So with a buck or boost driver, there is an input voltage range, and if you provide a voltage with in the range, it’ll out put a steady set current, right?
Borrowing some data from koef3:
A Nichia 519A at 2.2A has a Vf of 3.08V - so a buck driver will need a battery to be at 3.08V plus some overhead to output 2.2A, right? And I see that at the max current of 7.6A, the 519A has a Vf of 3.47V. So a regular lithium ion will run regulated for most of its life.
A Luminus SFT 25 has a way higher Vf. At 8.2A, it is at 3.84V. So with an 8 amp driver, it’ll run regulated at 8amps until the battery drops to 3.84V (+ whatever loss in the driver), and then it’ll run unregulated, right? Basically, it’s more efficient to run them in series and use a boost converter?
Now, these drivers, they will output the current regardless of the Vf, right (within the output range of the driver)? Like using an SFT25 with the same driver will use more power at the same current output?
Some yeah, firstly if the buck converter isn’t capable of 100% duty cycle, there will always be a voltage drop, for example convoy 8A/10A isn’t 100% capable, and will have a max duty cycle around 92% or something. So there is always at least 8% of Vin voltage drop (ex : 4Vin, drops to 3.68V, which is significant). But Convoy 5A driver is 100% capable, thus no such drop.
Secondly there is a voltage drop due to the resistance of the circuit : components, wires, springs, cell DCIR.
For example, guesstimating Convoy 5A at ~55mR, springs maybe 15 mR each, clicky switch 10mR? 18650 cell 25mR, 5mR wires/traces.. so about 100mR total.
Thus at 2.2A there will be 0.22V drop, so if the Vf is 3.08, then it will decreases below 2.2A when the cell voltage at rest is below 3.3V.
At 5A, 0.5Vdrop, 3.32Vf, fall out of regulation at 3.82V cell rest voltage. (About 60% SOC probably).
So minimising circuit resistance is pretty important with a 1S input to 1S output buck driver, as well as selecting a 100% duty cycle capable buck IC.
Efficient isn’t the right term, but yes it’s better as it will be able to maintain regulation regardless of the input voltage (unless pushed really hard and current has to be lowered at lower input voltages).
So because of the lower duty cycle ICs, there isn’t much difference in the 8amp and 5amp Convoy drivers? (Or the difference is reduced because of it?)
In the case of the SFT25, it’s high Vf means that a linear driver and a boost driver will have the same efficiency once the cell drops below the necessary voltage to maintain regulation? Also, with the drop because of the resistance in the light, coupled with the high Vf, it means that a convoy 8A light never runs in regulation at 100%?
Well, depends on the Vf of the led, also the convoy 8A driver has lower circuit resistance because better components (inductor, RPP mosfet, buck control mosfet).
Regarding the max duty cycle I said 92% from memory but looking at the datasheet again (MPQ8612) it’s dependant on the switching frequency chosen, which I dont know because I dont have this driver to measure it.
It’s the min off times that determines the max duty cycle at a certain frequency, and there a min typical and max value for the min off time so :
freq
min
typ
max
300kHz
99.1
97.8
95.5
600kHz
98.2
95.5
91
1000kHz
97
92.5
85
The min off time is quite variable, so it’s quite a wide range of max duty cycle, e.g at 600kHz it could be between 98.2 and 91%.
That ”92%" I recalled must have been the typical value at 1000kHz but I would hope that Convoy engineer has chosen a lower switching frequency to minimise the voltage drop, at 300kHz, it would be not too bad for the typical off time value (97.8% max duty cycle).
Anyway it’d be better to actually measure it.
In dropout linear is the most efficient, because the MOSFET is fully conducting, and it’s just its ON resistance and the sense resistor that incur losses.
Then a 100% capable buck driver, since the control MOSFET is ON all the time, it’s its resistance + inductor + Rsense+ RPP MOSFET, more stuff than linear so a bit more losses.
A boost converter will have lower efficiency because no matter the input voltage it’s always doing voltage conversion so switching losses + conduction losses. That’s assuming similar components characteristics, in practice the boost driver can achive higher efficiency if its conduction losses are lower (better MOSFETs, inductor).
Great discussion - I’m trying to understand but the technical level is surely above my pay grade. Just want to share simple measurements of the 2 drivers in case it helps.
“Turbo” output at turn-on, both with SFT-25R 5000k LEDs, highest output of multiple measurements using Texas Ace lumen tube, batteries fully charged:
Convoy M21E Buck 8A: Max level output 1645 lm, EVE 40PL 21700 battery.
Convoy T8 Buck 5A: Max level output 1370 lm, Vapcell H10 14500 battery.
Does the difference in output mean there is not a significant voltage drop with 8A driver?
If there is a drop, how does the battery current capability, tabless EVE 40PL vs Vapcell H10, affect this drop, or discussion? Thanks.