Very simple Comparator Power Supply.

[Flunze]

Ok, mates. I used over 2 hours, to create this nice piece of work. I guess, you could like it.

Picture 1
*** SCHEMATIC REMOVED BY ADMIN FOR SAFETY REASONS ***

And now the prepared Circuit board - this is, by the way, created for my Power Amplifier. You will see 4 Pins unconnected on the left side. This is, where a very small, week SMPS at 12V will be added, which is required by my amplifier. (It is completely MR insulated, so it will not short out anything in the circuit.).

Picture 2 ( Edited here!!! Now you see, what took me 2 hours of work - it was NOT the schematic! )
*** SCHEMATIC REMOVED BY ADMIN FOR SAFETY REASONS ***

If you want to say, the Ripple will be very high, you are right, but I also have another PCB with a filter bank, using a high value inductor and also a bunch of Capacitors, so this will not be a problem at all.


This circuit is untestet, but will be tested, as soon, as I get my components. Well: Why is this Method not used anywhere in reallife electronics? This should realy be working, so what is wrong with that? Maybe to high ripple currents for cheap components? o.O

Also, forget about the fact, that I wrote a wrong resistor value. R14 in picture 2 has to be 18M. (180V@C4 in picture will be split up to 170.5V and 9.5V as the trigger. The Op Amp will stay on, untill the voltage across R14 reaches about 9.5V, which is connected to the inverting input. --> As long as the Voltage on the Inverting input is lower than on the Non Inverting input, the Opamp will let current flow through T1).
Another mistake in the image: R10 and R13 also is not in M, it is in K Ohms.

Please feel free, to discuss about that and tell you opinion. But if you do, please talk about Picture 2, since this is more planned, than P1. Thank you very much.

[Alana]

But... what does it do?
From description and schematic i cannot tell what is the purpose of a device.

[Flunze]

An Opamp let's current flow, if the Voltage on the "-"Pin is lower than the voltage on the "+" pin. (Ground as reference). Using this knowledge, you can load up capacitors with that. If the Capacitor reached a certain Voltage, it turns off. If the voltage drops again, it turns back on. Works exactly the same as a Switch Mode Power Supply without a transformer. Just follow everything. All the resistors in series are used, to split the Voltage up, as wished, to set up the target voltage of the Output Pins. The rest (Cap in series, Resistor in series) is used, to give me a positive and negative Voltage from Ground. The resistors are used, to keep the Capacitor charges equal, even if the Capacity shifts away for a bit. (This will happen to any Electrolytic cap)



This is all, actually. Do you have questions about that, or did you get, what I am trying to explain? I hope, someone, who knows more about that, cann tell me about that.

[alm]

Both pictures look the same to me. With the current connections, IC2's output will always be close to the negative rail, so Q2 will be in the cut-off region. This would essentially make the negative output terminal floating. Even with the correct connections, I don't see the point. Do you need regulation at all with this much capacitance? What's wrong with power transformers, do you hate safe circuits? If you don't care about ripple rejection or line regulation, why not make a simple resistive divider followed by a source follower? Replace the resistive divider with a zener and you have an sixties-era transistor power supply with much better ripple rejection and regulation.

[Psi]

10,000uf 400V capacitor on the DC mains input side?   (C3)

That's going to be a pretty huge and expensive cap and will cause a massive inrush current when you plug it in.
May even pop the breaker or blow the bridge.

[Kremmen]

Assuming for the sake of argument that this is not a troll. And even if it is, there are lots of morons out there who might be suckered in...

Ok, mates. I used over 2 hours, to create this nice piece of work. I guess, you could like it.

Sorry to disappoint, but there little to like.

This circuit is untestet, but will be tested, as soon, as I get my components. Well: Why is this Method not used anywhere in reallife electronics? This should realy be working, so what is wrong with that? Maybe to high ripple currents for cheap components? o.O

Please forget about testing it. Go to your file Explorer, highlight the file, hit Delete and forget you ever drew it. Then go do gardening or something.
Why people generally shy away from bombs like that is their irrational affection to life. Need any more reasons?

