for new orders please refer to the 5. batch here [Finished] 5. Batch "TA" Bistro HD OTSM/Narsil drivers, 15-47.5mm size fit H03, MF01, MT03, Q8 1S or 2S, Clicky or E-switch
I hope the other AMC7135 suppliers claw type work as well as the old supplier ones
When they do I do another big order of parts from Digikey and Mouser
for discussions use this topic
Any update on the MF-01 boards, Lexel? Are you still waiting for Oshpark?
They are shipped, but it takes some time
Hi, I have Bistro HD OTSM and trying to install lighted tailcap in S41 with stock tailcap board. Iām adding resistors in the tailcap with LED. The problem is: if resistor value is to high (>220), green led does not work; if resistor value is around 220 and lower, green diode is working but I cannot change modes.
This sounds not right
Usually those green LEDs work with 20-50k Ohm resistors
Did you add a bleeder resistor on the OTSM pad near the single AMC7135?
Hm, that might be the reason 
What value should have bleeder? Could you mark the place where it should be soldered? 
Edit:
found it, marked BR: 
. Is it correct? According to this thread, 470-630ohm. Is better to have it larger or smaller?
I would go towards >1-2kOhm as the bleeder resistor value,
because it increases current when the light is on the moonlight current, this would be more than MCU plus main LED with 600 ohms
In tail you need resistors to even out the LED voltages with 2.2-30k depending which brightness you need
Well, it looks like I better get around to writing that taillight chapter
This could be tricky. No time right now, but I'll look at it soon. Basically though, raising the bleeder resistor goes the wrong way. The bleeder is parallel to R1 and R2, and is actually redundant with R1 and R2. It only serves to lower their value. Those three resistors together form one effective resistor Reff from the top of C1 (basically the battery +) to the case. Reff is in series with the LED resistance. Reff and Rled form a voltage divider (only sort of being that the LED itself is non-linear). Ok, the goal is to get the voltage across Reff low. It must be low for the mcu to detect power off and start timing the click.
To get the voltage across Reff low, you need Reff relatively low compared to Rled. So yes a low value of Rled will break OTSM.
The possible solutions are to lower BR (if it's infinite now, then yes that means add some resistor) or to use lower resistors for R1 and R2.
Using a 1K will reduce the Reff by about 3 or 4, and it might work. I'd probably try without with 500 though, and either way try to get that LED resistance back up some as well at the same time.
I've been meaning to try to calculate something for this, but in the end, trying is a good way, especially when led's are involved.
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@#$ donāt know why it quit working (fixed, crazy post editor)
What in the world is āaboveā the switch? There is no āaboveā the switch. The switch is in the tailcap. my BR IS āaboveā ie at more positive voltage than the switch. Itās in driver board and it must be. Iāve actually made these. Iām 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. Iām actually looking at one.
The only part of that circuit I havenāt actually seen is the tail switch bypass of the indicator LED, but thereās not just a whole lot that can happen there. If you want to add some other ābleederā there around the LED itself, fine, but thatās not what is usually called āTHEā bleeder resistor, and it wonāt change the picture current from batt to the led goes from bat ā¦ 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 canāt 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)