Ask the EE: power supplies and high brightness LEDs

[Bruce Frederiksen]

Mike had a question that I didn't know the answer to, so forwarded it to my brother along with some questions of my own.  I thought that the answers might be interesting to everybody, as they concern power supplies in general, so am posting it here!

Original Question:

Bruce,

I thought I would take you up on your offer regarding basic electronics and a circuit I am playing with. Currently, I have a high brightness LED with an IF of 1.6 A and a step-down switching regulator that is rated at 1.5 A (see detailed specs below). My question is do I run any problems with the switching regulator if I run it at max? At some point I will be getting a LED driver for full output current but right now I am doing some testing with dimming of the LED by attaching a PWM signal; from the Arduino to the switching regulator. Here are the detailed specs:

LED

10 W

VF: 8 V

IF: 1.6 A

Switching Regulator, Step-Down

Output Current: 1.5 A 

10 W

Input Voltage: 4 - 24 VDC

Output Voltage: 2.5 - 18 VDC

To power the switching regulator I have a wall wart around 12 VDC and 1000mA

As I said, this is just a temporary set up while I work out the circumstances of when and how I want to dim the LED. Let me know what you think.

Thanks,

Mike

Answer from my brother (Fred):

Hi Michael,

Wow - I've done some work with LED's b4 & never came across any rated @ 10W!  (showing my age??)   Does it come with it's own heat sink?

 

A power supply 'should' operate at it's maximum ratings...  surprisingly, many won't:  it's customary to derate power supplies by 25% - 50% as a 'safety margin' (against crappy designs; or not using it in the Arctic).  Be advised that most power supplies will provide a 'surge' current (typically 125% - 150% of maximum continuous rating), so steady state running of the power supply in current limit is dangerous (overheating).

To answer your question:  I would expect the power supply (& wall wart) to accomplish their maximum ratings, but would also expect each to get (very) hot.  The heat would be the thing to cause a failure, so I would recommend fan(s) to cool the power supplies & LED. 

Keep an eye on temperatures all around & abort if maximum temperature is exceeded!

(If it's too hot to hold your finger on it, that's about 70 deg. C)

Good Luck & Have Fun!

-  Fred

My first question:

I don't know anything about switching regulators.  Can they produce more current than they consume?  (since "wall wart" goes to 1A).

Answer:

YES: 

Power = P

by law of conservation of energy:

Pout = Pin * (efficiency ~= 1 {.8 to .9, but always less than 1 so Pout < Pin})

(or Pout = Pin - Plosses)

 

Vout*Iout ~= Vin*Iin

so, if Vin >> Vout then Iin < Iout

My second set of questions (in white) with his answers (in green):

So why does is high-brightness LED rated at VF 8, IF 1.6, but 10W (8*1.6=12.8W)?

? never seen anything like 10W LED - I'd bet u can't get near 10W!  (or not long) it would melt!

probably the peak current is 1.6A; the real limitation would be device temperature!  (maybe 2W; ice bath)

And does an LED turn on exponentially like regular diodes? yes Meaning that you can't just stick 8V from an unlimited power supply on it? not long Or does the light emitting part make them more linear, or ??? yes - crappy diode

What happens when you try to draw more than the rated current from a switching regulator?  overheat & or burn out Are most regulators current protected?  should be - most lab ps's would be expected; but many special purpose (cheap) ps's are not If the regulator is rated at 1.5A, can he just stick the regulator straight on the LED and let the voltage fall where it may? in theory, yes.  but many ps's provide 'surge' overcurrent in excess of continuous rating & / or many ps's will overheat in current limit Or would that have to be a special current limiting option on the regulator? ultimately u gotta test it 2 c...

