Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

LPF Donation via Stripe | LPF Donation - Other Methods

Links below open in new window

ArcticMyst Security by Avery

Sanwu Blue Laser - Silver Series Batteries?






WizardG

0
LPF Site Supporter
Joined
May 9, 2011
Messages
1,170
Points
113
The `protection' in the batteries is to protect the batteries. If you drain a lithium ion cell all the way flat it can never be recharged. The protection circuit, for all intents and purposes, turns the battery 'off' if its voltage drops below a setpoint that's usually between 2.5 and 3 volts. Discharge to any lower voltage causes permanent damage to the battery.

But you're not entirely wrong about the driver issue. Low voltage input to a boost type switching regulator can get those hotter than usual as the transistor(s) responsible need to switch higher current through the commutating inductor to maintain a constant output.

The diodes themselves have no issue at all with being underdriven.

As far as I know laser diodes are like transistors ( active components not passive ) when you underdrive them they create more heat which in turn cuts the their life expectancy down, or in other words strain the component. Could be wrong, somebody correct me if I am. :thinking: edit~~~~ you know thinking about this, it just might be the driver that gets undriven??? Curious, have to check on this ???
 

BobMc

0
Joined
Apr 23, 2016
Messages
3,685
Points
113
The `protection' in the batteries is to protect the batteries. If you drain a lithium ion cell all the way flat it can never be recharged. The protection circuit, for all intents and purposes, turns the battery 'off' if its voltage drops below a setpoint that's usually between 2.5 and 3 volts. Discharge to any lower voltage causes permanent damage to the battery.

But you're not entirely wrong about the driver issue. Low voltage input to a boost type switching regulator can get those hotter than usual as the transistor(s) responsible need to switch higher current through the commutating inductor to maintain a constant output.

The diodes themselves have no issue at all with being underdriven.

I was thinking I overstep the line with the diode being an active component. Thanks for the input, it's always nice to learn something new. :)
 




Top