Edit- to Kiriba-ru, I think it was an excercise in understanding something he intuitively knew but wanted to prove mathematically. It’s not an arguement promoting one over the other, just curiousity to find where the power is dissipated in each one.
I really don’t know what you mean here. The example at the end is using one set of parameters: duty cycle, voltage, etc. The calculation is general; you can choose any voltage or duty cycle you want.
Yes, rufusbduck, I’m definitely not trying to say one way is better than the other.
When you operate with values that are changing in time and values dont have time inside them, this is point values.
What is cell capacity? Watt×hour. Operating with power only can not show difference, we need to know how does it changes in the time.
How can we compare power consuption, if nobody can fix it with fet pwm? Comparing fixed current regulation with non-regulated fet? It can be usefull for some average current, but will show nothing in one point. And the results shows this.
Dear EasyB.., your analysis is pertinent and the principle of power conservation is demonstrated again , so it's safe with us...
I do not want to disagree with some of your statements but I have to , please do not take it personal , we are not talking about personal opinions or understanding of a process , but about physic laws as : P=U x I =square U / R = square I x R . I hope we all agree with this !
" In the linear regulation method, the driver itself dissipates heat. In the PWM method the driver dissipates no heat." quote .
This is the statement I want to challenge , because the PWM drivers do dissipates a lot of heat , due to their active internal resistance ( I'm calling it " active " because is changing with frequency and duty-cycle) . Just take a look at the efficiency of a buck driver and you 'll see that the difference from 100% to , lets say , 75% is power loss , transformed in...HEAT , in driver circuitry ( fet , inductor , fly back diode , resistance of copper traces , wires...).
On the other side , the DD drivers , made with quality components , should dissipate LESS heat , especially when the input voltage decrease and is closer to Uf of the led , because in the circuit we have only the resistance of the FET ( Rd-s in " on " state ) and resistance of the traces and wires.Logic , right ?
Lets take them, one by one and see what the drivers can do ( in terms of efficiency and subsequently , in terms of heat dissipation )
1. Linear Driver. Basic , is working like a variable resistance , between power source and LED. So , assuming that the current is CONSTANT, the efficiency will be :V led/V battery . Lets take V led=3.4Volts.
For a full battery ( 4.2 V) : 3.4/4.2=80% efficiency.
For a depleted battery (3.4V) : 3.4/3,4=100% efficiency . Between this values (4.2 V and 3.4 V) we'll have a gradual increase of efficiency , from 80% to 100 % , and a decrease of power loss (less heat).
Under the 3.4 V ,the efficiency will remain 100% , but with the dimming of led luminosity .
2. DD drivers . Is acting as a direct path from power supply, to the led. Trying to add a resistance in the path , for limiting the current , will make this driver less efficient than a Linear Driver , the current is greatly variable with the power supply voltage. Good for high currents , one battery , one led . Could be PWM controlled , for modes .
3. Buck Drivers. The stated efficiency of these drivers is from 70% to 90 % , depending on the design , quality of components , work frequency... The diversity of these are so big that is gonna be very difficult to state which is better , and in what conditions ( 2 to 4 batteries in series , 2 to 4 LEDs in series...) .To be short and very close to the experimental side , I'll present here ,one (considered one of the best) buck driver , from Skilhunt , as a prove of what means efficiency , in real world...
First set up : Three batteries in series ( Vbat=12.1V),one led XM-L ,Vf=3.18 V.
Current to batteries Ibat=1,47 A , Current Led Iled=3.87 A .Doing the math , efficiency almost 70 % ! Heating was big , hard to keep fingers on the driver !
Second set up.: Same 3 batteries in series , but 2 XM-L in series .
Ibat=2.90 A , Ubat=12.0V ( 34.8 W) ; Vf 2xled=6.76 V , I leds=3.87A ( as you can see , the current is constant to 3,87 A!) . Efficiency =75% ! Better...,heat was lower but still uncomfortable to touch...
