Driver test load confusion`

[rhd]

It looks like the operating voltage is 3.7-4.5 DCV, so I don't understand why I would need any diodes on a test load since my battery only supplies 3.7 volts.

Battery is irrelevant. Your test load attaches to the driver output, not to the battery. A test load has no idea what battery is attached to the other end.

Here's your calculation:

(a) Voltage you need the test load to drop
= 445's Vf at your current (say 1.8A)
= 4.8 V (roughly)

(b) Voltage your 1 ohm resistor will drop
= Current x Resistance
= 1.8 x 1
= 1.8 V

(c) Voltage left for your diodes to drop
= a - b
= 4.8 - 1.8
= 3 V

(d) Number of diodes you need
= c / (voltage drop of your diodes at your current)

** So, check the datasheet for your diodes. How much voltage do they drop at 1.8A ? Let's say they drop 1V when run at 1.8A **

= 3 / 1
= 3 diodes needed

Someone should check my presumptions / math. I'm really tired and my brain is foggy.

 

[sassafrass686]

Awesome! I appreciate all of your help!
Thank you much.

 

[benmwv]

Yes the schematic is correct ! You must connect your diy driver leads where the blue-ray should ! Also at 1,25A the lm317 will get very hot ! Do you have a heatsink for it ? I recommend you to use an lm350 which has maximum of 3A so you wont need heatsink at all !

This ^^ has been nagging at me since I first saw it.

Just because an lm350 can handle 3A doesn't have anything to do with the amount of heat it produces. They will both be to-220, at the same current, same load, and same input voltage. Therefore the heat produced will be exactly the same, and they will need the same amount if heatsinking.

Their max current is irrelevant.

 

[sassafrass686]

Ahaha. Just got home from work and assembled the load. I tested it with a linear driver (as I'm waiting for the buck driver to come in the mail) and it works perfectly! Thanks again for all of your help rhd.
+rep