Guys i need advice because i'm not sure how this is supposed to work!
I have these full wave boards:
30A Class A Amplifier Power Supply Rectifier Filter Kits Schottky Rectifier DIY | eBay
Im following these diagrams
from each board i get about 24v
Now i want to link them so i connected the positive of one board to the negative of the other:
Now the thing is if i measure the voltage of boards together i get around 53v, which i dont want!
Trying to do a dual rail voltage psu or to make this:
So where am i going wrong?
Thanks!
I have these full wave boards:
30A Class A Amplifier Power Supply Rectifier Filter Kits Schottky Rectifier DIY | eBay
Im following these diagrams
from each board i get about 24v
Now i want to link them so i connected the positive of one board to the negative of the other:
Now the thing is if i measure the voltage of boards together i get around 53v, which i dont want!
Trying to do a dual rail voltage psu or to make this:
So where am i going wrong?
Thanks!
i thought thats what i was supposed to do? Yes the red wire is the jumper wire joining the two rectifiers together. I thought looking at the diagrams above that they are supposed to be linked.
I haven't grounded the common connection, where do i ground it to? The earth coming into the transformer?
I haven't grounded the common connection, where do i ground it to? The earth coming into the transformer?
Do you want a single bipolar +25V/0V/-25V supply, or two of them?
Your transformer has two secondaries, so you can have only one bipolar supply.
Then you only need one dual rectifier board, and it should have a jumper to join
the two outputs on the board. That jumper IS the ground (more correctly, common).
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I'm not the best person to offer advice on modification of this circuit, as it is not my strong suit. So without building it like the schematic shows, I probably shouldn't offer too many ideas 
for troubleshooting. What I can say, is that if your jumper had a "test point" in the center of it (or just a carefully meter from the bottom of the solder pad), and if you placed your meter on that point, and each of the Vout points individually, you would read 24, or 28, whichever the transformer putting out through the rectifier bridge. I believe that you are metering from both ends right now, so you are reading both voltages at once(showing you twice the amount).
for troubleshooting. What I can say, is that if your jumper had a "test point" in the center of it (or just a carefully meter from the bottom of the solder pad), and if you placed your meter on that point, and each of the Vout points individually, you would read 24, or 28, whichever the transformer putting out through the rectifier bridge. I believe that you are metering from both ends right now, so you are reading both voltages at once(showing you twice the amount).A thorough and well prepared question.
First, the circuit marked with "B" is the one ordinarily used. Then, you only need one rectifier board (if the rectifier board can stand some 50V). It seems to be a 200V rated board. This way you only have one diode drop (loss) to each rail.
What you seem to be doing is to make two individual voltages (should be around 18V, not 24V) and then stack them. Also a possibility (option "A" above) but then you have higher diode losses.
As I see the problem, you believe exists, your probe ground is connected to the negative rail voltage while the probe tip is connected to the positive rail voltage. Then, you measure the sum of the two rail voltages.
If you move the probe ground to one of the terminals where the small red jumper is connected, you have a +24V rail and a -24V rail.
Load the rails a little and the voltages should drop to around +18V/-18V.
First, the circuit marked with "B" is the one ordinarily used. Then, you only need one rectifier board (if the rectifier board can stand some 50V). It seems to be a 200V rated board. This way you only have one diode drop (loss) to each rail.
What you seem to be doing is to make two individual voltages (should be around 18V, not 24V) and then stack them. Also a possibility (option "A" above) but then you have higher diode losses.
As I see the problem, you believe exists, your probe ground is connected to the negative rail voltage while the probe tip is connected to the positive rail voltage. Then, you measure the sum of the two rail voltages.
If you move the probe ground to one of the terminals where the small red jumper is connected, you have a +24V rail and a -24V rail.
Load the rails a little and the voltages should drop to around +18V/-18V.
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I don't see any capacitors in your otherwise excellent test. This will raise the DC voltage appreciably.
The transformer is two windings of 24VAC each.
One winding into rectifier with NO capacitor will give 22VDC. (May vary depending on meter.)
One winding into rectifier AND capacitor will give 35VDC.
Both may be ~~10% higher with NO load.
So for TWO such systems, 22V no-cap plus 20% un-sag does give just over 50V"DC" as you observe.
And you also observe that this is very UGLY "DC", not smooth steady DC.
What do you actually want? +/-28V as shown in the last figure of the first post? Then 24VAC is too much, you want 20VAC*2. Or re-figure the thing you are powering to accept +/-33V.
The transformer is two windings of 24VAC each.
One winding into rectifier with NO capacitor will give 22VDC. (May vary depending on meter.)
One winding into rectifier AND capacitor will give 35VDC.
Both may be ~~10% higher with NO load.
So for TWO such systems, 22V no-cap plus 20% un-sag does give just over 50V"DC" as you observe.
And you also observe that this is very UGLY "DC", not smooth steady DC.
What do you actually want? +/-28V as shown in the last figure of the first post? Then 24VAC is too much, you want 20VAC*2. Or re-figure the thing you are powering to accept +/-33V.
I suppose he's building this:
30w Jean Hiraga Power Supply Design, one large capacitor or several?
