Question - PWM effieciency??

[xbrite]

Dear Gents,

I have a couple of questions re PWM effieciency. The test conbfiguration is as below;

1) B+ connected one end of 6V 10 Watt Halogen lamp, the other end of lamp connected to a MOS FET Drain, FET Source connected to B-, Gate driven by a PWM source.

At 6V input, I applied 100% On duty. The current through the Lamp is 1.67A. I increased the B+ to 9V and for sure the current increased very much. I adjusted the PWM duty to make the current thru the lamp to 1.67A( I don't know if it is RMS value or not as I used a simple Multimeter).

a) In this configuration, is the power consumption at lamp 10W(6V X 1.67A) or 15W (9V X 1.67A)??
b) Should I reduce the the RMS current to 1.1A to make the Power consumption at lamp to 10Watt (9V X 1.1A)??
c) Is the efficiency 100% or what? Is it better than using 90% effciency DC-DC step down conveter??
d) What about with 3V lamp and 9V battery? Can the effieciency 100%??

2) The same but 5 Watt Luxeon LED is connected. Assuming Vf at 750mA is just 6V and RDS On of FET is Zero.. Adjust the PWM duty to make the current thru LED at 700mA at 9V input..

a) In this configuration, is the power consumption at LED 6V X 0.75A or 9V X 0.75A??
b) Should I reduce the the RMS current to 0.55A to make the Power consumption at LED to 5 watt(9V X 0.55A)??
c) Is the efficiency 100% or what? Is it better than using 90% effciency DC-DC step down conveter??
d) What about with 1 Watt(Vf=3.4V)LED and 9V battery? Can the effieciency 100%??

Thanks..

 

[markus_i]

I'll try to answer some of that:

0) unless you actually measure RMS (or pulse width and pause ratio), forget your measurements. And if your meter says 'RMS', be sure that your wave form is within the defined frequency and crest factor range, otherwise use a 'scope.

1)
a+b) Power consumption at the lamp is tricky. Especially if it's an incandescent and your PWM frequncy is higher than the lamp's internal time constant. Essentially, you can/should measure/calculate the power at several (a lot..) points during one PWM cycle and then integrate over that cycle.
For a rough estimate, source voltage times RMS current should work.
c+d)Efficiency will _never_ be 100%. You'll lose:
- power to generate the PWM signal
- power to drive the Transistor (yes, even if it's a FET)
- drop-off voltage in the Transistor (both when saturated and while switching)

The same goes for 2), with an additional nasty twist in the tail: while incandescents are pretty slow (i.e. take a few dozen msecs to light up/go from cold to hot) and are slightly tolerant to short peaks, LEDs are fast. So simply PWMing al LED without additional filtering will drive the full peak current through the LED. Driving an incandescent with the rated power at a pulse ratio of 10% might work - try that with a LED and you'll have it destroyed.

 

[Jonathan]

markus_i,

I believe that you are spot on with one small nit: the rough estimate of power supplied to the lamp will be the _RMS voltage_ times the _RMS current_.

Another way to figure things is to find the _peak_ current, multiply by the _supply_ voltage, and then multiply by the duty cycle.

As you suggest, the best way is to measure the voltage and current at many points on the PWM cycle, multiply together to calculate the power at each point in the PWM cycle, and then average to get the average power delivered to the load.

-Jon

 

[koala]

I don't understand how do you PWM at 100% ?

 

[xbrite]

Dear Gents,

Thanks for the replies. Let me further elaborate what I am doing;

I put a MCU varying the PWM duty ratio to the FET switch which is connected to the LED in series and it seemed working great. I could adjust the brightness from zero to max brightness, could maintain the same maximum current thru the LED with varying Battery voltage. No current limiting resistor used except 0.1 ohm current sense resistor put in between FET source and ground.

Even I turn on the FET constantly, the max current thru 5W LED was around 1.5A and I guess it is because of RDS On, 0.1 ohm current sense resistor, Battery internal resistance, wiring resistance.

I think I have to use the RMS current and RMS voltage applied to the lamp(or LED) and multiply them to get the power consumption. I should not multiply Battery voltage and RMS current from the battery. Is this correct??

My bottom line question is;

1) Is it more efficient driving 5W Luxeon from 2 cell Lithium Ion battery with PWM than using 90% DC-DC step down converter?? Assuming the power consumption for gate driving and RDS On loss is minimal. I actually used a FET with 0.045 ohm RDS On.

2) The same question in case of incandescent lamp, which is 6V 10 Watt Halogen.

3) One more question. Current vs light output from LED. Is the light output linear to the current thru the LED? If I apply 1.5A, will the light ouput twice as bright as 0.75A?? If the light output is saturated at 1A, applying 1.5A is not a good idea..

Re the question of Koala, 100% PWM means constatnt on..

Cheers..

 

[Jonathan]

[ QUOTE ]
I don't understand how do you PWM at 100%

[/ QUOTE ]

Think of 100% and 0% as the limiting cases of your modulation. At 0% or 100%, your output is no longer pulsing, but presumably your modulation system is _capable_ of pulsing the output and is still running some sort of internal time base. So 0% and 100% are simply what you get when your modulated 'off' period (or 'on' period) is the same as your total modulation period.

