bridge rectifier error?!

[evostars]

before modified, what was the voltage at the input of the 5V regulator IC20?

Around 18 but im not sure, could be 22

any voltage above 5V will do. the regulator will put the rest into heat anyways...

Yeah, the problem was the c rating of 16V.

[gompie]

Change 16v to 25v c's i meant.
Ac bleed: i mean its connected straight to the ac of the adaptor.

The first elco's after the rectifier are there to create a sort of DC made from the polarised ac coming from the diode brigde. They even out the polarised AC into DC. If on the elco (C5 or C3) there is still a ripple you don't like to see, you should increase the size of the elco, am I correct?

[evostars]

wow. I have not built an x0xb0x, but this does seem strange. are you referring to a classic 'diode bridge'? like this:

diode bridge w cap.PNG

if so, it would appear that you are correct in that the x0xb0x does not follow this design.

what voltages were you reading at top and bottom of C5?

Thats exactly what i mean. A diode bridge.
Maybe I'll makr a picture to explain what happens to the voltage.

The ac of the adapter is transposed upon the positive rectified signal from the diodes.

If I'm correct the 9v of the adaptor is actually plus and minus 4.5v rms

The rectified signal rides on top of this. 9x1.4=12.6V-1.1v(diode)=11.5v plus and minus 4.5V is a signal between 7 and 16 volts

[evostars]

Change 16v to 25v c's i meant.
Ac bleed: i mean its connected straight to the ac of the adaptor.

The first elco's after the rectifier are there to create a sort of DC made from the polarised ac coming from the diode brigde. They even out the polarised AC into DC. If on the elco (C5 or C3) there is still a ripple you don't like to see, you should increase the size of the elco, am I correct?

Correct. But the zener and the 7806 and 7805 stabilize the voltage(no more ripple)

[gompie]

Change 16v to 25v c's i meant.
Ac bleed: i mean its connected straight to the ac of the adaptor.

The first elco's after the rectifier are there to create a sort of DC made from the polarised ac coming from the diode brigde. They even out the polarised AC into DC. If on the elco (C5 or C3) there is still a ripple you don't like to see, you should increase the size of the elco, am I correct?

Correct. But the zener and the 7806 and 7805 stabilize the voltage(no more ripple)

cool, I thought you had an issue with AC bleeding through, like in: I see 50Hz somewhere where I don't expect to find it

[machine.cuisine]

I don't understand the way the AC-DC converter is set up in the x0xb0x schemes. C5 is across one of the diodes that is in the bridge. the 5V regulator taps off of the negative input of the bridge...AC connects directly to the 5V regulator!!! is this correct?

if this is some circuit-trickery that I obviously don't understand - I would appreciate to be enlightened, because I just don't get it.

[antto]

i don't understand any of this..
there's a nice online circuit simulator (java applet) here: http://www.falstad.com/circuit/
you might already know about it
if not, it could be useful, it visualises voltage and current, and the circuit can be exported to "pastable" text (and then imported back)

[evostars]

I don't understand the way the AC-DC converter is set up in the x0xb0x schemes. C5 is across one of the diodes that is in the bridge. the 5V regulator taps off of the negative input of the bridge...AC connects directly to the 5V regulator!!! is this correct?

if this is some circuit-trickery that I obviously don't understand - I would appreciate to be enlightened, because I just don't get it.

The 5 volt regulator isn't connected to the negative input. Its just the "other input" with ac the pos and neg change... But the ground is connected to the negative output of the bridge.

If those guys at banned did design this on purpose, why didn't they use the right voltage ratings for the C's?
Hence... I think its wrong. A mistake, even at banned...

Im going to order the 25v 2200 uF's and than Im done with it...
For now...
Because I want to change the PS even more, but that is a diffent post

[evostars]

maybe I'm wrong, but it should explain it a bit.

A= ac from the adaptor at input 1
B=ac from adaptor at input 2 (out of phase, ow am I wrong?)
C=after the diode rectifier at C5 (negative side is added to the positive side, thus rectified).
D= B+C which ads up to E
E=double the rectified voltage, but with gaps. The 2200uF condensors takes this voltage and smooths it out to a positive dc voltage with a ripple on top.

[evostars]

what they basically did:

[machine.cuisine]

yes sir, I understand what a rectifier and diode bridge is (the diodes rectify the AC wave, and in effect, flips the "negative" portion of the wave up to what we would call "positive side". The capacitors then try to hold this voltage, making a crude DC from AC.) and I know that when it comes to AC there is no "negative" - that was just a figure of speech related to the schematic.

I just don't understand why they (the x0xb0x designers) tapped off of the converter the way they did. I don't see a reason. But mostly:
1. I have never seen a capacitor placed across AC like they did.
2. I have never seen AC connected directly to a DC voltage regulator

I admit I have not seen a lot in my life and that I could be totally missing something, though.

[evostars]

yes sir, I understand what a rectifier and diode bridge is (the diodes rectify the AC wave, and in effect, flips the "negative" portion of the wave up to what we would call "positive side". The capacitors then try to hold this voltage, making a crude DC from AC.) and I know that when it comes to AC there is no "negative" - that was just a figure of speech related to the schematic.

I just don't understand why they (the x0xb0x designers) tapped off of the converter the way they did. I don't see a reason. But mostly:
1. I have never seen a capacitor placed across AC like they did.
2. I have never seen AC connected directly to a DC voltage regulator

I admit I have not seen a lot in my life and that I could be totally missing something, though.

Haha. Same here. Can't seem to find a reason.
Still think it's a fault... Mostly because of the voltage rating of the C's.
But to think some guys at banned did it.... Hmm makes me wonder

[evostars]

I understand the opamp that makes the 5.333 V &12 V needs 2v headroom. So 14V should be enough. Actually the tr303 had 12 V feeding the opamp i believe. Correct me if I'm wrong....
So with the modded bridge rectifier, the opamp gets powered like the TR303. That's another plus. Anybody like warble?

[antto]

if you think your mod makes the power supply section close to the TB-303 then it's a good thing to put it in the x0x wiki

[gompie]

yes sir, I understand what a rectifier and diode bridge is (the diodes rectify the AC wave, and in effect, flips the "negative" portion of the wave up to what we would call "positive side". The capacitors then try to hold this voltage, making a crude DC from AC.) and I know that when it comes to AC there is no "negative" - that was just a figure of speech related to the schematic.

I just don't understand why they (the x0xb0x designers) tapped off of the converter the way they did. I don't see a reason. But mostly:
1. I have never seen a capacitor placed across AC like they did.
2. I have never seen AC connected directly to a DC voltage regulator

I admit I have not seen a lot in my life and that I could be totally missing something, though.

Haha. Same here. Can't seem to find a reason.
Still think it's a fault... Mostly because of the voltage rating of the C's.
But to think some guys at banned did it.... Hmm makes me wonder

Check http://en.wikipedia.org/wiki/Rectifier for an explanation on retifiers.
C5 is charged from the correct point of the rectifier (Full-wave rectification)
C3 is charged from the odd point of the rectifier (Half-wave rectification)

Half-wave rectification will creat a bigger ripple on C3 ofcourse, but the regulators behind it (like the 5v regulator with the heatsink) don't suffer from that and create a stable output voltage.

The reason of this design is to be able to charge C3 and C5 seperately, so they can then function as a voltage doubler (what I tried to explain earlier). The doubled voltage (measured from the + of C5 to the - of C3) is intended to power the op-amp (which can handle a voltage up to 30v). I figure this is indeed intentional in the design, to compensate for power fluctuations, so that the output of the op-amps (a and b) will remain stable under all sircumstances)