Besides the RC/LC arrangement, this is essentially the same layout as what is used in the first watt amps so you could study that. If you take diyaudiostore boards as example, the grounds of each rail are connected (when used as dual rail psu) to form the common ground for both rails. That is, the terminals you jumpered should be connected both to the common ground.
30w Jean Hiraga Power Supply Design, one large capacitor or several?
Besides the RC/LC arrangement, this is essentially the same layout as what is used in the first watt amps so you could study that. If you take diyaudiostore boards as example, the grounds of each rail are connected (when used as dual rail psu) to form the common ground for both rails. That is, the terminals you jumpered should be connected both to the common ground.
I see, bit of a double take then. The added Rs are to bring the volts down with 24v secondaries I suppose.
Hi All
As @silasmellor said i'm attempting to build a Jean Hiraga Le Classe A 30W amplifier.
I want to use what i have which is:
2 x 500va R Core Transformers 2x24vac
Looking to build dual mono amplifiers each one will have 0.6F capacitance.
I realise this is an overkill and that the trannies are probably too much but as
@Alan4411 said i plan to use resistors to drop the voltage if needed.
I also have 12 x 0.11F capacitors
I would like to use chokes like the lundhals ll2733's i have two available, i was thinking and also advised to try a common mode approach or buy two more ( but budget is tight) hoping i can use one for each mono PSU.
http://www.lundahl.se/wp-content/uploads/datasheets/2733.pdf
For the moment i'm just practicing putting things together, measuring voltages, learning to use an oscilloscope to check ripples and voltages etc. If any of you can aid me in using a handheld oscilloscope and what to check for that would be awesome.
So just playing around with CRC circuits to practice on! But ultimately it's to build a CLCRC with all the right values in resistors, chokes and voltages.
excuse the basic drawing below, just trying to figure out visually what it would look like.
Now how does that connect to one amplifier board?
As @silasmellor said i'm attempting to build a Jean Hiraga Le Classe A 30W amplifier.
I want to use what i have which is:
2 x 500va R Core Transformers 2x24vac
Looking to build dual mono amplifiers each one will have 0.6F capacitance.
I realise this is an overkill and that the trannies are probably too much but as
@Alan4411 said i plan to use resistors to drop the voltage if needed.
I also have 12 x 0.11F capacitors
I would like to use chokes like the lundhals ll2733's i have two available, i was thinking and also advised to try a common mode approach or buy two more ( but budget is tight) hoping i can use one for each mono PSU.
http://www.lundahl.se/wp-content/uploads/datasheets/2733.pdf
For the moment i'm just practicing putting things together, measuring voltages, learning to use an oscilloscope to check ripples and voltages etc. If any of you can aid me in using a handheld oscilloscope and what to check for that would be awesome.
So just playing around with CRC circuits to practice on! But ultimately it's to build a CLCRC with all the right values in resistors, chokes and voltages.
excuse the basic drawing below, just trying to figure out visually what it would look like.
Now how does that connect to one amplifier board?
Positive rail (A) to V+, negative rail (B) to V-, ground (blue) to gnd.
FIY PSUD2 returns 26V with 3.5A current draw using your CLCRC with L=0.4H and R=0.1. I don't know what the draw of your circuit is, but you may want to check what your voltage is under load before adding the R infront of the filter section.
Edit: not sure about the Vsl, but the documentation provided by Jim's audio should tell you that.
FIY PSUD2 returns 26V with 3.5A current draw using your CLCRC with L=0.4H and R=0.1. I don't know what the draw of your circuit is, but you may want to check what your voltage is under load before adding the R infront of the filter section.
Edit: not sure about the Vsl, but the documentation provided by Jim's audio should tell you that.
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yes, measuring each rail to gnd (blue) will give you the rail voltage with opposite sign. From that, logically measuring from rail to rail gives you twice the rail voltage.
Sorry I derailed the thread a bit there, I think the choke vs resistor discussion is already ongoing (or over) in your other thread.
Sorry I derailed the thread a bit there, I think the choke vs resistor discussion is already ongoing (or over) in your other thread.
it may not doing what you want, but it's doing what you connect it for !!!
you connect two output in serie , so you double the voltage , it's totally basic , and that's what the Hiraga needs
also , remenber that the voltage you'r measuring is without load , so it's higher than on load
building those kind stuff need some electrical basics
.
you connect two output in serie , so you double the voltage , it's totally basic , and that's what the Hiraga needs
also , remenber that the voltage you'r measuring is without load , so it's higher than on load
building those kind stuff need some electrical basics
.
OK!
You again measure from one rail to the other rail. Then, you do not get + and - 37V but 37V+37V= 74V.
Move the probe ground to the midpoint between the two capacitors and check each rail voltage.
33uF is not much when checking the output voltage of a power supply.
Please tell if you do not want +/-37V?
You again measure from one rail to the other rail. Then, you do not get + and - 37V but 37V+37V= 74V.
Move the probe ground to the midpoint between the two capacitors and check each rail voltage.
33uF is not much when checking the output voltage of a power supply.
Please tell if you do not want +/-37V?
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