-Forward voltage of- -ZLLS410- S4 diode. Edited- This is NOT a Zener mod

This is a mod I worked out originally almost 2 years ago(minus the FET and divider correction in 7135 drivers with higher voltages and yes it requires some caps and jumpers. Here is a pic of how that mod was done. It ran 3 XML’s in series from 12V input. I ended up just stacking the caps on the pins of the regulator. This was before I knew of the MTG (before Gen1) and I never used it because of the lack of low voltage detection. It’s kind of exciting to finally be almost in position to use this idea after so long.

Edit-oops! Here’s a pic with the regulator in place sans caps. It appealed to me because of the sot89 package.

At the time I was also toying with better heat sinking for 7135’s.

Your drawing shows D1 as “remove”. That’s fine, but you’ll have no reverse voltage protection.

You’re still talking about a diode though, so I suppose you plan on putting it back… somewhere?

You seem to have it backwards. With the divider attached like a stock 105c of course you need to know the drop across the diode. That’s why it’s accounted for in our existing firmwares! Once you connect the divider to BAT+ you no longer care about drop across the diode.

The reference voltage produced by your ATtiny13A is completely unaffected by D1 - it will always be 1.0v /-0.1v. (yeah, that’s a/-10% tolerance… eww). The voltage coming out of a divider fed directly from BAT + will be whatever the divider ratio dictates.

Yes, I realized later that I could use D1 as the jumper for the regulator.

I probably have this all wrong but with the diode dropping .25V and a cell at 2.8V and with R1 and R2 at 19.1k and 4.7k pin 7 will see (2.8-.25)(4700/23800)= .5V

To get low voltage regulation to work I need to be able to show pin 7 - .5V when Vb=5.6V.

If the voltage drop across the diode is some value other than .25v then .5V is the wrong target for the divider.

Do I ignore the diode in the calculation? 2.8 x .1975= .55V.

5.6(4700/R1+4700)=.55

5.6(4700)=.55(R1+4700)=.55R1+.55(4700)

(5.6(4700) - .55(4700))/.55 = R1 = 43155ohms

Using R1 = 43155 and accounting for the same .25v diode drop pin 7 should actually see .5V. Let’s try it.

(5.6-.25)(4700/(43155+4700)=.53V not .5V.

This was assuming the voltage divider was left connected after the diode and the voltage drop across the diode ignored.

Try again with a target of .5V and connecting again after the diode.

(5.35(4700) - .5(4700))/.5=. R1 = 45684 ohms.

And one more before the diode

(5.6(4700) - .5(4700))/.5 = 47940 ohms .

I have the feeling I’ve missed something crucial but at this point I figure I’ll just try some different values for R1 and see what happens but to me it looks like with R2 at 4.7k and

Divider connected after the diode, R1 = ~ 45.7k ohms.

Divider connected before the diode, R1 = 47.9k ohms

Whether you put the divider before or after, I don’t see how the diode can be ignored.

I had some trouble following your post, but let’s wade into this mess:

This is how the stock divider on a 105c works when the battery hits 2.8v:
4700/(19100+4700)*(2.8 - 0.25) = 0.503v

Exactly as expected. Of course, this leaves you in a situation where changing the diode potentially changes the voltage reading. This is what I was commenting on earlier.

In order for a 2s battery LVP to work correctly without the diode we just need change a resistor. Swap the 19.1k for a 48.1k to get 0.498 out of the divider before the diode like this:
4700/(48100+4700)*5.6 = 0.498v
This will cause LVP to kick in at 5.617v

When I say “before” the diode, I mean that the diode is connected between MCU’s VCC and BAT. I consider BAT to be “before” VCC. I think that’s pretty standard phrasing, anyone can feel free to correct me on that though if they know better. Once you have connected Vin for the divider to BAT+ (“before” any diodes or any other things) the diode is no longer part of the calculation for the divider circuit.

