board from Amazon that has a fuse built-in. Originally it was 6 A; I replaced it with 1 A. This is hidden under "220 V 50 Hz AC" in the diagram.

 

[Averagesupernova]

Yes, I agree. Especially when the SMPS benefits aren't really necessary. However, National's Simple Switchers are really easy to use. They are highly integrated ICs and they've done the design for you for common applications, including layout and BOM.

I have noticed that switchers are becoming easier to design due to the reasons you mentioned.

 

[gary350]

I need to create a 5 V stable DC supply out of 6 V AC from a transformer. By "stable", I mean that I do not want the ripples in the voltage that one gets from the output of a full-wave bridge rectifier. I have to feed this to two ATmega 8A μCs.

I am going to create the following circuit:

View attachment 275252

6 V RMS means ≈ 8.5 V peak. The capacitor will hopefully smooth out the ripples. The LM 7805 voltage regulator will bring down the voltage to 5 V. I can attach a heat sink to it if it tends to get heated up too much.

I don't yet have all the materials; once I get them, I can use the Arduino analogRead() function to have a look at the waveform of the output of the circuit because I don't have an oscilloscope.

Does this circuit look okay? Any suggestion for improvement?

You need another 100uf capacitor on the output to reduce ripple, I would use 330uf or 500uf on the output. If you have no scope test it with a digital meter. If meter reads 5VDC and 0VAC your good. If meter shows .01VAC you still have a tiny amount of ripple. If meter reads any amount on AC it means you need a larger output cap. I use 7805 in some of my circuits I think it needs a voltage difference of 1.5v to work. I bought a hand held $25 scope from china it works great. You should be ok, 6VAC x 1.414 = 8.484 volts peak to peak. Put a heat sink on the 7805.

 

[Averagesupernova]

Any time you need 100uF on the output of a 7805 drawing minimal current, something is seriously wrong. You can't lose going by the data sheet on capacitor recommendations. I've never seen a 7805 data sheet specifying 100 uF on the output.

 

[davenn]

I've never seen a 7805 data sheet specifying 100 uF on the output.

maybe not, but it's standard practice when using regulators whose initial source is AC rather than a battery supply

 

[shjacks45]

I need to create a 5 V stable DC supply out of 6 V AC from a transformer. By "stable", I mean that I do not want the ripples in the voltage that one gets from the output of a full-wave bridge rectifier. I have to feed this to two ATmega 8A μCs.

I am going to create the following circuit:

View attachment 275252

6 V RMS means ≈ 8.5 V peak. The capacitor will hopefully smooth out the ripples. The LM 7805 voltage regulator will bring down the voltage to 5 V. I can attach a heat sink to it if it tends to get heated up too much.

I don't yet have all the materials; once I get them, I can use the Arduino analogRead() function to have a look at the waveform of the output of the circuit because I don't have an oscilloscope.

Does this circuit look okay? Any suggestion for improvement?

"two ATmega 8A μCs." So a 16 Ampere supply? The 1N4007 are 1 Amp max and LM7805 versions are 1 Amp max. Transformer standard voltage 6.3 VAC typically max out at 10 Amps. Or pay for custom transformer. It will weigh a lot and cost $$$.
8 volts is typical computer power supply from 6.3 volt transformer, rem diode forward voltage drop which increases with current. Actually old time computer power supplies used a center tapped 12.6 volt transformer and 2 diodes (not a bridge as you show) so only one diode drop not two.
I'm puzzled by your "8A" requirement as Atmega products are usually low power.
8 Amps is more 5V power used than desktop computers with a billion times more computing capability than Atmega chips.

 

[Wrichik Basu]

"two ATmega 8A μCs." So a 16 Ampere supply?

Where did I say 16 A?

I'm puzzled by your "8A" requirement as Atmega products are usually low power.

ATmega 8A is a uC. It is not the same as ATmega 8 ampere (which doesn't make sense anyway).

 

[Wrichik Basu]

You need another 100uf capacitor on the output to reduce ripple, I would use 330uf or 500uf on the output.

If you see this post, where I have posted the updated circuit diagram, you will find that the transformer has been changed to 9V, and the electrolytic capacitor has been increased to 4700uF.

 

[shjacks45]

Where did I say 16 A?

ATmega 8A is a uC. It is not the same as ATmega 8 ampere (which doesn't make sense anyway).

ATmega8A is micro, "ATmega 8A" is just more bad spelling.

 

[DaveE]

If you see this post, where I have posted the updated circuit diagram, you will find that the transformer has been changed to 9V, and the electrolytic capacitor has been increased to 4700uF.

I think he meant after the regulator, but I wouldn't bother. It wouldn't hurt, but shouldn't be unnecessary.

I would do this if I had a load that intermittently drew more current than the regulator could source, but for a duration that exceeds the ability of the other downstream caps (bypass caps, mostly) to supply. I don't think 2 uPs are that sort of load. If they were, then the caps need to be fast (low esr, ceramic, etc.) and very close to the uPs. When digital circuits want lots of current the di/dt can be huge (but usually isn't); they want that current faster than an electrolytic located back at the power supply can deliver. If your uPs came on a PCB designed by someone else then they have already solved that for you with local storage.

 

[Averagesupernova]

...they want that current faster than an electrolytic located back at the power supply can deliver.

This was my whole point. I'm not saying a single 1uF across the complete 5 volt supply rail is adequate. Capacitors that exist to dump current on sudden demand need to be placed as close as possible to the device that demands it. This means often times several in parallel as @Baluncore described.

 

[Wrichik Basu]

After a really long wait, I could finally gather all the required components (including the oscilloscope proposed earlier). But trying to put the Schottky diodes into a breadboard drained away my enthusiasm. Man, I just couldn't push them in — the leads were so thick! Even the large capacitor was creating trouble: the positive lead was so long, that when I pushed it in, the shorter negative lead couldn't be pushed in. Finally, made a compromise:

That diode got bent when I was trying to put it into the breadboard. Probably one of the ugliest circuits I have ever built.

Will test it tomorrow morning, after rechecking the diode bridge connections.

 

[DaveE]

After a really long wait, I could finally gather all the required components (including the oscilloscope proposed earlier). But trying to put the Schottky diodes into a breadboard drained away my enthusiasm. Man, I just couldn't push them in — the leads were so thick! Even the large capacitor was creating trouble: the positive lead was so long, that when I pushed it in, the shorter negative lead couldn't be pushed in. Finally, made a compromise:

View attachment 283086View attachment 283087​

That diode got bent when I was trying to put it into the breadboard. Probably one of the ugliest circuits I have ever built.

Will test it tomorrow morning, after rechecking the diode bridge connections.

Be careful about forcing fat leads into the breadboard, you can stretch the springs and cause connection problems later with really thin leads.

Honestly, I hate those things and will only use them for something I know for sure I won't be using tomorrow. OTOH, most everyone else doesn't agree with me!

95% of my breadboards look more like this:

 

[Wrichik Basu]

95% of my breadboards look more like this:

With experience, you can directly solder components. But I don't have that, and have to depend on breadboards.