Hi;
I'd like to make sure on something.
I was told long ago, that when you rectify AC with a full wave bridge, the voltage goes up 1.41 (root 2) because of the RMS concept.
Example, say you have a transformer. It goes from 110VAC primary to 24 VAC secondary. You full wave rectify it, then smooth the output with a properly sized electrolytic capacitor. Rather than having 24 VDC, you actually end up with 33.8 Volts DC, minus about 2 volts for the full wave bridge. (24 x 1.41)-2=31.8 VDC
So far, correct?? I think I verified this with a volt meter on one of my power supplies.
Next example. Let's say we don't use a transformer at all. Let's simply rectify and smooth the 110VAC to DC.
Would we get (110 x 1.41) - 2 = 153 VDC ??
Or am I missing some concept that is different because we didn't use a transformer to reduce the VAC?
Just to inform you, where I am going with this, is to create a DC voltage source for large numbers of LEDs. If the voltage didn't have to be reduced, there may be advantages to this (such as avoiding small wall-warts that operate at around 85% efficiency, and therefore waste power).
The 2 GU-10 type 72 LED bulbs I have, do have a full wave bridge, a smoothing cap, but no step-down transformer. Therefore, they avoid wasting power on an unwanted component, ie, transformer. Since there are 36 LEDs in series (and 2 of these "banks" in parallel) I assume the voltage is shared by the 36 LEDs and current limiting resistor, which is about 500 ohms.
I like the fact that there is no small transformer because of the power waste issue.
Thanks. You'ave all been very helpful this far. Tom.
I'd like to make sure on something.
I was told long ago, that when you rectify AC with a full wave bridge, the voltage goes up 1.41 (root 2) because of the RMS concept.
Example, say you have a transformer. It goes from 110VAC primary to 24 VAC secondary. You full wave rectify it, then smooth the output with a properly sized electrolytic capacitor. Rather than having 24 VDC, you actually end up with 33.8 Volts DC, minus about 2 volts for the full wave bridge. (24 x 1.41)-2=31.8 VDC
So far, correct?? I think I verified this with a volt meter on one of my power supplies.
Next example. Let's say we don't use a transformer at all. Let's simply rectify and smooth the 110VAC to DC.
Would we get (110 x 1.41) - 2 = 153 VDC ??
Or am I missing some concept that is different because we didn't use a transformer to reduce the VAC?
Just to inform you, where I am going with this, is to create a DC voltage source for large numbers of LEDs. If the voltage didn't have to be reduced, there may be advantages to this (such as avoiding small wall-warts that operate at around 85% efficiency, and therefore waste power).
The 2 GU-10 type 72 LED bulbs I have, do have a full wave bridge, a smoothing cap, but no step-down transformer. Therefore, they avoid wasting power on an unwanted component, ie, transformer. Since there are 36 LEDs in series (and 2 of these "banks" in parallel) I assume the voltage is shared by the 36 LEDs and current limiting resistor, which is about 500 ohms.
I like the fact that there is no small transformer because of the power waste issue.
Thanks. You'ave all been very helpful this far. Tom.