Basically the voltage divider R1 and R2 wont change the voltage when the the switch is shut down, as they still got the same ground as the C2 and MCU
They discharge the C2 when the the switch is off, so the MCU voltage drops while the reference keeps the same
Or am I wrong?
while the BR LED resistor and tail LEDs have still the battery ground, which is separated from the C2 and MCU ground
This should not affect the MCU at all
Yes, you're wrong. Sorry, that didn't mean to sound blunt, but you asked. Without a tail light, the current through R1/R2 goes to zero so V=IR goes to zero, ie the cap fully discharges. This creates a poweroff detection
With a tail circuit, there is current through both BR and R1/R2 and V=IR is NOT zero. It's higher than zero. So there's a voltage across R2. If that voltage is too high, OTSM doesn't work. In fact it always means voltage will drop at a slower rate and OTSM won't trigger quite as fast, but that shouldn't be a big deal so long as it gets low enough pretty fast.
This wouldn't work at all if not for the fact that the diode itself has what, 2.1V. So for a full battery the total voltage across all other resistors when off is only 4.1-2.1 = 2.0V. That helps. We only need to get to, er, about let's say 0.8V. (technially 0.3*Vcc, but Vcc has also dropped a little by then, and you want to dive below spec a bit to get there faster, 0.8 is a good ballpark for a full battery.
So you can actually calculate all this, but it's tricky because of the non-linear diode. If you just assume a fixed diode Vf then it gets easier, but you probably want to target a certain diode current to get that. Then use Vbatt-Vf as the voltage over the resistor network, then work out the current in Rled, then iterate, or course use algebra, which I've been meaning to do. If someone can toss out a good number for a target Vf and diode current that match, it will help me make a start on that algebra.
There's a reason I never wrote this chapter. It requires extra thinking.
Anyway, the simple version is you need either lower BR or lower R1/R2 or both and/or the LED resistor higher. If he was close before, then just trying something significantly lower should be good. And adding a 1k BR does at least significantly lower the C1 parallel resistance.
I have to say though, I was a little surprised that a higher LED resistor won't light the LED, even without a bleeder. I call for what about 4K of R1+R2 resistance?
So a 1K LED resistor would make 5K total in series. Ok, so at the lowest Vbat, let's just say 2.7, there's 0.6V left between batt and the diode. 0.6/5K is 120uA... enh, ok that's pushing at, with a low battery. With a 1K bleeder and 1K tail resistance, that's up close to 300uA , should be ok for lighting the LED.
For a high battery though, that puts the BR voltage drop at half the 2.0V total, so 1.0V (just a bit less when correcting for the R1+R2 contribution). R2 brings that down another 30% so right about 0.7V on R2. It should work... probably.
Bottom line: Ok, so try a 1k series LED resistor and a 1k BR. See what happens. Test with full and empty battery. If it works, I won't need a chapter after all, just the components.
order arrived,already install in courui ,thanks.
Usually people use Tail LED resistors between 4.7 and 15k per LED
You claim that for example a 10k Bleeder resistor would kill the Timing totally, which makes electrically no sense at all
I am an electritian and do not see what a 100-300uA through a not common ground parallel wired should affect the MCU and R1+R2 at all
As soom as you half press the switch you separate the C2 voltage on the ground by the switch and Batt+ with the diode, so it becomes a separate voltage source like a second battery
R1 and R2 wont conduct any current through the bleeder and tail LEDs as there is no electrical path or potential difference
LED high-current circuit not shown.
Batt+ Batt-
|--| |-----|
| |----*<--Rled---| (tail cap LED,shorted by switch)
| |---------/ ----| (tail cap switch)
| |
| |
|---------BR----------------| (case ground plane, not equal to
| | batt- when tail cap led is on)
R5 |
| C1 |
|---------------||----------|
| |
|-------R1----------R2------|
| | |
| Vsense pin |
LDO---- |
| |
(Schottky) V |
| |
| |
|----------mcu---------|
|--------<(zener)------|
Vcc |----------||----------|
C2
(Generally either the LDO or the zener or neither will exist but not both at once.)
or for reference, it can be found here: https://budgetlightforum.com/t/-/44344#TAcircuit
But I added the Rled in the version above, in the way I >>believe<< it's wired (pretty sure, but never saw one).