The little 3 pin regulators aren't switching regulators, right? yes, although there exist mini switching regulator IC's... So they have Iin == Iout and just dissipate the difference in V (*I) as heat? yes Which is going to limit them to low power applications? yes: linear = low power & clean; switcher = hi power & noisy

And then if I want more power I either need something fancier, like a switching regulator, or a big heat sink. yes & fans I remember Dad having some power supplies with big heat sinks and those big diamond shaped power transistors TO-3 (now use TO-247) back in the day...  But I don't think those went much more than an amp or two.  But variable voltage, so just a transformer down to Vmax, and then dissipate the rest down to Vout?  Isn't that essentially what the 3 pin regulators are doing (you provide the transformer, rectifier diodes and beer can cap)? yes, but can't get so much heat out of 3-pin (TO-220) - 3-pin's come with variable output...
It's shocking to me how often heat management is neglected in designs (exploding batteries; all flavors of failures...)

My own 2 cents (from the peanut gallery) to Mike:

Sounds like you need current limit resistors.  If you had a 12V supply, then a 4V drop on the resistor (to 8V for the LED) with E = I*R becomes 4 = 1.6*R, so R would be 2.5 ohms; less if using a lower voltage supply.   But this resistor would be dissipating 6.4 watts (4 * 1.6, but also V^2/R == 4^2/2.5, or I^2*R == 1.6^2*2.5)!  So you'd probably want a 10W resistor (buy these at Mouser) which will probably be as big as the end of your little finger to dissipate the heat.

And be careful about the supply voltage, because if the LED actually runs at less than 8V, you'll get more of a voltage drop on the resistor and more current through the LED.  If the LED actually runs at 7V with a 12V supply, that's 5V on your 2.5 ohm resistor == 2A through the LED (and 10W in the resistor).

Now if you went with a 10V supply, then you'd figure 2V drop at 1.6A == 1.25 ohms (3.2W in the resistor).  But in this configuration if the LED actually runs at 7V, there would be a 3V drop on 1.25 ohms == 2.4A through the LED (and 7.2W in the resistor).  So having a smaller difference between the supply and the LED VF makes the current more sensitive to VF.

I have no idea when they say 8V for VF how accurate that number is, so don't know whether this is a real problem or not.

Bottom line, you might want to get an assortment of small resistor sizes (at high watts) and some extra LEDs (if they're not too expensive, in case you burn one or two out).  Hopefully you have a simple DC voltmeter to check this out with?  If not, I have one and can bring it to a meeting.

Also, I'm not sure how much the switching regulator is going to do for you.  If the wall wart gives 1A at 12V, you're not going to get any more than that at 12V.  You'd get 1.2A at 10V and 1.5A at 8V if the switching regulator was 100% efficient (which it isn't), so by the time you're getting 1.5A, you don't have any extra volts left to drop across a resistor.  I don't know what controls the regulator has on it though...

-Bruce

[Fred^2]

So, it seems that regulating current in the LED by using voltage
controlled power supply is awkward!

It'd be nicer getting a DC/DC converter that had a knob to adjust
current output & directly regulate the brightness by regulating the
LED current; but, the ucontroller doesn't have fingers to be able to
automatically adjust this knob...

So, cleanest solution (no hot series resistors & self compensates for
temperature & input voltage fluctuations) would be to build your own
current controlled regulator using the PWM output.

See "LED.jpg" in files area.
where:
D is fast recovery
R is ~ 0.1 ohm to sense current
(may amplify Is to fit A/D converter)
Q is rated > 24V
L is sized for ~ 10% current ripple

Steady State:
look @ L: Von*Ton = Voff*Toff
('on' & 'off' refer to Qon & Qoff)
(12V - Vled)*Ton = (Vled + Vd)*Toff

Set base counter PWM to Ton
Set Iref to desired brightness
Measure Ipk just b4 PWM switches off
Ierr = Iref - Ipk
new counter PWM = base + Ierr (*k)

You now have instantaneous control over current (hence brightness) in
LED.
I'd recommend a temperature probe directly on the LED tied into
another A/D input so you don't burn it out...

