Li-ion power mod

[upsss]

I have a different module.

https://www.ebay.nl/itm/USB-Lithium-Li- ... 978851987?

Unless you you perform some delicate surgery on this module and wire your On-OFF switch between the charging circuit and the step-up converter (as described by MorganFlint on the 1st page of this thread) this module is a poor choice for this application. The reason is that this Charging/Step-Up module as is, draws almost 0.5mA from the battery in standby compare to only about 20uA if you separate the charging circuit from the step-up converter.

Also, by having a separate step-up converter, it allows you to select a better converter like the one I used which has an additional LC filter on its output with a very low switching noise which is a bit of a problem with the standard step-up converter.

[RobNL]

I change it like this https://www.jyetech.com/forum/viewtopic ... 1153#p4076

I disconnect the batterie complete from any circuit in off position.

[alphaseek]

Back on the subject of a low battery indicator, I ran across references to two ICs that specifically perform this function. They are three-pin SOT chips, and all you would need in addition is the LED and resistor. They are sold pre-trimmed at the factory for specific voltages, induding around 3V. The are the MAX809 and the STM1061, both around 50 cents at Digikey. So if you were planning to buy stuff there anyway, it might be good to buy one of these for the scope. The advantage in addition to low parts count is that they only draw 1 uA when the LED isn't on.

[RobNL]

TL (TM) 431

https://www.flitetest.com/articles/diy- ... ge-warning

[LupinIII]

This low battery indicator is a very bad idea, because it will drain the battery continuously itself (even when the LED is not lit).

[RobNL]

When the batterie full it consume 0.00028 A......OMG, my poor batterie

[alphaseek]

Those interested in doing the battery mod should consider which analog board they have when ordering a battery. I've ordered the BH6X 1800 mah battery that upsss used, but should have been more careful comparing my scope to his picture. He has the new "E" analog board, but I have the older "C" board, which exends about 0.6 cm further than his. I have 3.5 cm of room for the battery, but the battery is 4.1 cm. When it arrives, I'll have to see if I can still make it fit with some creative carving of the side support tabs, and will report on that with pictures, but it would have been better if I had just measured my scope in the first place instead of assuming the analog boards were all the same size.

I decided not to carve on the case to make the BH6X work with my old-style analog board, and plan to replace it with a different battery that looks like it will fit a lot better:

https://www.ebay.com/itm/390980174321

The battery is an AB663450GZ, which I'm hoping is a generic copy of the original Samsung. I think if it's a copy, it's more likely to be newer. As the battery's name implies, it should be 50mm long, 34mm wide, and 6.6mm thick. The width appears to be perfect, and while it's a bit thicker than the BH6X, which is about 5mm, as far as I can tell there's still plenty of room for it. And at 1300 mAh, it should give good time on the battery. But I'll have to solder wires to those little tabs. I'll report back when it arrives.

[RobNL]

https://www.ebay.nl/itm/Accu-Akku-Batte ... SwnNBXTtZF

[alphaseek]

Looks like that would be a good battery for use with the new analog board, but it won't fit with the old board.

[alphaseek]

Yet another warning about bad modules. I bought one of the MT3608 boost modules from Banggood, the kind that has the little multi-turn pot on it to adjust the output voltage. The pot is marked as 100K, which is the right value, but it actually measures as 35K. And that means the maximum output voltage is 10.1V. For mod purposes that should be ok because we only need 9V or a little less. But if you test the module on arrival, be sure you don't apply an input voltage greater than 10V. There's a Youtube video of MT3608 modules catching fire when applying 18V to the input, which normally would be fine if the pot was really 100K.

Earlier here someone reported replacing the pot with fixed resistors, mainly to reduce height. I think that's the best answer. The output voltage is:

Vout = 0.6V (1 + R1/R2)

Since R2 is already on the board as 2.2K, the pot replacement should be 28.6K to give a nominal output voltage of 8.4V. Not sure how you would get there with standard resistor values. Two 56Ks in parallel, then in series with 560R, would be pretty close. Or 33K and 220K in parallel would get you an 8.43V output. These would be nominal values. Actual would be subject to some variance.