18-03-2017, 11:14 PM
I think you're forgetting the diode drops there, Mike. I assume 1V per diode, which is generous but not far off the mark. So I think the rectified DC will be nearer 6 or 7 volts in practice. But I'm of the "build it and see" school of thought, as trying to predict this stuff with any more precision is a lot of complication for relatively little reward. The biggest problem is the current taken by the rectifier+capacitor will be short "gulps", and these peak at some 4 to 5 times the average current (so 8 to 10 amps here). That's why the diode drops are higher than you might expect, and why you get surprising voltage drops in any resistance in the primary wiring (I-squared-R loss).
The downside of putting the 2 in series is that now we have about 14V DC after the rectifier. This means 16 watts of heat to lose in the regulator, which requires a surprisingly big heat sink for continuous operation.
A 9V transformer is a much better bet. After the bridge, you'd have around 10V DC, and around 8 watts of heat dissipation in the IC. When the mains is 10% low, you'd still have ~8.5V, which should be just about enough for the regulator.
9-0-9 with 2 diodes rather than a bridge will half the losses in the rectification, and is what I usually prefer at these low voltages.
Interestingly, for some applications, the rectifier losses are still too much, even when using Schottky diodes. The solution is to replace the diodes with MOS-FETs. Of course, you need to turn them on and off appropriately, so here's an IC just for that: http://www.linear.com/product/LT4320
This means that it's just the RDS(ON) of the MOS-FETs that cause losses, and typical values of that might be 0.05 ohms or less. I'm not necessarily advocating such a thing for this, but it's worth mentioning because it's quite interesting, I feel. In the drive for higher and higher efficiency, even the humble diode is on the radar!
The downside of putting the 2 in series is that now we have about 14V DC after the rectifier. This means 16 watts of heat to lose in the regulator, which requires a surprisingly big heat sink for continuous operation.
A 9V transformer is a much better bet. After the bridge, you'd have around 10V DC, and around 8 watts of heat dissipation in the IC. When the mains is 10% low, you'd still have ~8.5V, which should be just about enough for the regulator.
9-0-9 with 2 diodes rather than a bridge will half the losses in the rectification, and is what I usually prefer at these low voltages.
Interestingly, for some applications, the rectifier losses are still too much, even when using Schottky diodes. The solution is to replace the diodes with MOS-FETs. Of course, you need to turn them on and off appropriately, so here's an IC just for that: http://www.linear.com/product/LT4320
This means that it's just the RDS(ON) of the MOS-FETs that cause losses, and typical values of that might be 0.05 ohms or less. I'm not necessarily advocating such a thing for this, but it's worth mentioning because it's quite interesting, I feel. In the drive for higher and higher efficiency, even the humble diode is on the radar!
