Is this "BMS" powerfull enough for 6 x 21700 ?
https://www.amazon.com/Anmbest-Charger-Protection-Lithium-Battery/dp/B07XCGRD7V

As long as you need no more than 15A peak, surely. (And I surely hope that is the case.)

[EDIT: fixed a dropped “long” and added commentary]

 

I was implying using a 12 volt regulator instead of 5 volts. You still need a regulator for the ICs because 24 volts is too much. Sorry if I was not clear enough.

 

I was implying using a 12 volt regulator instead of 5 volts.

I don't think there is any real advantage there. The max current drawn off the 5 volt supply is appx 1 ma.
That's a dissipation of only 19mw on the LP2950

 

Is IRLZ44N and IRLZ44NPBF the same ?

 

Is IRLZ44N and IRLZ44NPBF the same ?

PbF means Lead free!
Pb (from the Latin plumbum)

 

PbF means Lead free!
Pb (from the Latin plumbum)

Thanks and CD40106 vs. CD40106B/BE/BM ?

 

Thanks and CD40106 vs. CD40106BE ?

Use CD40106BE

 

CD40106BE is PDIP

 

Is it ok if I post mail #62 in the Lighting Forum ?

 

Is it ok if I post mail #62 in the Lighting Forum ?

I would think so, it's your post and photos correct?

 

Yep. Just want to make sure I am not starting a duplicate thread.

What's the Lighting Forum?

 

Why is there a 4.7k resistor on the red LED in the curcuit ?

 

Why is there a 4.7k resistor on the red LED in the curcuit ?

To limit the current.

 

To limit the current.

Yes, but is that not a rather high value ?

 

Yes, but is that not a rather high value ?

Depends on the LED. The ones I use are super bright and a 4.7K resistor is sufficient.

 

R=(Vs-Vf)/If=(5-2)/0.01=300 ohm ?

 

R=(Vs-Vf)/If=(5-2)/0.01=300 ohm ?

Each inverter in the 40106 is not designed to handle more then appx 4ma without loading.
Even using a 330 ohm resistor the current is only 4.7ma because the voltage drops to 3.4 volts.
You can parallel inverters for more output or use a split arrangement.
This parallel arrangement yielded 10.5 ma.
Do you really need 10 ma? As I said before it depends on the LED used.

 

Each inverter in the 40106 is not designed to handle more then appx 4ma without loading.
Even using a 330 ohm resistor the current is only 4.7ma because the voltage drops to 3.4 volts.
You can parallel inverters for more output or use a split arrangement.
This parallel arrangement yielded 10.5 ma.
Do you really need 10 ma? As I said before it depends on the LED used.
View attachment 286497

I suggest checking the specifications of the DC4049. I believ they have an impressive SINK current rating. So then you could use 4 in parallel And if you power them with 10 or 12 volts the capability is even greater.

 

I suggest checking the specifications of the DC4049.

I prefer the 40106 because of the Schmitt-Trigger inputs.

 

I prefer the 40106 because of the Schmitt-Trigger inputs.

For driving a visual display why does it matter? Really. It is not critical And after the first inverter they snap from one state to the other. And if it really made a difference a bit of feedback across two inverters will make the transition very fast.