Disclaimer: I've never actually did it this way, but it would be neat
if it worked!
Potential problems:
1) A/D conversion of 'moving target'
2) stability (this would be easiest possible, but may not be brain
dead - start with small k {shift})
3) CPU must compute next PWM counter during off time (speed) - may
have to slow switch rate?
(higher switch rate = smaller L)

On Jul 2, 1:40 pm, Bruce Frederiksen <dangy...@gmail.com> wrote:

> I thought I would take you up on your offer regarding basic electronics and a circuit I am playing with. Currently, I have a high brightness LED with an IF of 1.6 A and a step-down switching regulator that is rated at 1.5 A (see detailed specs below). My question is do I run any problems with the switching regulator if I run it at max? At some point I will be getting a LED driver for full output current but right now I am doing some testing with dimming of the LED by attaching a PWM signal; from the Arduino to the switching regulator. Here are the detailed specs:
>
>
>
> LED
>
> 10 W
>
> VF: 8 V
>
> IF: 1.6 A
>
> Switching Regulator, Step-Down
>
> Output Current: 1.5 A 
>
> 10 W
>
> Input Voltage: 4 - 24 VDC
>
> Output Voltage: 2.5 - 18 VDC
>
>
>
> To power the switching regulator I have a wall wart around 12 VDC and 1000mA
>
>
>
> As I said, this is just a temporary set up while I work out the circumstances of when and how I want to dim the LED. Let me know what you think.
>
>
>
> Thanks,
>
>

> And does an LED turn on exponentially like regular diodes? yesMeaning that you can't just stick 8V from an unlimited power supply on it? not longOr does the light emitting part make them more linear, or ???yes - crappy diode
> What happens when you try to draw more than the rated current from a switching regulator? overheat & or burn outAre most regulators current protected? should be - most lab ps's would be expected; but many special purpose (cheap) ps's are not If the regulator is rated at 1.5A, can he just stick the regulator straight on the LED and let the voltage fall where it may? in theory, yes.  but many ps's provide 'surge' overcurrent in excess of continuous rating & / or many ps's will overheat in current limit Or would that have to be a special current limiting option on the regulator?ultimately u gotta test it 2 c...

[Bruce Frederiksen]

Fred^2 wrote:

So, it seems that regulating current in the LED by using voltage
controlled power supply is awkward!

It'd be nicer getting a DC/DC converter that had a knob to adjust
current output & directly regulate the brightness by regulating the
LED current;  but, the ucontroller doesn't have fingers to be able to
automatically adjust this knob...

So, cleanest solution (no hot series resistors & self compensates for
temperature & input voltage fluctuations) would be to build your own
current controlled regulator using the PWM output.

See "LED.jpg" in files area.
where:
     D is fast recovery
     R is ~ 0.1 ohm to sense current
        (may amplify Is to fit A/D converter)
     Q is rated > 24V
  

Q > 24V is just standard rule of thumb 2x fudge factor?  (I only see 12V on Q)??

     L is sized for ~ 10% current ripple

Steady State:
look @ L:  Von*Ton = Voff*Toff
('on' & 'off' refer to Qon & Qoff)
   (12V - Vled)*Ton = (Vled + Vd)*Toff
  

Ton/Toff ~independent of Iref, except for slight(?) change in Vled vs I.

Set base counter PWM to Ton
Set Iref to desired brightness
Measure Ipk just b4 PWM switches off
Ierr = Iref - Ipk
new counter PWM = base + Ierr (*k)
  

seems like a couple of options here.

1.  I think you're thinking base is variable.  I.e., last line is: new counter PWM = base = base + Ierr (*k).  Then have to worry(?) about stability.
2.  But you could hold base constant at Ton (since Ton is ~independent of I).  Then k is L/(12V-Vled) to correct I in one cycle.  No stability problem, right?  It just self corrects periodically for error in calculated Ton.

You now have instantaneous control over current (hence brightness) in
LED.
I'd recommend a temperature probe directly on the LED tied into
another A/D input so you don't burn it out...

Disclaimer:  I've never actually did it this way, but it would be neat
if it worked!
Potential problems:
1)  A/D conversion of 'moving target'
  

Atmega ADC has automatic sample and hold on input, so this shouldn't be a problem.

