I have an elementary quesion regarding development of a simple AC to DC converter. I was hoping that I could get some input on the components that I've chosen before I go off an start buying transformers.
The input is US 110-120VAC @60Hz. The output needs to be ~2.8VDC. I plan to use an LM317T for regulation after the rectifier. The application will source < 300mA at this voltage. My question is regarding the transformer and bridge rectifier. Given the overhead voltage requirements of the linear regulator (~2V), and the possibility that I may want 3.3V or 5V DC at some point (to drive a MAX233A, for instance). I was considering this transformer:
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(sorry, I don't know where to get a datasheet for this one), and a KBP04M Bridge Rectifier (which I have on hand), the datasheet for which can be found here:
If anyone has time to look over these components or suggest an alternative to what I've mentioned, I would appreciate the input. Anything I can buy from Mouser, Digikey, Jameco, etc will work. Sorry for the newbie question. I didn't think a schematic was necessary, but I can convey this if it helps.
I think I would use a transformer with a center tapped secondary and use two diodes in a half bridge (perhaps Schottkys like 1N5817 or SR102 to minimize voltage loss). Then if you decide to raise the voltage to 5 volts, you can change to a full bridge and double the raw DC voltage. A
6.3 VCT secondary would put out a peak voltage of about 4 volts with the half bridge. This is a bit low, unless you get a transformer with a very conservative current rating (say, more than 1 ampere), so you may need to go with a 10 volt CT or dual 5 volt windings that you can series as a center tapped winding.
There is also a way of getting two voltages out of a center tapped transformer without extra parts. Put a bridge rectifier across the whole winding, with the negative side of the bridge output grounded. There will now be rectified DC at 1/2 the voltage at the center tap. The voltage at the center tap will only be one diode drop down. Add the usual capacitor at the bridge output, and another from the center tap to ground. This works because there is no AC voltage at the center tap. To get 2.8V from 4V would probably require an LDO, or a discrete circuit with a PNP pass transistor. I have done this with a 12.6VCT transformer to get +12 and +5.
Thanks for the detailed feedback, I appreciate it. Would it be considered wasteful to regulate 12.6V down to 2.8-5V using an adjustable linear regulator? I could certainly use something like this:
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Also, with respect to using both outputs in parallel, would this require two rectifier circuits or just one with the primary and secondary in parallel ahead of the rectifier? I would also like to figure out how to determine approximate load on the transformer when the DC load is sourcing (I think that's the way to say it) 200mA through a linear regulator. I'm sure the datasheet on the LM317T provides information on losses, etc. Is this easy to sort out?
Here's something better, at about 1/2 the cost: MPJA stock # 12254 PD is a 9 V DC at 1 amp wall wart for $3.50. That's enough current for all three regulators so you can have 2.8, 3.3 and
5 volts simultaneously. Use a 7805 for the 5 volts, and a 317 for the 2.8 and 3.3 volt supplies. The Jameco transformer would be marginal for the 5 volt supply, and would require a rectifier and filter cap, and power only one regulator at a time, while costing close to twice the MPJA DC supply.
I'd drop the voltage between the wall wart and the regulator chip with a 5 watt resistor so that you have close to 3 volts headroom at the input of the regulator when the current is 300 mA. This will keep the maximum heat in the regulator to roughly .9 watts. To figure the value of the resistor, measure the output from the wall wart with a
300 mA load attached, then subtract 3 volts for headroom, and then subtract the desired regulator output voltage. Divide the result by .3 to get the value in ohms, and select the closest standard resistance value equal to or lower than your figure.
For example, say the wall wart output with a 300 mA load is 15 volts, and you want 2.8 volts: 15-3 = 12, then 12-2.8 = 9.2, then 9.2/.3 = 30.6666 ohms. So you would use a 30 ohm, 5 watt resistor. At 300 mA, it would drop the voltage at the input of the regulator to 6 volts, leaving 3.2 volts to be dropped in the regulator. At 300 mA, that would be about .96 watts
Shown below is a diagram of all three regulators, based on an assumed 15 volts output from the wall wart with a 300 mA load.
Since power dissipation in the regulators will be about
1 watt worst case, they need to be installed on heat sinks. The power resistors - 22 ohms, 27 ohms and 30 ohms need to be 5 watts each. Those values will need to be re-computed if the wall wart output, when supplying a 300 mA load, is not
15 volts.
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