Just bleeds off the charge from the capacitors (over several minutes) so you can work on the power supply with no external load attached.
>50V is a bit zappy so you want to drop the voltage down quite quick without unduly impacting efficiency.
Hi Andrew,
I used your figures to try and confirm what transformer to specify in order ensure that I get close to 55V DC @ 240V at idle.
I assumed the regulation on your big transformer was 4.5% and your figures were off-load. At idle with 245V mains, your 230V:40V transformer would have say 44.5V secondary voltage. If you observed 58.5V then your total rectification losses would have been:
44.5 x 1.414 - 58.5 = 4.4V
Does that sound right? I measured my equivalent losses to be 3.1V using dual bridges. If you used a single bridge, why is the loss higher? Are you including losses across your fuses/fuseholder? What have I missed?
I know it seems like small beer, but if I pay out for decent transformers I don't want to end up 2-3V off target, and I definitely want to stay away from the 63V cap rating.
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Measure your 230:dual 35Vac transformer.
It will show what effect varying the input voltage has.
It will show the effect of taking transformer regulation into account.
It will show the effect of applying an increasing load on both average output voltage and on rail ripple voltage.
Then apply a factor to convert all your new found data to a 40Vac transformer.
It will show what effect varying the input voltage has.
It will show the effect of taking transformer regulation into account.
It will show the effect of applying an increasing load on both average output voltage and on rail ripple voltage.
Then apply a factor to convert all your new found data to a 40Vac transformer.
Just hooked up 2nd bank of capacitors, this time using one bridge rectifier.
Rail voltage with dual bridge = 48.8V
Rail voltage with single bridge = 49.4V
So... each diode drop is worth 0.6V at idle. Of the total 3.1V loss then in the dual rectifier setup, 1.2V is attributed to diode drop. Hope this might help someone else.
Rail voltage with dual bridge = 48.8V
Rail voltage with single bridge = 49.4V
So... each diode drop is worth 0.6V at idle. Of the total 3.1V loss then in the dual rectifier setup, 1.2V is attributed to diode drop. Hope this might help someone else.
popchops, just to get a clearer understanding...
Are you using a transformer with two secondary windings?
Each winding is then connected to a bridge rect. and 2 x 10,000uF (100v) smoothing caps,
(2 in parallel on each supply rail) ?
Got my final pieces today from Maplin UK, so once ive finished drilling the holes for amp i/p, o/p and mains i/p, mains switch, wiring will begin.
Delange have you still got the fire extinguisher?
Are you using a transformer with two secondary windings?
Each winding is then connected to a bridge rect. and 2 x 10,000uF (100v) smoothing caps,
(2 in parallel on each supply rail) ?
Got my final pieces today from Maplin UK, so once ive finished drilling the holes for amp i/p, o/p and mains i/p, mains switch, wiring will begin.
Delange have you still got the fire extinguisher?
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I didn't needed one; amp works perfectly.
I'm sure yours will work fine too. Just double check everything before applying power the first time. And use power resistors as fuses for the initial startup and adjustment.
This is how I did it:
* use a power resistor of 47 ohms / 5 watt resister in place of the fuses in each power rail (you need two)
* Turn P1 completly to the left
* apply power and meassure the voltage accross one of the power resistors. The current should be around 30 mA (if I remimber correctly). In any case, the current should be very low. Leave this on for a few minutes so that the power resistors get to their "operating temp". Be sure the meassure the current with the spreaker relais activated !
* disconnect power and meassure the resistance of one of the power resistor. This is the resistor you will use the set the iddle current.
* apply power again and meassure the voltage drop accross the resistor. Using ohm's law you calculate the current. (make sure that the speaker relais is activated!).
* Turn P1 until you increase the current by 30 mA - in other words, the current you have meassured in the previous step + 30 mA. (by again meassuring the voltage accros the power resister and using ohm's law to calculate the current).
I used this method because the current value is too low to get accurate readings from my (very old) Fluke multimeter. If you have a very decent current meter then you could use that instead.
That also works.
I used the power resistors because you need them anyway for a safety check at first startup. If there is an issue in the amp then the power resistors will smoke.
Great! Thanks Delange! How can I be sure the speaker relay is activated? Just listen or elee do I need to check continuity whilst the pcb is live?
You mention +30 mA... from your words it sounds like that figure is per rail? Is that correct? 30mA through each temporary resistor? Did you do setup using just one or with both pos and neg resistors?
I also cannot measure milliwatts accurately so this suits me perfectly. Many thanks.