Please feel free, to discuss about that and tell you opinion. But if you do, please talk about Picture 2, since this is more planned, than P1. Thank you very much.

Since you asked, I did.

[Bored@Work]

If anyone wants to know why this is rubbish, start with checking the wiring of R90. That should give you a pretty good idea of the "quality" of this two hour "invention". Remember, Kirchhoff's voltage law wins all the time.

[T4P]

The moment i saw the 10mF cap i knew this is a moronic attempt at a power supply
Guess how much current required to charge up the cap? You might not only trip the breakers, you will almost immediately blow the diodes into pieces and probably the larger 100amp breaker ( my house has one )

And ... kirchhoff won again

[Stephen Hill]

The moment i saw the 10mF cap i knew this is a moronic attempt at a power supply
Guess how much current required to charge up the cap? You might not only trip the breakers, you will almost immediately blow the diodes into pieces and probably the larger 100amp breaker ( my house has one )

And ... kirchhoff won again

The ESR would stop this from happening.

[T4P]

The ESR would stop this from happening.

Yeah right. 
The bigger your cap the lower the ESR, i just checked a Cornell Dubiler ( which costs 251$ WTF ) DCMC103M400 has 22.5mOhms ... IT WILL DEFINITELY TRIP THE BREAKERS

[Stephen Hill]

The ESR would stop this from happening.

Yeah right. 
The bigger your cap the lower the ESR, i just checked a Cornell Dubiler ( which costs 251$ WTF ) DCMC103M400 has 22.5mOhms ... IT WILL DEFINITELY TRIP THE BREAKERS

Yes, your right.

[Flunze]

Oh my god. Why the fucking hell? I inserted the wrong link? Arw! My Firefox sometimes does not want to copy text. I will edit the first post and you will see more realistic capacitors in the 2nd image. Sry, mates, that was not wanted. Ignore the first image, it was just meant for understanding it better.

To all the people, that say, that it's just crap: Tell me why. How is it supposed, not to work? I mean, the Opamp checks for the correct voltage and will load the cap depending on, if it is required, or not.


To all the people, that say, that the current will break down something: Already thinking about a good solution. Maybe like in one of Dave's video: The caps load through resistors and than a relay closes and allows higher currents to flow.

[mianchen]

swearing is cool, says the care bear.

[SeanB]

Sorta like a live version of the Darwin awards?

[Flunze]

Oh my god. Why the fucking hell? I inserted the wrong link? Arw! My Firefox sometimes does not want to copy text. I will edit the first post and you will see more realistic capacitors in the 2nd image. Sry, mates, that was not wanted. Ignore the first image, it was just meant for understanding it better.

doesnt matter it changed nothing. tell us why you make a voltage divider R90 and R91 and then tie the middle voltage to -ve rail and then to opamp (comparator?) noninverting input? and how is your logic try to regulate using low side switching and use the high side as -ve voltage output tab?

Please let's just think about Picture 2. I will explain it completely.

R10 and R13 Split the 325V at the Caps up, to power the OpAmp, because it only needs the low voltage of R13.

Than you need a fixed Voltage, that will go to the -InputPin of the OpAmp, to set a Voltage level, where the OpAmp should begin to amplify, if a Voltage higher than that is on the +InputPin. To get a Reference Voltage, I split up the 325V again with R11 and R14.

And than there is a Resistor on the Mosfet (R15), which will discharge the Gate, if no voltage is applied. If Voltage is applied, it will let current flow into the Capacitor C4. But when does it need to stop? Very simple! I divide the Voltage of C4 with R9 and R12, to get the correct Voltage. The Voltage at R12 will be compared in the OpAmp with the Voltage at R14.

Only 2 more things to know: What does the Opamp do? Well, every OpAmp is some sort of a comparator, that can be used, if a Voltage is higher than the other Voltage to ground. So if the Voltage at C4 through R12 is not higher than the fixed reference Voltage at R14, the Opamp will allow current to flow through the Mosfet's Gate and turn on the Charging process of C4. If it reaches a certain Voltage, the Voltage across R12 is higher than the Voltage across R14 (Reference Voltage), so the OpAmp will not allow Current to flow. Since that, the Mosfet will turn off with the help of R15 and the Cap will not charge up more.

