voltage regulation question

[lessing]

I have an efficiency question. I have said before that I do not do current regulation well. I usually work with mains supplied power and efficiency is not that important in theatrical lighting. My question is how efficient is it to run a 6-9volt supply voltage from batteries and voltage regulate down to ~3.5 volts. I am speaking of course of running luxeons and other leds. If you start with say 9 volts and regulate to the voltage of the led, is there a current overrun issue, and if so what about regulating just above the led voltage and adding a resistor?

If you start at 9 volts regulating to 4 volts should not be much of a problem until the batteries are nearly dead I think.

Am I even close or am I spoiled from running 5 amp transformers from unlimited power supplies?

 

[MrAl]

Hello,

There are basically four ways to drive LED's:
1. linear voltage regulator
2. linear current regulator
3. power converter
4. series resistor

The voltage regulator 'regulates' the voltage down to a lower level.
This isnt the recommended method, but it has been known to work
with LED's that have a rather high equivalent internal resistance.

The current regulator 'regulates' the current through the LED to a level
the led can operate at safely.

The power converter converts the input voltage to an output voltage
that the LED can work with. The power converter has the advantage
of usually providing better efficiency with the drawback of having
extra circuit complexity.

The series resistor drops voltage so the
LED can operate at it's normal voltage. It has
the disadvantage of using extra power in order
to accomplish this task.

The only time you get a fairly decent efficiency with current or voltage
regulators is when the input voltage is only slightly over the actual
LED operating voltage, or when the input voltage is only a little over
the operating voltage of a series combination of LED's and you can connect
them in series.

The formula to determine efficiency is quite simple:
eff=Pout/Pin
where
Pout=Vout*Iout
and
Pin=Vin*Iin

For example, you have 9v in and 3.5 volts out, and
since the LED draws 350ma you have 0.35 amps out and
the same in, so:
Pout=3.5*0.35=1.225
Pin=9*0.35=3.15
now to calc eff:
eff=1.225/3.15=0.389 (to three significant digits)
So the eff with a 9v input is about 38.8 percent, which isnt good.

Now at 6v input, we have:
Pout=3.5*0.35=1.225 (same as before)
Pin=6*0.35=2.1
now to calc eff:
eff=1.225/2.1=0.583 (to three significant digits)
So the eff with a 6v input is about 58.3 percent, which isnt great but works.

Lets see how a 4v input would work out...

At 4v input, we have:
Pout=3.5*0.35=1.225 (same as before)
Pin=4*0.35=1.4
now to calc eff:
eff=1.225/2.1=0.875 (to three significant digits)
So the eff with a 4v input is about 87.5 percent,
which is MORE then just acceptable.

From the above we can see that the efficiency of a simple voltage or
current regulator can be either good or bad, depending on the
relationship between input voltage and output voltage.

Now if we could put two LED's in series, their total voltage drop
would be twice that of a single led, or about 7v.
Calculating the eff using a 9v input now would show a different result:
Pout=7*0.35=2.45 (twice as before)
Pin=9*0.35=3.15
now to calc eff:
eff=2.45/3.15=0.778 (to three significant digits)
So the eff with a 9v input and two LED's is about 77.8 percent,
which is really not bad at all!

Anytime you wish to know the eff of a different combo of LED's and
input voltages, just use that simple formula.

BTW, it doesnt change anything to use a voltage
regulator at a slightly higher voltage and then
also use a series resistor as the efficiency
still suffers. You can verify this with the
simple formula for efficiency above.

Good luck with your LED circuits,
Al

 

[Tsao]

Hi Mr.Al
Is there a circuit ragulated the current through the LED? Not voltage.
Thanks

 

[INRETECH]

Yes there is

Use the LM317 in constant current mode:

LM317:
1) Input: Your Input voltage
2) Adj: Output to the LED
3) Output: Resistor to pin 2

The LM317 wants to "see" a 1.25v drop from Output to ADJ, so calculate the Resistor as:

R=1.25/Current

Your input voltage needs to be at least 1.5v higher than your LED voltage

The nice thing about this circuit, is that no matter what you put in the LED socket, it will ALWAYS be set for the same current

 

[Doug S]

Originally posted by INRETECH:
Your input voltage needs to be at least 1.5v higher than your LED voltage

<font size="2" face="Verdana, Arial">Actually with this circuit you need the input voltage to be at least 3V above the LED voltage. In addition to the 1.25V across the resistor, the LM317 needs at least 1.7V across in to pass much current. If you have ample input voltage and the means to get rid of the heat, this is a very easy/inexpensive means of current regulation.

