How do I select the proper diode for my solar panel battery charger?

I have a solar panel that puts out 3V at 40 mA. I have seen the Instructables on the site and read them all, and while they explain why you need a diode, they do not tell how they decided on which one to use. What goes into this? I would like to know how to calculate these things for myself.

110100101108 years ago
the diode should stand the voltage and current most diodes stand well 3 V 1N4001 diode can stand up to 1 A (way more than 40 mA) diode dumps 0.7 V and you still have to have atleast some 1/2 V spare (so that panel - diode > battery + 1/2 V). is your battery lower than 1.8 V ? if no then you need special diode or use no diode at all (you risk discharging the batteries into the solar panel and wasting their charge)
phalanx447 (author)  110100101108 years ago
So when I read a diode's voltage reading, is that the maximum value, or is it +/- 5% of that voltage? Is that 0.7V drop standard for all diodes? Because the game plan was to use one panel to charge two 1.2V rechargeable batteries, rather than have a 1 to 1 ratio with panels and batteries. Because if that 0.7V drop is standard, I need to thing of a way around that. Where is that .5V coming from? I'm trying to wrap my head around this math and where the numbers are coming from- I can't find anything about this in my circuits textbook. So if each 1N4001 diode will only be able to handle a max of 1.8V, what if I configured my circuit like this? This way, each diode is only dealing with 1.2V rather than one diode with 2.4V. What do you say?
you need only 1 diode

the voltage the diode takes depends on the materials it is built of. silicon diodes (by far most common) like 1N4001 take 0.7 V. there are others that take 0.2 - 0.3 V

with 0.2 V diode you get 3 V - 2.4 V - 0.2 V = 0.4 V. a bit under the 0.5 but still will work

if you charge the batteries one on each link you have much more space to play (3 V - 1.2 V = 1.8 V. you can use 0.7 V diode and you still have more than a volt spare - 2 X the 0.5 V wish. thats good) but less efficient when you charge more than 1 battery

the circuit has resistance of few ohms at least (often way more) - most of it is the solar panel. to get decent current to charge the batteries you want some spare voltage (think of it ike energy) to push the current thru the resistance. i guess 0.5 V is a good compromise between charge current (we want high) and how much voltage we can sacrifice (we dont have much anyway). if you can add panels to get more volts then this voltage can be raised and itll be better
kostya8 years ago
Try a germanic diode. It has a voltage drop of 0.2-0.3V.