-Jon

 

[Jonathan]

If your measure of efficiency is 'electrical power delivered to the LED/electrical power supplied by the battery', then IMHO the PWM system will be more efficient. The power losses in the MCU and the FET are almost negligible as compared to your load.

But if you measure efficiency in terms of 'light output versus electrical power supplied by the battery', than the question is much more complex.

As you note, if the LED at 1.5A is not twice as bright as the LED at 0.75A, then clearly operating a LED at 1.5A for 50% of the time is less efficient than operating that same LED at 0.75A continuously. And if you look at the datasheets for the LEDs, you will clearly see that the 'lumen per amp' falls off as the current goes up. The problem is that a large portion of this fall is caused by the die getting hotter, and if you are modulating to the same power level, the die temperature should be similar. I am pretty certain that a Luxeon 5W at 1.5A is less than twice the brightness of a Luxeon 5W at 0.75A, but I don't know by how much.

As an additional issue, the Vf at 1.5A will clearly be higher than the Vf at 0.75A. So operating at 1.5A for 50% of the time will clearly require more power than operating at 0.75A continuously.

So I state with certainty that operating at 1.5A for 50% of the time will require higher power and produce less light than operating at 0.75A continuously, but I do not know if the magnitude of the efficiency decrease would make it worth using the less electrically efficient DC-DC converter.

-Jon

 

[koala]

Yeah.. how can PWM be constant on? /ubbthreads/images/graemlins/naughty.gif abit confusing here..

 

[LED-FX]

[ QUOTE ]
Yeah.. how can PWM be constant on? abit confusing here..

[/ QUOTE ]

Think about it for a second...
Zero on PWM will be a flat line on a `scope at 0V with NO pulses.

100% will be a flat line at supply V on a scope, its all pulses with no gap.

Adam

 

[Doug Owen]

[ QUOTE ]
Jonathan said:

So I state with certainty that operating at 1.5A for 50% of the time will require higher power and produce less light than operating at 0.75A continuously, but I do not know if the magnitude of the efficiency decrease would make it worth using the less electrically efficient DC-DC converter.



[/ QUOTE ]

This is true enough. However, putting a suitable inductor in the circuit changes things a lot. While the *average* voltage across it is zero, relative to ground there is no change in voltage on the output end (that is 'all the AC is on the drive side') and the current through it (and therefore the LED) is *constant*. Granted there are fairly minor losses in the inductor, overall it's a winning addition.

Doug Owen

 

[MrAl]

Hello there,


With your setup:
The power P of the bulb is 10 watts
The operating voltage Vo is 6 volts.

When using a 9v power supply:
The peak pulse voltage Vpeak is 9 volts

First, calculate the hot resistance R of the bulb:
R=(Vo^2)/P
R=6*6/10
R=36/10
R=3.6 ohms


Next, calculate the duty cycle DC:
DC=P*R/Vpeak^2
DC=10*3.6/(9*9)
DC=36/81
DC=0.444444

Next, calculate the average voltage Vavg:
Vavg=DC*Vpeak
Vavg=0.444444*9
Vavg=4
(This is the voltage you will measure across the bulb
with a regular meter set to measure dc voltage when
the duty cycle is adjusted to supply full power, but
only when using a 9v power supply.)


Here's a possible way to go about it:

1. Set the duty cycle on low (less then 10%)
2. Connect the bulb
3. Measure the average voltage across the bulb
4. Turn up the duty cycle until you measure 4v
across the bulb using a 9v power supply.
If you are using a 12v power supply, turn
up the duty cycle until you measure 3v
average voltage across the bulb (see below).


When using a 12v power supply:
The peak pulse voltage Vpeak is 12 volts

First, calculate the hot resistance R of the bulb:
R=3.6 ohms (same as above)

Next, calculate the duty cycle DC:
DC=P*R/Vpeak^2
DC=10*3.6/(12*12)
DC=36/144
DC=0.25

Next, calculate the average voltage Vavg:
Vavg=DC*Vpeak
Vavg=0.25*12
Vavg=3
(This is the voltage you will measure across the bulb
with a regular meter set to measure dc voltage when
the duty cycle is adjusted to supply full power.)

Note that when using a 12v power supply the average
voltage measured when full power is supplied will
be lower then when using a 9v power supply.


If you dont know if your meter is measuring average
voltage but you know it's got a 10megohm input
(or greater) you can use a 10k ohm resistor in series
with a capacitor of value 0.1uf to measure average
voltage. Connect the meter across the cap to get
a reading (meter set to dc volts). Connect the
loose end of the resistor to the output of the
pulser, connect the loose end of the cap to ground.
This puts the resistor/cap network across the bulb.


Here are the relationships between rms values and
average values when dealing with pulses...

P=DC*Vpeak^2/R
Vrms=sqrt(DC)*Vpeak
Arms=sqrt(DC)*Apeak
Vavg=DC*Vpeak
Iavg=DC*Ipeak

where

DC is the duty cycle (0.0 to 1.0)
R is the resistance of the bulb
Vpeak is the power supply dc voltage,
Ipeak is the current pulse maximum=Vpeak/R
Vrms is the rms voltage
Arms is the rms current
Vavg is the average voltage
Iavg is the average current


Take care,
Al