Third set up : Same 3 batteries in series , but 3 XML in series !
I bat=3.20 A , U bat=11.62 V (37.18 W) ; Vf3xled=9.71 V , I leds=3.98 A (almost constant!). Efficiency=96% !!! .Wow , this was unexpected , the driver was working slightly warm , as this was the point where he was designed for...! I have to admit that I changed the inductor with a bigger one , but was the same one, for all 3 setups . Nothing else was changed !
So, as a conclusion , the drivers has to be chosen accordingly with their destination , and the heat will be a measure of their efficiency !
I hope I didn't upset anybody , and if I did , I do apologize ...
Adrian,
Yes of course the FET driver will dissipate some heat. I pointed out the perfect switching as an assumption, and I think that is fine for the purpose of the analysis.
I think my take-away message that I learned is that you can’t just consider the driver efficiency by itself. You have to consider the whole system. The FET driver itself is more efficient than a linear driver because it produces less heat, but if you consider the whole system, modulating a FET driver using PWM is less efficient at producing light than using constant current with a linear driver.
OK, that’s fine. One can certainly calculate more things. What you want would just be a continuation of the analysis, using the same concepts but integrating over the discharge of a cell.
That’s no surprise to me, because i.m.o. it’s the obvious thing to do, whereas PWM is (i.m.h.o.) a bit of a lame route to take…\
My obvious question is: Where can i buy it??
And i’d be more than happy with 3 modes like 10% 30% 100%.
Those Atmels have enough legs for it, no? 
There was also someone here who made, or was making, a driver with individual 7135 control back in 2014. I think it was a driver with a USB interface made by someone with a baby as avatar (could be mixing different projects though). At the time I didn’t design drivers and was interested but I don’t think it was ever made available here. When I started designing myself I took his idea of individual 7135 control.
I made drivers and firmware available where everything in the firmware could be programmed by the switch, modes, number of modes, voltage step-down and cut off levels, temperature protection etc etc, but I don’t think anyone ever built one.
[quote=Jerommel]
That’s no surprise to me, because i.m.o. it’s the obvious thing to do, whereas PWM is (i.m.h.o.) a bit of a lame route to take…\/quote]
I was unable to eliminate PWM altogether though, I use PWM on a single 7135 for the steps in between. I was experimenting on a true constant current driver but I’m no electronics guy and all that datasheet reading gave me a headache so I gave it up.
I’ll be making some available and also do a giveaway of a few (which has already been sponsored by BLF) in this thread: Mike C drivers: v8 series, ATtiny1634 based.
Without mucking about with the reset pin these little things have 11 I/O pins. I’ve already used every single one of them in a headlamp driver I built
I’m still not sure what question you are interested in answering. How the power consumption will change with the FET driver and constant duty cycle? You could certainly do this calculation for a specific cell, and it’s pretty simple to estimate it for different states of charge. But this is an additional calculation and beyond the scope of the present analysis, the purpose of which was to determine where power was dissipated with each dimming method. And the analysis answers this question for any combination of emitter, cell voltage, and average current
You have taken strange inputs and get nothing in output.
Your calculations are right for one full pwm period. For realistic results, it is better to avoid power value, and use energy value. This is the only way to get what you want.
Have just recognized very big mistake in first post. Would like to wait a little until smb else will found it.
Please explain more precisely how I “get nothing” in output.
You calculated two areas. One is 4.2x3=12.6, second is 4.2×(0.5×6)=12.6. And you have commented that I can put any voltage and results will be right. Im trying to say very simple thing - difference between power consumption will make much bigger influence that power loss and its dividing between components.
OK, the calculations are not of interest to you. That’s fine. But the calculations make sense; your earlier criticisms did not.
Honestly I did the calculation because I found it interesting. I mentioned in the OP it wouldn’t change how we do things, just that it was interesting.
Great post