How do you calculate the LM317's inputs?

[KevinL]

How do you calculate the LM317\'s inputs?

In a continuation of the desk lamp project..

I am using an LM317 to power a number of stars (anywhere from 1-3, design not finalized) from a constant current source and I'm interested in finding out how you calculate the INPUT of the LM317. The output side has been covered very well.

The circuit will have the LM317 driving 3 x Lux3 stars in series at 700mA nominal. Given an approx. Vf of 3.6V each, plus 3V for the LM317, I get 13.8V needed.

Let's say the DC input is rated for 20V, 700mA, 14 watts. The string of stars requires 700mA at 14V, so the LM317 'burns' the rest of the 6V as heat. My DC input cannot provide >700mA.

What happens if I adjust LM317 to provide 1 amp to the stars? Will it intelligently 'step down' the voltage to provide increased current or will it instead demand 1A from the power source while still flaring off 6V as waste heat? In other words, I'm aware an intelligent converter will perform the appropriate step-down and supply 14Vx1A (14W) out from 20Vx0.7A(14W) in. Will the LM317 do this or will it require 20Vx1A (20W) and still burn 6W as heat?

 

[evan9162]

Re: How do you calculate the LM317\'s inputs?

The 317 will try to pass the required current and drop the voltage difference by dissipating it as heat. The thing that will take a high voltage/low current and convert it to a lower voltage/higher current is a switching converter. The LM317 works as a linear regulator, which means in its simple setups, it simply drops voltage as heat.

If you try to draw too much current from your power supply, several things could happen: 1) It'll power off because of an overload condition. 2) It'll drop the output voltage so it's output current doesn't exceed spec. 3) Some part in the power supply will overheat and/or pop.

 

[MrAl]

Re: How do you calculate the LM317\'s inputs?

Hi there Kevin,

If i understand you correctly you want to know how much
heat to expect and how much input current...

I think you were already close to the answer.
If you have three LED's in series each 3.6v that comes up
to 3.6 * 3 which is 10.8 volts. The voltage across the
LM317 is therefore Vin-10.8 volts.
To find the current, simply take the output current,
because it's the same!
To find the power dissipation, simple multiply the
current times the voltage across the LM317, which is:

Pd=Iin*(Vin-10.8)

This means with 20 volts input and three LED's you'll
have
V=20-10.8=9.2 volts
and with 700ma output you'll have
Iin=0.700
so
Pd=0.7*9.2=6.44 watts

With the same input (20v) and 1 amp output,
you'll have
V=20-10.8=9.2 (same)
and
Iin=Iout=1.000 amps
and
Pd=V*Iin=9.2*1=9.2 watts

Both of these will require a heat sink.

If, on the other hand, you were to use a switcher
you could get away with perhaps no heat sink or a
much smaller one.

Take care,
Al

 

[KevinL]

Re: How do you calculate the LM317\'s inputs?

Thanks guys, that's exactly what I wanted to know - now I understand the difference between a switcher and linear reg (and it only took me what.. 8 years*.. /ubbthreads/images/graemlins/smile.gif)

In a nutshell the LM317 isn't 'smart' enough to convert extra voltage to amps to power the light. My DC power supply cannot handle >700mA, that's why I asked. A setup like this would not be able to drive the Luxeons at my desired 1A current.

* Dealt with the issue of switching VRMs vs linear VRMs a long time ago on computer motherboards, always knew that linears were inefficient and put out a lot of heat, but HOW inefficient and WHY never occured to me... yet another education brought to me by CPF /ubbthreads/images/graemlins/smile.gif