Need help in lowering voltage

[notsmoking]

I could use a little help in what would be the easiest method to lower my 5 volt PV to about 4.2 or 4.4 volts. I was thinking a resistor. If that is the right way. Could you tell me about what resistance value and where would i connect it. Thanks for any help you can give me.

 

[Rocketman]

3 amp or higher rated silicon rectifier forward biased will drop about 0.7 volts at 2 amps

 

[TomCatt]

Just out of curiosity, would it be possible to reduce the voltage by 0.2 or 0.3 volts?

 

[notsmoking]

Rocketman.... Will two 1.5 amp silicon rectifiers in series work? Also would I solder them in just after my lm7805
Thanks again

 

[mamu]

As Rocketman said, a diode has a typical forward voltage of 0.7 volts. That's pretty close to the voltage drop you're asking for.

Splice the positive wire of the atty connector and insert the diode there.

Sumfin like this:

I've used these diodes in a few mods and get a 0.6-0.7 voltage drop: DIODE SCHOTTKY 2A 80V DO41

 

[Rocketman]

Parallel them when thinking current. two 1.5 amp diode in parallel is (almost) 3 amps, and good enough.
They will drop the voltage delivered to the atty/carto, lower the current flowing through the whole circuit because of that, and take a little of the "heat" off the regulator.

 

[slimest]

If you want voltage drop about 0.2 - 0.3 - 0.4 volts, you can use Shottky diodes. They have more current density.

 

[CraigHB]

Check the data sheet for the part you want to use. Forward voltage (the voltage a diode drops) with respect to current flow is normally shown on a graph in the data sheet. It can vary quite a bit depending on the diode and current flow. Schottky diodes can be as low as .3 Volts at 2 Amps. Standard diodes can be as high as 1.1V at that current. In any case, it's best to drop voltage with a diode. The drop is less dependant on current flow and a diode is going to be a lot smaller than a resistor that can handle the same power.

It's better not to parallel diodes if possible. What happens is that one ends up carrying the brunt of the load. It's because forward voltage drops as a diode heats up. So the first one to heat up carry's more of the current since it's the path of least resistance. This is also true for MOSFETS and becomes a real problem for ultra-high current applications like electric vehicles. Paralleling them is tricky. You have to make sure they are perfectly matched and all heat up at the same rate.

 

[notsmoking]

thanks everyone... that helps me understand more