Using Diode(s) for Voltage Drop

You can use a single diode.  The voltage drop will vary a bit with current, but it'll start out close the .5 volts with very little current.

If the accessories only draw .5 amps, you don’t need 20 amp diodes!

Just because the power supply can deliver as much as 20 amps does not mean you need 20 amp diodes. The accessory will only draw what it needs to work...in this case .5 amps. You could connect it to a 4.5 volt, 1,000 amp power supply and the accessories will still only draw .5 amps.

2 to 5 amp diodes would be plenty big enough.

Thanks for the responses.

Does your 5V 20A power supply look like this?

This was the first "hit" on Amazon when I googled 5V 20A power supply.  Most, if not all, power supplies of this ilk have an adjustment control that is typically  +/-10% or 4.5V to 5.5V.  So you can just set the power supply to 4.5V DC output rather than fussing with diodes.

@The GN Man posted:

Paul, if you were to push 20A through a diode with 0.5V drop, it’s dissipating 10W!  My point is, silicon diodes are inexpensive and using several will allow you to spread out the dissipation so nothing is too hot (as well as avoid a single point of failure).

You cannot “push” amps through anything. Read my post above. Whatever accessories are connected to the power supply will only draw whatever power (amps) they need to work!!

As I said above, you could connect these accessories to a 1,000 amp power supply and the accessories will still only draw the .5 amps they need to work.

As further illustration of this principle, think of your car. When you open the door, the little dome lights go on. What is supplying the current to those lights? A 12-volt battery capable of delivering 900 to1,000 amps! But the lights only draw what they need to light up. And since they are likely LEDs, they are only drawing a small fraction of one amp.

He is asking if he should use multiple diodes, one on each accessory, or one diode where all the power for all accessories goes through.

John’s reply referenced one diode so I assume it needs to be 20 amps.

Rich’s reply sounds like one diode for each accessory, so the diode only needs to be large enough to cover the amps the accessory needs.

Since no one stated which is better, I assume either way is acceptable?  Seems one 20 amp would be easier (not sure about that heat issue).

Either method will work.  However, if you're truly using 20 amps of current, splitting the diodes into a couple might make sense as the single diode would be dissipating around 12 watts and getting pretty warm.  Obviously, a 20A diode would be up to the challenge or it wouldn't be rated at 20 amps.

Personally, I'd probably use one diode and if the load started to get to the point where the diode was getting hotter than I liked, I'd just split the load between two or more diodes.

Thanks again for all the responses.  I guess either way works well as long as the total current requirements for all the accessories on a diode to not exceed the rated value.  The other factor when splitting or not splitting the accessories is what the power supply is capable of providing current-wise.

Hey Stan2004... Thanks for the info on the power supply from Amazon.  I just ordered on and it should be here tomorrow!  It will solve a number of problems.

I would use individual diodes on each accessory. This would allow you to tweak the voltage for each accessory, if necessary. They do NOT have to be 20 amp diodes.

Not to muddy the water any further, but bear in mind you can also use TWO diodes, hooked up elephant style to each other (a.k.a. in series), and drop 0.5 -0.7 v on each, 1.0 to 1.4 v total for both, reducing the voltage to the accessories to about 4 volts.  Your lights will be a touch dimmer, and perhaps more realistic.

Using two diodes in series is one way to get a constant intensity light using a 1.5 v lamp across two diodes in series.  This was a common practice back in my HO days.

Dale

Do you have a voltmeter? Measure the output of the supply with the load attached, and you may find that you already have 4.5 volts or thereabouts. using the adjustment pot will get you within a tenth of a volt of 4.5. Close enough.