 

[INRETECH]

You are correct, I forgot about the electricity to run the device

 

[MrAl]

Hi there Tsao,

As INRETECH and Doug were saying, the LM317 drops
quite a bit of voltage in order to provide
good regulation, but it is a very simple
IC to use. If it's your first regulator circuit
i'd highly recommend trying it out. Of course
you want to test it by measuring the current
output BEFORE connecting it to a real LS.

To do this, you could use the LS simulator that
was developed in the Zetex thread.
This is nothing more then three diodes connected
in series with a 1 ohm resistor.

If you are thinking of regulating current to a
small Nichia type LED, you could get away with
using the smaller SM packages.

Although more complex, you could also use an
LM334 and external transistor to get a nice
near-zero voltage drop current regulator.
This would be better for operation from a
battery who's voltage isnt too much different
from the LED (or series LED's) voltage drop(s).
I would have to recommend going with the Zetex
transistors as they have such a low sat voltage,
not just any run of the mill transistor.

Good luck with your LED circuits,
Al

 

[Tsao]

Mike, Doug S, Mr.Al
Thanks
You've gave me a quite different way to use this 3 terminal regulator. I could not express how I surprice.
You guys are great!!

I think it will be very useful in the car interior light mod.

 

[PsycoBob[Q2]]

I've heard of LM317 components that have a lower dropout voltage (the minimum voltage difference between Vin and Vout), but I don't remember which CPF thread they were in.

 

[INRETECH]

Yes, always test your circuit with some "cheap" parts before trying the real thing - using a bunch of diodes in series to simulate the part is a very good idea

I also bring up the voltage on a new design slowly using a variable power supply and watch the current for any expected things; a little amount of testing can save a expensive part

The LM317 is just one of those "gotta have" parts in your junk-box, just like the LM555, LM324, LM741

Remember, fuses are your friend !

Its nicer to have a cheap $0.10 blow out than a nice expensive LED

Yes, you can get more eff by using a couple more parts; the LM317 is just the most easiest to use parts ever made - and I am a digital person !

 

[georges80]

Originally posted by Tsao:
Mike, Doug S, Mr.Al
Thanks
You've gave me a quite different way to use this 3 terminal regulator. I could not express how I surprice.
You guys are great!!

I think it will be very useful in the car interior light mod.

<font size="2" face="Verdana, Arial">The 3 terminal regulator will get very hot for a 12V input when driving a 1W luxeon - about 3W being dissipated in heat.
I have a circuit description of a switcher at web page . I do sell complete units, but I have schematics there too - if you want to build your own.

cheers,
george.

 

[MrAl]

Hello there tall paul,

http://www.national.com/ds/LM/LM134.pdf

look on page 8, the circuit
labeled "Higher Output Current".

Although you will want to use a PNP Zetex transistor
with very low Vsat and able to handle enough
power too.

referring to that page:
Although i havent tried it yet, you may be able to get
away with using a 0.01uf cap between the pins
marked +Vin and V+ instead of the R1 and C1.

Rset would be equal to 0.064 divided by the desired current,
so for an LS this would be:

Rset=0.064/0.35
which equals about 0.18 ohms.

The amount of power the transistor has to dissipate
(via some kind of heatsink)
is equal to:

Pd=(Vin-Vled)*Iout

Using three batteries in series (about 4.5 volts) and
using an LS this comes out to (using 3.3v for the LED voltage
and 350ma for the LED current):

Pd=(4.5-3.3)*0.35

so

Pd=0.42 watts

so to drive an LS at full current using a supply of 4.5 volts
you would need a pnp transistor that can handle about 1 amp
and dissipate at least 0.5 watts. The Vsat should be one of
those nice low vsat zetex transistors like the 717.

If you go to the Zetex site look for pnp transistors ending with the
number 717, or look for other transistors with low sat like
40mv or so.

If instead you decide to drive it from a higher voltage like 6v,
then you have to calculate the Pd again:
Pd=(6-3.3)*0.35

which equals about 0.95 watts, so you would
need a transistor that can dissipate at least 1 watt.