2)  stability (this would be easiest possible, but may not be brain
dead - start with small k {shift})
3)  CPU must compute next PWM counter during off time (speed) - may
have to slow switch rate?
(higher switch rate = smaller L)
  

Couldn't you use fast PWM rate with slow CPU by sampling every Nth cycle, (taking N cycles to compute new PWM)?  (The ADC time (100uSec) probably dominates compute time).

Final questions:

1.  The initial spike on I graph is diode recovery?
2.  Imax on led is just Wart's Imax (no current amplification)?

[desNotes]

I can't seem to find the proper symbols for the components in my LED
circuit so didn't want to fake it and turn the focus on the symbols
rather than the circuit itself. Maybe what I can do is draw it out and
scan and provide it later.

What I have found is if I start with the LM350 voltage regulator (3A)
and then I use a NPN Transistor GP 800 ma 45V that connects to the LED
and provides the control of the current using the PWM of the Arduino.
I am using a 10 watt resistor connected between the output and the adj
of the LM350 and a 1K resistor as input to the PWM on the Arduino.

I haven't had a chance to hook it up yet and have a LED with a much
smaller wattage spac to start with before I go with the 10 watt one. I
have plenty of heat sink to use also. When I test it out I will
provide pictures.

Mike

[desNotes]

I've added a scan of the two circuits I've been working with
(LED_Circuit01.jpg). The only difference is I am now using the LM350
instead of the LM317 due to the higher current capability.

Mike

[Fred Frederiksen]

GREAT!
Solid concept that is guaranteed to work without problems.

NOTES:
1) T1 rated @ 0.8 amp where LED rated @ 1.6 amp => set LM350 to < 0.8 amp
2) T1 thermal resistance ~= 200degC / W so 0.8 amp * 1 Vce = 0.8W => 160deg
C temp rise!
Do you have a heatsink for T1?
3) T1 absolute maximum rating is 625mW @ Tc = 66 deg C (finger test)

[desNotes]

Yes, I will be providing some type of heat sink to T1, LM350 and the
power resistor.

Mike

On Jul 12, 2:19 pm, Fred Frederiksen <controlengineerin...@gmail.com>
wrote:

[Bruce Frederiksen]

I'm seeing that the LM350 has a minimum load current of 3.5-10mA (didn't check the LM317).  I guess with no load (transistor off), Vout just rises and who cares?

[Fred Frederiksen]

----- Original Message -----

From: Bruce Frederiksen

To: tampa-bay-microcont...@googlegroups.com

Sent: Sunday, July 12, 2009 9:03 AM

Subject: Re: Ask the EE: power supplies and high brightness LEDs

Fred^2 wrote:

So, it seems that regulating current in the LED by using voltage
controlled power supply is awkward!

It'd be nicer getting a DC/DC converter that had a knob to adjust
current output & directly regulate the brightness by regulating the
LED current;  but, the ucontroller doesn't have fingers to be able to
automatically adjust this knob...

So, cleanest solution (no hot series resistors & self compensates for
temperature & input voltage fluctuations) would be to build your own
current controlled regulator using the PWM output.

See "LED.jpg" in files area.
where:
     D is fast recovery
     R is ~ 0.1 ohm to sense current
        (may amplify Is to fit A/D converter)
     Q is rated > 24V
  

Q > 24V is just standard rule of thumb 2x fudge factor?  (I only see 12V on Q)??

yes: 12V + Vd + 12V ringing; 

over rating Q (V, I, & P) will make circuit robust (note that there are different classes of transistors - you want a 'switching' transistor; not an 'amplifier', 'signal', 'RF', etc.  It should be easy to find a cheap switching FET 30-50V; <0.10 ohm; TO-220 {I've got handfulls of D560's that should work OK up to ~5A @ 5Vgs; but will current limit @ ~10A without more gate voltage?}).