Now some other components:
-C1 will give enought power, to even allow charging the Caps, because a 100hz DC half-sine-wave would not work good at all.
-D1-D4 will bridge rectify the 230V AC to 325V DC.
-But wait? What is R1-R8 used for? 2 Things actually: It will discarge the caps as safety feature, if power is turned off AND also fix the maybe not perfectly split up Voltage across C2 and C3. Electrolytic Caps are never even close to perfect, so it would not be 90V on each side. The power resistors will fix that problem.

If you think, there is no such good Cap, that fits on the PCB, you are right. There will actually be a thick wire, to connect to external Capacitors - 3300uF each one. A friend found them and will give it to me.

I guess, that actually is everything, that could be told about that.


I think, you might have noticed a mistake. What is it? Well, if the MOSFET is turned off, how does current flow from R12 to ground? Well, it is not a mistake at all. You don't need to take care of that. Why? Well, the Voltage only can rise, if Current flows through the MOSFET. It only needs to turn off the OpAmp, if Voltage rises to high. So, when the Voltage rises, the Mosfet will work as a connection to ground and current flows through R12 and can be compared to R14.

To anybody, who wants to say, that this cannot work, explain why.

The high ripple current wont be a problem, I got a solution for that. The coper on the PCB will also allow enought current to flow, I also got a fix for that problem. So, what do you think? This must work - I mean, how are the Components supposed, to not do, what I want? Impossible, if they do, what they where designed for.

[T4P]

I don't see how a simple PCB can take you 2 hours and if you didn't know a 400V 1mF cap is pretty expensive still, use a freaking transformer PLEASE?!
And a 1mF cap will still trip the breakers

[poptones]

To anybody, who wants to say, that this cannot work, explain why.

Because it is directly coupled to the mains and there is no isolation in the event of a fault. It will put everything connected to it directly to the mains. It will be a shocking experience, and you will probably fry any preamplifier connected to it.

[ptricks]

Very simple, but very dangerous design. If any resistor or the FET latches. or the opamp fails there is nothing to stop the whole thing from going up in smoke. Also a 400v capacitor is not the correct voltage for a circuit like this, you need much higher and that is going to be very expensive.

[Flunze]

I don't see how a simple PCB can take you 2 hours

Arw, was just playing with the design in Gimp. Pixelwork.

use a freaking transformer PLEASE?!

Oh, of corse. You are totally right. Than, if you say that, you sure will pay a transformer with such high values at several amps for me? Of course, thanks for that money.

Very simple, but very dangerous design. If any resistor or the FET latches. or the opamp fails there is nothing to stop the whole thing from going up in smoke. Also a 400v capacitor is not the correct voltage for a circuit like this, you need much higher and that is going to be very expensive.

Dangerous, that's unfortunately right.
Well, I got the Caps for 15 bucks.

To anybody, who wants to say, that this cannot work, explain why.

Because it is directly coupled to the mains and there is no isolation in the event of a fault. It will put everything connected to it directly to the mains. It will be a shocking experience, and you will probably fry any preamplifier connected to it.

And that, mate, is one thing, I just thought about... But! isn't there a way, to simply use an audio transformer on the input, to insulate the Input from anything else? If the speaker outputs etc. are at such high voltages, it is no problem at all, but what about an audio transformer? Are there any with high quality? Will google.


Well, so this simple supply should actually work, but kill preamps on the amp's input and the caps are the problem?

Well, I got power caps for that reason, no problem at all. And if I can insulate the input with an audio transformer, there shouldn't be any other problem left.

So for that, that I get the capacitors very cheap, I would say, that this method is the one, with the lowest possible costs for me at all. I will try to wind my own transformer, to step down the voltage for that, hope, that this works, than this method must not be. But I am not sure (I actually doubt it), if the Core of a microwave oven transformer is large enought, for pulling that power out.