Also, if your battery never runs down (car battery)
or rarely does so then you can get away with a wider
range of transistors that have higher Vsat specs.
Check the specs before you build it up to make sure
your transistor doesnt drop too much voltage (some are
as high as 2 volts at 350ma).
If your transistor drops that much the circuit wont work
very well if you are driving with a 4.5 volt power source
because the LED voltage (3.3v) plus the transistor drop
(2.0v) will be higher (5.3v) then the supply voltage (4.5v).

As a last note, you have to be aware that the
LM334 is a bit temperature sensitive, so that
means the current will vary a little bit
with heating or cooling of the flashlight.
The amount wont be that much if you keep the
light within about 30 degrees C of the temperature
at which you set the output current (via Rset).
You may wish to adjust Rset slightly higher
to compensate for this (maybe 0.20 ohms or so).

Good luck with your LED circuits,
Al

 

[Doug S]

About that LM334 circuit. I've used it and like it. I do find that the RC combo is more reliable than just a C to prevent oscillation. Try R = 470 ohms and 0.1<C<1.0uf. If your input voltage can drop so low that the voltage across the transistor falls below about 0.6V, the polarity across the C reverses so C should not be a polarized type. Also, if in your application the input voltage can drop to where the load cannot draw enough current to maintain regulation, you need to add a resistor at the LM334 V+ terminal to keep the LM334 from drawing too much current. Maximum allowable in 10mA. I endorse the use of the Zetex transistors if you need really low dropout voltages: Good stuff.

 

[tall paul]

MrAl,

Could you please elaborate on you LM334 driver circuit?

thanks,
Paul

 

[MrAl]

Hello again,

I did a quick simulation of the circuit and
found some interesting things...

1. The single C to V+ didnt work at all
(as Doug S was talking about). This
means one R and one C will be required
for compensation as shown on the schematic.

2. Setting R to 100 ohms and C to 0.01uf
seemed to work very well.

3. Making C too high (greater then 0.1uf)
causes a longer turn on surge time.
The longer the turn on surge, the more
dangerous it is for the LS.
Since 0.01uf seems to work, i'd try that
first.

4. Making C too low (less then 0.002uf) doesnt
compensate enough and the circuit still
oscillates somewhat.

5. R as low as 20 ohms still seemed to be ok,
so the value of the resistor doesnt look
critical.

6. Making Rset equal to 0.20 ohms allowed
an output current of 320ma at 25 degrees C.

7. A variation in temperature of 25 degrees C
causes an output current variation of about
8 percent. This means if you set the circuit
up at 25 degrees C to put out 350ma then
if the temperature rises to 50 degrees C then
the output current will rise to about 380ma
(1.08 times 350ma).
if the temperature decreases to 0 degrees C
then the circuit will put out about 320ma
(0.92 times 350ma).

8. There is apparently no easy way to compensate
the 'Higher Output Current' circuit for
temperature variation, but the way it is
will probably be ok if it doesnt get too
hot or the initial output current is set a
little lower (like using a 0.2 ohm resistor
for 320ma output).

9. The 'Temperature Compensated' circuit
(also shown on the data sheet) will introduce
another 0.5 to 0.7 volt drop across the
regulator and so would no longer comprise a
'near zero voltage drop' current regulator.

To check for oscillations if you dont have
an oscilloscope, you could use the ripple
detect network from the Zetex thread.
The network is also shown here:
http://hometown.aol.com/xaxo/page2.html
near the bottom of the page. Also shown is
the LS simulator which is used to test circuits
before connecting a real LS.

Good luck with your circuits,
Al

 

[tall paul]

Originally posted by MrAl:
Hello there tall paul,

http://www.national.com/ds/LM/LM134.pdf

look on page 8, the circuit
labeled "Higher Output Current".

Al

<font size="2" face="Verdana, Arial">Thanks for the info, but with that ciruit, the battery is connected to the Vin+ and Vin- points, right? Where is the LED connected?

Thanks,

Paul

 

[tall paul]

Nevermind, I'm retarded but I think I have it figured out. the positive terminal on the battery connects to the Vin+, with the Vin- connecting to the positive terminal on the LED's. then the negative LED terminal connects to the negative terminal on the battery, through a switch.

 

[Doug S]

Originally posted by tall paul:
Thanks for the info, but with that ciruit, the battery is connected to the Vin+ and Vin- points, right? Where is the LED connected?

Thanks,

Paul

<font size="2" face="Verdana, Arial">The LED goes between the collector of the transistor and Rset.