I have no idea what makes you think that a diode that isn’t in-circuit with the divider can make a difference, but it simply isn’t true. If you make a complete circuit with a voltage divider (it has Vin, Vout, GND) and the diode isn’t in there… it DOES NOT NEED to be accounted for in your math.

I’ve lost my skills at solving equations, so I don’t do it. I use a solver such as this one: http://www.raltron.com/cust/tools/voltage_divider.asp

It was the first thing we both did to get the .5V we both started with. That’s what I mean by accounting for it. Yes, of course if the divider is tapped into the circuit before the diode then it’s not part of the circuit but to get that .5v we both started from the assumption was a diode drop of .25v and if that is incorrect then the new value of R1 will be incorrect.

My appologies since my intent was to gain a better understanding rather than argue over a misunderstanding. I think we agree but are looking at different ends of the elephant.

Please avoid using phrases of the type “I have no idea what makes you think…” As they are a bit antagonizing and just show that you did not understand me. Just say you don’t understand. Just about everything in your last post was covered above in mine so I don’t think we really disagree. There was a typo between my last calculated value of 47940 and the 47.k. 47.9k and 48.1k aren’t too different.

Here’s another pic of a 105c from back then. I think it was the last one I did and shows the diode turned and used to connect to the input of the voltage regulator. The white wire goes from Vreg out to Vcc and you can see the caps stacked on the regulator pins.

I’ll show a board that’s been cleaned up and has the proper traces cut in a week or two when I get back to this part of the project.

Sorry, I shouldn’t antagonize. Once I read “Whether you put the divider before or after, I don’t see how the diode can be ignored” I became really frustrated with this situation.

Now, upon looking very carefully back over the thread (more than once), I think I’m starting to see where we went wrong. I think you’ve been assuming that I meant to attach an unmodified divider (19.1k / 4.7k) to BAT+ and then use an unmodified firmware. When you kept saying that the diode can’t be ignored, I suppose you meant that the diode makes a difference - when present it reduces voltage and when absent it does not.

If my new understanding of the situation is correct, I’m still at least a little frustrated! When I jumped into the discussion you were already talking about changing resistor values to suit different diodes. So all that needed to be done when you get the divider Vin from BAT+ is to change the resistor values to suit that.

The bottom line remains that if you hook divider Vin to BAT+ and divider GND to BAT- then the diode is not present in the math for the divider.

EDIT: slightly less antigonizing on this post too. :wink:

Nice work squeezing the regulator, 2 caps, and diode all into that tiny space btw. If you hadn’t specifically said that the diode was there I wouldn’t even have seen it.

I was sorry to see your frustration growing since all along I had been assuming I was going to connect the divider before the the diode. My typo was in the number I typed at the end. My original question was “what’s the voltage drop of this diode?” which I needed to know to calculate the (.5v) number and for nothing else. I’m fine with solving a simple equation for x or locating which trace to cut where. I kept picturing you banging your head saying “who is this idiot?” while with each post we were operating with different assumptions. You have been very helpful typing long posts affirming things I already knew while I couldn’t seem to make clear my question.

This second is the sole reason for my post. We assume a certain value for the voltage drop across the diode to make the first calculation of the voltage on pin 7 on boards as they are now configured.

In post 5 I listed 4 different new values for R1 to show how a different value for D1 in the first calculation affects the value R1 in the second even though the diode is bypassed. Please note that we agreed then and still agree now that if the diode drop is ~.25v then R1 should be ~48k when bypassing the diode.

Is the .25v assumption a correct one?

Again, I apologize for your frustration in this as it’s obvious you are trying to help. I’ve seen a number of threads go south out of frustration and have no desire to see that happen here. Thanks for spending some time on this.

If this works out the way I hope then a dedicated board would definitely be nice and much easier to do with a single FET on the board as opposed to a bunch of 7135’s. This is a test board showing the FET and gate resistor on the pwm pin. You can see how all the room I need for the voltage regulator is there. I’ll be swapping out the processor and R1 so I’ll cut the trace to R1 when I do that.

It might be possible to gerrymander the traces on the backside to jumper the divider.

