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How to convert 48VDC to 15VDC@300mA?

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Junior Member level 2
Apr 21, 2002
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34063 high voltage 60v

Some modules are too expenssive,so I want to built a simple circiut to
do this.Who can give me some tips?Thanks anyway.

lm317 with current 300ma

Use LM317HV

input 48 VDC output 1.2V to 45V

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mc33063 60v

Actually you can use a standard LM317. Max allowed difference between input and output voltage is 40V.
In your case the difference is only 48V-15V=33V
So there is no need to use the special high voltage type (LM317HV), just use standard LM317.

LM317 data sheet: **broken link removed**

If R1 is 237ohm 1%, then R2 should be 2.55kohm 1% for 15V output.

voltage regulator 48vdc

It is much higher than the power capability of LM317 alone, even using a heat sink regulation maybe affected by high chip teperature.
also as u know theinput voltage may be higher than 48V, therefore it must be concerned.
for your application there is 2 solutions:
1. analog regulator such as LM317 plus a power zener at input for voltage reduction. u can use several lower power zener in series also.
2. a better and efficient way but alittle more copicated method is using integrated switchers from several manufacturers like:
or even if u want u can make your own SMPS, if so refer to:

Thanks a lot!

How about LM317 which is MetalCan package,maybe it has high power output?

as i mentioned it is possible, but the power disspation is acceptable for your application?

again: it is better to use a zener at input to reduce power dissipation of regulator, obviousely power will be dissipated in zener so the zener must capable of dissipating it without damage. power zener are inexpensive and can be used in series for higer power.

You can also use a series resistor at the input of LM317.
Min Vin is Vout+3V=18V
Max voltage over resistor (or zener): 48V-18V=30V
Max resistor value @ 300mA max: 30V/300mA=100ohm

If you use a 80 ohm resistor you will get this dissipation in resistor @ 300mA: 300mA^2*80ohm=7.2W

The rest will be dissipated in LM317: 9.9W-7.2W=2.7W

You can use more resistors in series or parallel to spread the power dissapation.

standard TELCO

48 V is a standard power source in telephone central offices. It may be overall cheaper to buy a ready made switching supply module.

Use a simple "pre-regulator" with a cheapest bipolar transistor ( e.g. 2N3055 or equiv.)+zener to reduce the voltage down to about 20VDC and then use LM7815. That's it.

Use LM3578A in step down configuration, total cost is arround U$D2,50 and is capable to drain 300mA.

**broken link removed**

Your best option would be to use an ST type L4960 switching type of power regulator.
Combines compactness and efficiency.
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Put some 12V high-current zener diode in series to the LM317 input pin

It's easier to use the fixed 15V voltage regulator LM7815 rather than the variable voltage regulator LM317

LM78XX data sheet: **broken link removed**

ether you use LM317 or LM7815 etc, the power on

iberia said:
Use LM317HV

input 48 VDC output 1.2V to 45V

**broken link removed**


the power on chip is too high ,plse use DC/DC switch,

many dc/dc chip can be used, for example: MC 34063 ,is simple and cheap

if the current is small (34063 can output 500mA),you can add a transistor to improve current

use mc33063.
this provide 1.5a at output.

48V --> 15V

You can buy a module for that kind of specs. Easier and cost-effective. Look at the link from PowerOne below.


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