 

If anything 'bad' happens, Q will normally be the first to blow, but if Q is over rated combined with a power limited source (wart), the circuit should be (reasonably) bulletproof.

     L is sized for ~ 10% current ripple

Steady State:
look @ L:  Von*Ton = Voff*Toff
('on' & 'off' refer to Qon & Qoff)
   (12V - Vled)*Ton = (Vled + Vd)*Toff
  

Ton/Toff ~independent of Iref, except for slight(?) change in Vled vs I.

& Vin & Vr & Vq & Vd & temperature & gremlins & ... = reason to use feedback

Ideally, duty cycle is independant of current.

Set base counter PWM to Ton
Set Iref to desired brightness
Measure Ipk just b4 PWM switches off
Ierr = Iref - Ipk
new counter PWM = base + Ierr (*k)
  

seems like a couple of options here.

1.  I think you're thinking base is variable.  I.e., last line is: new counter PWM = base = base + Ierr (*k).  Then have to worry(?) about stability. Always worry about stability... this would integrate Ierr

2.  But you could hold base constant at Ton (since Ton is ~independent of I).  Then k is L/(12V-Vled) to correct I in one cycle.  No stability problem, right?  It just self corrects periodically for error in calculated Ton.

stability problem if you overcorrect in one cycle, so correction should be to 50%-80% of error in one cycle.

 

error can be made vanishingly small by integrating Ierr ie:

     new counter timer = base + 0.5 * Ierr + 0.25 * sum(Ierr)

        Note:  * 0.5 = shift right 1 bit;  * 0.25 = shift right 2 bits

       There may be issues with integrator wind-up;  more CPU time

       is it necessary? how well can the eye distinguish brightness?

I bet that clever selection of R will set k to 1. (based on slope of Is = dI/dt)

You now have instantaneous control over current (hence brightness) in
LED.
I'd recommend a temperature probe directly on the LED tied into
another A/D input so you don't burn it out...

Disclaimer:  I've never actually did it this way, but it would be neat
if it worked!
Potential problems:
1)  A/D conversion of 'moving target'
  

Atmega ADC has automatic sample and hold on input, so this shouldn't be a problem.

so it's doable !!

2)  stability (this would be easiest possible, but may not be brain
dead - start with small k {shift})
3)  CPU must compute next PWM counter during off time (speed) - may
have to slow switch rate?
(higher switch rate = smaller L)
  

Couldn't you use fast PWM rate with slow CPU by sampling every Nth cycle, (taking N cycles to compute new PWM)?  (The ADC time (100uSec) probably dominates compute time).

VERY unorthodox (probably for a reason).  Generally one wants iron-clad control over current pulses.  This design is unusual in that it just observers current for correction on next pulse - assuming that conditions remain constant.  It could be argued if conditions are constant for 1 cycle, then they are likely to be for 2 cycles, BUT more assumptions allow more errors!  Cycle skipping MIGHT work - try it! 

I would have much less problem moving the sample point from the end of Is to the middle of Is (as measured from the end - hence assuming constant current slope), but in any case the switching period should be > 100uSec (painfully slow => big L).

Final questions:

1.  The initial spike on I graph is diode recovery?

yes + Q turn-on current + charging capacitance in L

2.  Imax on led is just Wart's Imax (no current amplification)?

Imax (traditionally) occurs when Q is continuously on: so Iled = Iwartmax.

(the power limit foldback in the wart will interfere with this)

 

NOTE:

1)  at less than Imax, Iled > Iwart ,  so wart will run cooler

     Why?  Assume the current (I) in L is constant (we choose an infinite inductor)

    during Ton, Iwart = Iled = I

    during Toff, Iwart = 0; while Iled = I (through D)

    Total Iled   = (Ton*I + Toff*I)/period  = I

    Total Iwart = (Ton*I + Toff*0)/period  = I*Ton/(Ton + Toff) = I*DutyCycle

2)  you don't need Iwartmax to run the LED at it's maximum rating!

    Homework:  you can verify this by deriving duty cycle from Steady State [above] & apply to Iwart [in 1)].