[poptones]

So to save buying a $30 power transformer, or clipping one from an old piece of junk, or winding a light, efficient switching transformer, you're going to invest in audio coupling transformers? Do you have any idea how much decent performing audio coupling transformers cost?

You don't think a 1KW transformer is enough to "get the power out?" How much power are you planning on getting from this amp? You're trying to build a KW amplifier and you're worried about saving a few bucks on a proper power supply?

You're trolling, aren't you?

[Flunze]

I guess, you don't understand that 2 facts:

Where on the universe do you get a Transformer, that gets 100V,90V,GND,-90V (after rectifying), that pulls that out? Such a transformer is not payable. Also, I never saw any Transformer, that has pinouts like that. The price, to buy one with that power in Germany would be over 200€, if you let them make one, you would maybe pay over 300€. A Transformer with 2x20V at 5 Amps costs over 15€. If you want 10 Amps at 64VAC (90VDC) and an additional 10V Rail, you would pay over 100€, if there would have ever been any mass produced Transformer. The parts for my supply (Including the big Capacitors) cost 20€ + a small audio transformer. The Audio Transformer, I'll use, costs just 2,22€. This makes 22,22€ for my Supply. Now, how can it be any cheaper?

[T4P]

A "small" audio transformer is limited to it's wattage dickhead. Audio transformers are WAYYY more expensive per watt
So you want to save 100$ on safety? And again, your cap is going to trip the breakers, don't say I didn't warn you!
Don't forget that the mains is a essentially UNLIMITED power source before the breakers trip, no resistance at all.
Let the other forum people educate this idiot
Simple but ...

[poptones]

Where on the universe do you get a Transformer, that gets 100V,90V,GND,-90V (after rectifying), that pulls that out?

Pulls what out? Not once have you stated the goal of this. How much power ARE you trying to get?

By all means: build that. I'm sure it will sound just fabulous with a 3 dollar input coupling transformer - right up to the point where your circuit becomes imbalanced and shorts a couple hundred volts DC across your speakers. Do be sure to keep the camera going, I'm sure the youtube video will be popular enough to provide you a little income to help rebuild the house after the fire.

[Flunze]

Er, you just remember, that I want to use the Audio Transformer to seperate my INPUT from the Power Supply and not the Output. So I only need very, very low power.

1mW will be enought, to run hundrets of Opamps, since they are theoretically Voltage driven and not Current driven.

...dickhead....
...So you want to save 100$ on safety?...
...Let the other forum people educate this idiot...

Learn, to read correct... I sayd, if a transformer in that power rating would be a MASS PRODUCT, it would cost 100$, but they are NOT. You call me a dickhead? Than suck my dick, because it is just rude, to call me an idiot. You are just one of the useless conservative people, who can not say, they can not help, but still want to look like the best in a forum. If people like you cannot help, they just say: Don't do it, don't do it! You will die! OMG!

Man, I hate this conservative crap. If you don't like it, you can say this, you can give me hints, but NEVER try to stop me, because I got enought of this conservative forum crap! Say "I can't help you on your methods." or STHU! And everybody else reading this in the next 5 million years: Don't try to sound like the most intelligent person on the stupid world. If you can't help, it is ok, because nobody will kill you. Let me say, that only the stupid ones are smart alecks. The intelligent person does not write like that, because he is unassuming. Please understand, I have NOTHING against you, but I can't stand it, if somebody does not want me to do something, I will do at any point on my own risk. You may be able to stop others, but if I want to kill myselfe, you won't be able to stop me in years.

Where on the universe do you get a Transformer, that gets 100V,90V,GND,-90V (after rectifying), that pulls that out?

Pulls what out? Not once have you stated the goal of this. How much power ARE you trying to get?
.....
shorts a couple hundred volts DC across your speakers.

900W out of the amp, so I want a supply with more than 1kW.
It isn't an amplifier with a bipolar design, so it will not do that at all.

[Flunze]

You know? You are all just saying: You are stupid! Buy a transformer.

THAN GIVE ME A LINK TO A TRANSFORMER, THAT GOES ONLY 50% CLOSE TO THAT AND COSTS LESS THAN 100€!
Prove, that this even exists!