 

[MrAl]

Hello again tall paul,

Actually there are three ways to connect the LED to that circuit, i should have mentioned
this, but there are advantages and disadvantages to these.

Again referring to that circuit in the LM334 data sheet (Labeled "High Output Current"):

#1
The first way is to connect the anode (+) terminal of the LED to the point labeled
"-Vin" and the cathode (-) terminal of the LED to the battery negative (-)
terminal, with the battery positive (+) terminal connected to the point labeled
"+Vin" on the schematic. This puts that circuit in 'series' with the LED.
The LM334 is a 'floating' regulator so it doesnt necessarily have to be
connected to ground.

#2
The second way is to make a break in the circuit and insert the LED there.
The break is made between the transistor collector and the sense resistor Rset
as Doug pointed out.
Note that capacitor C1 is left connected to the collector also.
Note also that the sense input to the LM334 (labeled "R" ) is left connected
to the sense resistor Rset also.
After making that break, the LED anode (+) terminal is connected to the junction
of the collector and C1, while the LED cathode (-) terminal is connected to the
junction of the sense resistor (Rset) and the sense input (R).
This i believe is the way Doug was talking about.

#3
The third way is to connect the cathode (-) LED terminal to the point labeled
"+Vin" on the schematic and connect the anode (+) LED terminal to the
battery positive (+) terminal. The negative (-) terminal of the battery
then connects to the point labeled "-Vin" on the schematic.

The disadvantages to method #1 is that if you are running from a low power
supply (such as three batteries in series) then the circuit wont be able to
regulate to as low a voltage as method #2 will allow.
The advantages are:
* Slightly higher efficiency (because all the current from the
battery flows through the load)
* Better stability with less input surge because the LED reduces circuit gain
* The LED can be connected directly to ground
When the LED has to be connected to ground, this of course is the preferred
circuit.

The only disadvantage to method #2 is:
* somewhat higher instability with larger input surge
The advantage is that #2 can regulate down to a much lower voltage
then with #1. By 'much lower voltage' i mean about 0.5 volts difference.
Now while 0.5 volts doesnt seem like much, it can be when operating from
three batteries. This in itself means #2 is the overall preferred circuit
except when operating from voltages that are way higher then the LED
voltage drop (like 6v, 12v, etc) and that it wont matter
much in run time if the battery voltage does drop lower, or
when the LED has to connect directly to the system ground.

The disadvantages to method #3 are the same as for #1.
The advantages to method #3 are the same as for #1, except of course that
the LED can connect directly to the battery positive (+) terminal.

The above leads us to configure the LED depending on
the other requirements, so you have to ask these questions:

1. Does your LED have to connect to ground?

If you answer yes to this question, then you HAVE to use method #1.
If you answer 'no' then you proceed to the next question.

2. Does your LED have to connect directly to the battery positive terminal?

If you answer 'yes' to this then you HAVE to use method #3.
If you answer 'no' then proceed to the next question.

3. Is your fresh total battery voltage above 5 volts?

If you answer yes to this question, then you can use either
#1, #2, or #3.
If you answer 'no' then you should use only circuit #2.
As mentioned, #2 is the preferred circuit when operating from
three batteries in series (total voltage about 4.5v).
If you are using NiCd's the situation is even more critical and
you can only use circuit #2.

As a final word, if you decide to use circuit #2 (probably the
best choice whenever possible) then be careful not to disconnect
the capacitor C1 from the transistor collector. This connection
should always be exactly as before the LED was introduced.
Ditto for the LM334 sense terminal (labeled "R" on the schem) and
the Rset resistor. This LM334 terminal should always be connected to
the Rset resistor.

Good luck with your LED circuits,
Al

 

[**DONOTDELETE**]

ok, please excuse my ignorance, but i've built the high current lm334 circuit as described (at least as far as i can tell), but i'm having problems. i used the lm334z and the fcx717 zetex transistor, c1=.01uF, R1=100ohms, and Rset=.2ohms. i'm using 3 alklines to get me to 4.5 volts. i'm using a HD 1WLS that i have inserted in the circuit between the collector and Rset (the junction of the LM334 R pin) per the instructions of MrAl and Doug. the current output keeps coming out real low, like around 20ma (very dim). i've played with the R1 values (20 to 1000 ohms) but i still didn't get the target output. the only time i got the required voltage across Rset (.064V) was when i shorted across C1. if anyone has suggestions or at least point out what i did wrong i would greatly appreciate it.

thanks,
paul