Barrel Battery Replacement Options

[Chuck(G)]

Depends on the setup. If the battery connects through the external 4-pin header, it's not charged by the motherboard. So take one of these widgets and hook one side across the battery and the other to +5. If you're dealing with the on-board charging circuit, measure the applied voltage without a battery present. If it's under 4.2V, you're good as-is. Otherwise use a blocking diode and a widget powered off the +5 motherboard rail.

 

[tonata]

Depends on the setup. If the battery connects through the external 4-pin header, it's not charged by the motherboard. So take one of these widgets and hook one side across the battery and the other to +5. If you're dealing with the on-board charging circuit, measure the applied voltage without a battery present. If it's under 4.2V, you're good as-is. Otherwise use a blocking diode and a widget powered off the +5 motherboard rail.

OK for the external 4-pin header.
Sorry, I could not understand in the case of onboard charging circuit (where the old battery was).

Case 1: Charging in socket is < 4.2V

Case 2: Charging in socket is > 4.2V
battery socket ---------- diode ---------------- battery -------------- widget -------------- 5V mother board

So what is the idea in case 1? I do not understand how the widget is connected (it has two sides).
Please verify case 2.

 

[Chuck(G)]

If the Case 1 voltage is less than 4.2V, then the Li-ion battery will never be fully charged, as the voltage from a 3.6V rechargeable cell is around 4.2V.

You can see cordless tool manufacturers playing games with this number, advertising 20V tools in an attempt to gain an advantage over the 18V tool manufacturers. In fact, the output of a Li-ion cell fully-charged is 4.2V, but rapidly falls to 3.6V.

Does this make sense to you?

 

[tonata]

If the Case 1 voltage is less than 4.2V, then the Li-ion battery will never be fully charged, as the voltage from a 3.6V rechargeable cell is around 4.2V.

You can see cordless tool manufacturers playing games with this number, advertising 20V tools in an attempt to gain an advantage over the 18V tool manufacturers. In fact, the output of a Li-ion cell fully-charged is 4.2V, but rapidly falls to 3.6V.

Does this make sense to you?

And how do I connect the circuit in Case 1?
Is case 2 correct the way I explained it?

 

[Chuck(G)]

In case 1, just connect the battery as usual, since the charging circuit <4.2V is sufficient to keep the battery charged. You have case 2 correct, as far as I can tell.

Note that a charged Li-Ion battery typically loses only about 0.5-1.0% of charge per month.

 

[tonata]

In case 1, just connect the battery as usual, since the charging circuit <4.2V is sufficient to keep the battery charged. You have case 2 correct, as far as I can tell.

Note that a charged Li-Ion battery typically loses only about 0.5-1.0% of charge per month.

I see. In case 1 you think there is no danger of overcharging the battery, so no protection is needed.
In case 2 is it OK just providing 5V to charge Lithium-ion battery? I need to check.