Separated? The only way Bat+ is getting to Rled is through BR and in parallel also through R1/R2. That's current, through the divider resistors. Current on a resistor = voltage. (also through R5, but that's tiny) Yes, the mcu itself is on a completely isolated circuit. The mcu power circuit, once the switch is off has nothing to do with the BR-R1-R2-Rled circuit. The voltage for shutdown detection is not the C2 voltage, it's the C1 voltage, but more specifically the C1 voltage divided through R1 and R2.
image got broken, trying to fix it again.
In you picture above you made a mistake
The BR has to be above the switch, then you can see that it is separated from R1+R2 as soon as you hit the switch
The BR in your diagram would not work at all, just as it would be wired
The bleeder should be 1/5 of the tail resistor
If you calculate the relative voltage of the common ground with that relation you get 0.18-0.3V at 4V main battery source when the switch is open
And 0,93V when the switch is closed
d@#$ don’t know why it quit working (fixed, crazy post editor)
What in the world is “above” the switch? There is no “above” the switch. The switch is in the tailcap. my BR IS “above” ie at more positive voltage than the switch. It’s in driver board and it must be. I’ve actually made these. I’m not making stuff up. On a TA boar the bleeder resistor goes from the top of R5 to the case ground pin on the 7135. I’m actually looking at one.
The only part of that circuit I haven’t actually seen is the tail switch bypass of the indicator LED, but there’s not just a whole lot that can happen there. If you want to add some other “bleeder” there around the LED itself, fine, but that’s not what is usually called “THE” bleeder resistor, and it won’t change the picture current from batt to the led goes from bat … to the case (by way of current through resistors creating a voltage from batt+ to case, and across C1) and then from the case to the LED. You can’t get from batt+ to LED without getting to the case first through a resistance. No way.
An yes, the LED will work just fine in the circuit shown.
Yeah I calculated the relative BR ground voltage with 1/5 resistor ratio to be 0.18-0.3V
With the R1/R2 voltage divider this means the switch OFF level at pin 7 with 1k to 4.7k lighted tailswitch and 2.2-3V drop on the tail LEDs means 40-70mV at pin7, so this is almost not voltage at all
Of course the lower the BR is this voltage drops, but you increase the Bleeder current which means lower runtime in monn mode
I have no idea what you're talking about, because you're not even talking about the circuit I posted apparently, which is the correct one. (posted at same time, now you have a diagram that is identical to mine, but you mislabeled the resistances on R1 and R2, see next post)
If you call BR-R1-R2 a single effective resistance from batt+ to case (it is), that's Reff.
You now have bat+-----Reff--Case---RLed--2.1Vdiode---batt-
Where 1/(Reff)= 1/BR +1/(R1+R2)
we can remove the LED and write it as:
+2.0V-----Reff---Case--RLed---- gnd
Where the 2.0 V is just 4.1 minus the 2.1V LED voltage (roughly, for demonstration of course)
Obviously the voltage drop across Reff is 2.0*Reff/(Reff+RLed)
And the voltage across R2 is then about 70% of that. (edit: for clarity R2/(R1+R2) which is about 70%)
mV? without a bleeder Reff is about 4kOhms. Even if Rled is 4kOhms, you've got a 1 volt drop across Reff.
Ok, a that point the current may be so small that the led drop is less than 2.1. That only makes the answer bigger.
Your circuit now does at least look identical to mine. But you have R1 and R2 upside down. The 1K goes on top and the 3.3K goes on bottom (did I mislabel them in the manual?). The point is to make the sense voltage about 0.7*Vcc when the batt is on.
(Edit: They are labeled correctly in the manual and match with the circuit diagram in the TA page, R1 is on top, and I call it out at 1k. R2 is on bottom, and I call it out at 3.3K)
Ok then the off voltage between R1 and R2 is is 0.14-0.23V when off
Look in my example the voltage over the BR is 0.18-0.3V depending on red or blue Tail LED, as most of the voltage drop is on the LED and Tail resistor
I made the current flow now visible, maybe you now understand, why the affect on R2 is so low
The relative voltage, always depends on the current flow, not on the resistor values and supply voltage
Ok, nice red arrows but your mcu current path is wrong. Pin 7 has mega-ohms of resistance. The electrons flow back to the mcu through the mcu's own case conection, not through pin 7. The C2-mcu circuit is completely isolated and just floats on the case ground. It has nothing to do with anything. You can just remove it from the picture. You can also remove C1 as it has no affect in DC when the led is powered.
