Re: AC->DC converter question

James wrote:
Hi,

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:

http://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?langId=-1&storeId=10001&catalogId=10001&productId=221292&pa=221292PS
(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:

http://www.ortodoxism.ro/datasheets/GeneralSemiconductor/mXrruqz.pdf

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.

James


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.


------
| Wall |---------+---[22R]-------+---[7805]---+---> +5
| Wart |---+ | | | |
------ | | [.33uF] | [.1uF]
| | | | |
Gnd | +------+-----+
| |
| Gnd
|
+---[27R]--+---[317]-------+---> +3.291
| | | |
| [.1uF] +-[240R]--+
| | |
| | [392R]
| | |
| +-----+
| |
| Gnd
|
+---[30R]--+---[317]-------+---> +2.781
| | |
[.1uF] +-[240R]--+
| |
| [294R]
| |
+-----+
|
Gnd

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.

Ed
.

Quantcast