R
rickman
Guest
On 3/22/2016 10:00 PM, DaveC wrote:
Perhaps I am missing something. The battery terminals will be at the
battery voltage, no? So why would you use 4 volts which is at the
output? It would be 4 volts times the LED current or the battery
voltage times the input current.
--
Rick
On 22 Mar 2016, rickman wrote
(in article <ncssb4$c6l$1@dont-email.me>:
Didn't you say it used a *single* 1.5 volt AA cell? I think that would
make it 412 milliwatts not counting the conversion efficiency, so more
likely a 1/2 watt LED.
1.5v boosted to 4v (measured) output from the converter. 275 mA (measured) at
the battery terminals. I make that to be 1100 mW. If presume 75 percent
efficiency, 825 mW.
No?
Perhaps I am missing something. The battery terminals will be at the
battery voltage, no? So why would you use 4 volts which is at the
output? It would be 4 volts times the LED current or the battery
voltage times the input current.
--
Rick