You only need BR R1 R2 Rled and the led. But I already wrote the simplified circuit above.
You can do it either using voltage ratios or current, or both, and it all better get you to the same answer. I worked out the math fully rigorously above without direct need to ever reference current but of course the voltage I imply for a divider are derived from "ohms law" as it gets called. And the led voltage is probably closer to 2.1, not 3 and could even be less at these currents.
And your R1 and R2 values are still swapped.
Anyway, with the 1K BR and 4.7K Rled in your image, you get a total series resistance of 5.5K, and yes OTSM should work in THAT configuration, and the LED probably should too. The q8 is using 15K series resistance and still getting their two green LED's to light usually.
The OP started with no BR and said the LED only lit with Rled down at 220ohms. That's a total series resistance of 4.5k, but with too much of the voltage drop above the case, which is why OTSM is not working and is where all this started.
5.5k isn't much more than 4.5k. So yes your configuration could also work, probably should usually. Will it work for g_damien's LED? If not then a 1K bleeder and 3.3K LED resistor certainly will, because that puts him back below 4.5K where he was and still keeps the voltage drop situation ok for OTSM.
Traditional lights needed the bleeder for different reasons than this OTSM light does. They needed it because R1 was 20k. In this light the LED can work with no bleeder at all and the LED resistor only fine tunes brightness or also isn't needed at all. You need the large led resistance in this light and compensating BR exactly because of this OTSM voltage issue. Although OTC has a similar issue, and probably wouldn't work well with no LED resistance at all, but being analog and calibratable it has a higher range of tolerance. Downside being it required calibration.
Anyway,
Bottom line: Try lexel's 1k BR and 4.7K led resistor, OTSM will work. If the light doesn't light up, try a 1K BR and a 3.3K led resistor. That should certainly get it all.
You got it wrong
He did not use a bleeder at all, he used a 220Ohms tail resistor to get the LED light up, so the full tail current flew through the R1 and R2, of course that lifts the voltage above R2 over 0.8V and OTSM stopped working
Quoting myself, one post prior to that:
"The OP started with no BR and said the LED only lit with Rled down at 220ohms. That's a total series resistance of 4.5k, but with too much of the voltage drop above the case [ie across R1+R2 just as you're saying], which is why OTSM is not working and is where all this started."
Yes, he did not use a bleeder but he still had 4.5K resistance in series with the LED. I agree that that configuration causes a bigger voltage drop from Bat+ to case (it matters not how that voltage drops, through R1/R2 only or through an effective combination of R1+R2 with Reff, only how much voltage drops, as determined by the ratio of that combination with Rled, you can achieve the same adjustment by adding a BR or by lowering R1 and R2). Anyway, you seem only able to think about things in terms of current in separate paths. There are other, simpler ways to get there. In the end you aren't even disagreeing. The voltage drop across the BR is identical to the voltage drop across R1 and R2 and it is determined by the Ratio of the total effective resitance, 1/(1/(R1+R2)+1/BR) to the Led resistance, regardless of how you create that effective resistance. You can think of currents if you want, but you get the same answer, as you did. If instead of adding a 1K BR you make R1+R2 total to 1K, ALSO all of the current will go through R1 and R2, but the values of R1 and R2 are lower, so the voltage drop is still the same. Same total effective resistance makes the same voltage drop, regardless of what fraction of the current goes which way.
No bleeder can light the light equally well, but causes voltage problems on R2 that breaks OTSM. Which is exactly what I started out saying, and what I said again two posts up. :facepalm:
So we agree, you can light the led different ways, but some of those ways will cause voltage problems that break OTSM.
The case voltage is irrelevant as you have 2 separate voltage sources, I tried to explain it to you multiple times
You have to calculate either the voltage drop ratio between BR and the Tail board
This leaves you to 0.18-0.3V in my example, this voltage also applies to R1+R2 as it is parallel to BR
or
You have to calculate the current flowing through the tail LED, this includes all involved Resistors and the Diode
Then you calculate how much of that current flows through R1+R2, which is obviously less than through the bleeder, so you can calculate how much voltage drops over R2 U=R*I
In both cases you have to calculate the relative voltage lift between R1 and R2 by the current flowing through them or total voltage drop for the parallel part of the voltage divider BR,R1,R2 and tail board
The more voltage drops on the tail board the less is R2 affected