Heheh. Now I get it. :smiley: Looks like the proper amount of bold print can get the message across to me. Sorry for missing that key point, it’s easy to see now how we could be so close and yet so far away!

I’ve got to dig through my mess for a 105c to test with the diode in place. For the time being I used my $13 buck PSU to put 10mA across a diode I recently stripped from a FT 105c. Initially it read 0.28v but by the time I got my meter on it it read 0.30v. The meter agreed and it stayed there, I suppose that’s from the warmup at 10mA.

FWIW, my opinion is that you need to target a cutoff of 2.8-3.0v. The cell is effectively empty at 3.0v or greater, just not damaged yet. There’s no reason for precision for the actual low voltage cutoff. Step down is a little trickier and calls for more precision of course. If actual cutoff is the only concern, you have a huge target.

Thanks, wight. Glad we’re sorted.

I was looking at comfychair’s excellent pictures of both sides of a 105C board and I think I can use the output trace on the spring side to bring B+ through one of the vias to Q4 and use that position for the regulator instead of Q3. This is possible because I’m stripping all the 7135’s from the board and using an FET instead. It replaces two long jumper wires with simple solder shorts. :slight_smile:

I started working on the board this afternoon. I started with a stripped 105C. I save all sorts of parts if they’re not cooked. First I went over it with solder wick then wiped it clean with alcohol.

I need to cut several traces in order to reroute the mcu power through a voltage regulator, replace all the 7135’s with an LFPAK56 FET, and change the resistance on R1 to ~48k ohms.

First was cutting the power trace to the mcu.

Then I need to isolate the pads at Q4. This is where the voltage regulator will go. The 7135 pwm pin corresponds to the input pin of the regulator but since Vin will use the output trace I’ll cut the pwm trace to that pad.

The output of the regulator will jumper back over to the Vcc pin on the mcu(output pin of the regulator is the same as the output pin of Q4) so since I’m using the 7135 outpu trace for Vin I need to cut that as well.

Next up is the prep for the FET. It straddles Q1 and Q2 but the pads are mostly wrong. The FET gate pin lands on the pwm pad for Q1 so that’s ok but I need to sever the pwm trace to Q2.

The large output tab (drain) of the FET sits on the rest of the pads for Q2 so I need to isolate those pads, including a section of the ground ring, from the rest of the board.

Since I’m using the output trace for Vin I need to sever that as well.

Then I scraped the traces under the FET clean of solder mask and tinned them.

Oh yeah, severed the pwm trace for the FET gate resistor on pin 6 of the mcu. D1 will be the jumper from the former output trace to the input pin of the regulator.

Lastly a shot of the other side. Vin will jumper from the center pad to the output pad of Q6 and the rest except for C1 get covered with JB weld.

Check back next week for the reload.

The cut traces side makes me :smiley:

Cool thread! Gotta love BLF.

That's a lotta work though when we already have working FET and 7135 boards implementing the diode bypass for the divider circuit. Link. I have tested them both.

I happened to test the same physical diode (not a similar one, identical one, or something else) with the same DMM today, but I used diode test mode instead of putting a load across it. In diode test mode it shows 0.248-0.249, so I find that interesting. That’s a difference of 0.05v going from no-load to 10mA load. I doubt it sees 10mA in our circuit, but I bet it sees 5mA.

Pic of the board with everything in place:

Hole in center for led+ wire

FET installed with gate resistor on pin 6 of mcu.

Voltage regulator chip with input and output caps, jumper to Vcc pin, and D1 on the input pin.

Additional 28.7k resistor in old D1 location to bring total R1 resistance up to 47.8K for lvp.

Finished this up too late for testing but the wires are all on including 3 solid wires in the ground vias one of which connects directly to the 3 ground pins of the FET.

This had to be some of the worst soldering I’ve done in awhile. The fluorescent bulb in my magnifying lamp died and made this much more difficult. I may break down and get a better model with higher magnification and a longer arm.

Pretty crazy all the same. Keep us posted.