Re: 85-265VAC and 24VDC combined power supply

On Aug 10, 2:25 am, Christian Walter <wo...@xxxxxx> wrote:
Hello,

I am currently working on a device where I am evaluating options for the
design of the power supply. On the secondary side we need two power
supply output with 5V/4A and 13,8V/2A (uncommon voltage). The input
voltages are the standard 85V-265VAC range and 24VDC.

We want every unit to be capable of being used in both environments.
Therefore the power supply should allow both inputs. Because of 24VDC
input I will need two DC/DC converters at the output side. One from
24VDC/5VDC and one from 24VDC/13,8VDC. These converters are easy to
design and can have a high efficiency (>90-95%).

What I am unsure about is how to handle the transformation from the
primary 85V-265V to the intermediate internal voltage of 24V. A
regulated switched mode power supply is certainly a overkill because I
don't need the 24V to be regulated. So I thought I could use a standard
50/60Hz input stage with a simple full bridge rectifier and a good
transformer. Size is not the problem but power efficiency and costs are.
Or are there any better options? I searched for quite some time in
goggle for possible AC input stages but did not come up with a good
solution.

265/85 = 3.12

It is not very hard to make a DC-DC converter work with in input that
varies over that much range. If your quantities are low, you won't
want a custom transformer in the mains to 24V section but lets assume
you are free to pick the tranformer you want.

Many parts max out at 35V so lets assume that 270V gives you just
about 30V DC
This makes the secondary have

(30V+0.7) / sqrt(2) = 21.7VAC

At 85VAC you would have:

(21.7 * 85/270) * sqrt(2) - 0.7 = 9.66VDC

I have left out the losses in the resistance of the windings at this
point.

Now you need to make a first cut on the design of the DC-DC
converters. I assume they don't have to be isolated so a simple
bucker can be used for the 5V. The 13.8 needs something like a SEPIC
converter.

Once you have the first cut on the DC-DC design you need a good
estimate of its efficiency at the 9.66V end of the span. You then
will know how much current the circuit will draw at that end and can
figure out a corrected value for the 9.66V and see if you are in
trouble.


.

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