3)  Pwart - Plosses = Pled     (P = power)  {alternate verification}

    there's about 1W of losses in R + D + Q + L, so 12W - 1W = 11W available to LED (92%)

4)  in linear solution, there's 12-8 = 4W losses, so 12W - 4W = 8W in LED (67%)

5)  losses in switching solution are independant of Vled as in different (test) LED (new Vled just changes duty cycle)

[Bruce Frederiksen]

Fred Frederiksen wrote (red):

 

----- Original Message -----

From: Bruce Frederiksen

To: tampa-bay-microcont...@googlegroups.com

Sent: Sunday, July 12, 2009 9:03 AM

Subject: Re: Ask the EE: power supplies and high brightness LEDs

2.  Imax on led is just Wart's Imax (no current amplification)?

Imax (traditionally) occurs when Q is continuously on: so Iled = Iwartmax.

(the power limit foldback in the wart will interfere with this)

 

NOTE:

1)  at less than Imax, Iled > Iwart ,  so wart will run cooler

     Why?  Assume the current (I) in L is constant (we choose an infinite inductor)

    during Ton, Iwart = Iled = I

    during Toff, Iwart = 0; while Iled = I (through D)

    Total Iled   = (Ton*I + Toff*I)/period  = I

    Total Iwart = (Ton*I + Toff*0)/period  = I*Ton/(Ton + Toff) = I*DutyCycle

2)  you don't need Iwartmax to run the LED at it's maximum rating!

    Homework:  you can verify this by deriving duty cycle from Steady State [above] & apply to Iwart [in 1)].

3)  Pwart - Plosses = Pled     (P = power)  {alternate verification}

    there's about 1W of losses in R + D + Q + L, so 12W - 1W = 11W available to LED (92%)

4)  in linear solution, there's 12-8 = 4W losses, so 12W - 4W = 8W in LED (67%)

5)  losses in switching solution are independant of Vled as in different (test) LED (new Vled just changes duty cycle)

I think my confusion here is that there are 2 Iwartmax values: continuous and peak.  Your item 1), "during Ton" line says Iwart = Iled, thus Iled can't be > Iwartmax.  But this is max peak, while Iwartmax in item 1) "Total Iwart" and item 2) is max continuous.  And this is how the "cheating" is done.  Meaning that Iledmax is limited to Iwartmaxpeak in your solution (so long as item 1) "Total Iwart" does not exceed Iwartmaxcontinuous).  But Iledmax is limited to Iwartmaxcontinuous in linear solution.

Am I on the right track?

[Fred Frederiksen]

The rating of the wart is 1 amp (rms).  Let's all agree NOT to try to pull more than the rating out of the wart!

When you do go past the rating, as pointed out earlier, the wart voltage will start to collapse (foldback) & failure is imminent ...

Your item 1), "during Ton" line says Iwart = Iled, thus Iled can't be > Iwartmax. 

Requiring the wart to provide the 'peaks' would be too severe.  Even if the wart was capable of the 'peak', it would not respond fast enough to match the switching speed of Q.

That's what C is for - it averages the current out of the wart & provides 'peaks' (above 1 amp) to LED.

But this is max peak, while Iwartmax in item 1) "Total Iwart" and item 2) is max continuous. 

Wart provides small ripple current into C.  (infinite C -> ripple goes to 0) (this is (part of) the noise in switchers)

item 1) applies for average wart current <= 1 amp. (heating of the wart would be in spec)

And this is how the "cheating" is done. 

NO CHEATING is done!  this is all real.

Meaning that Iledmax is limited to Iwartmaxpeak in your solution (so long as item 1) "Total Iwart" does not exceed Iwartmaxcontinuous). 

The 'peaks' come from C.  (you could add an inductor in series with wart to emphisize the smoothing of C)

But Iledmax is limited to Iwartmaxcontinuous in linear solution.

yes! which would be 1 amp... (the magic of switching regulators)

Am I on the right track?

hope so...