Diode circuit isolation?

[iduff]

Hey, Cruisers, any electrical geniuses on here willing to give me a steer?

I’m trying to power one 12v 10A load from either one of two 12v 15A sources on the boat, without back-feeding the other loads on that source when that second source is powered down.

Specifically, I’d like to be able to power up either the masthead tricolor sailing navigation light circuit, or the deck-level under power navigation lights circuit, and use the power from whichever one is energized to power some instrument lights.

Turn on the masthead tricolor sailing navigation light, and the sailing instrument lights and compass light come on, but the deck-level under power navigation lights don’t get back-fed.

Turn on the deck-level under power navigation lights, and the sailing instrument lights and compass light come on, but the masthead tricolor sailing navigation light doesn’t get back-fed.

Pretty sure I need some sort of diode between each source and the loads, but the devil is in the details, dontchaknow. Zener, Schottky, rectifier, and what spec? I’ve fired up the EE degree I got from the University of Google, but can’t come up with the answer.

Anyone on here that could give me a steer? Is a diode of some sort what I need? If so, what kind, and what spec?

[smac999]

10a is a big load for normal diodes.

I’m gurssing that’s the breaker size and not the inst light draw

[Bycrick]

Almost all diodes will work as a “rectifier,” I.e. conduct current in only one direction. You can buy up to about 6 amp axial-lead power diodes usually for less than $1. If you really need to draw 10 amps, you’ll need a higher current rated diode, which will require some kind of heat sink. But they’re readily available for a couple of dollars.

All diodes have a forward voltage drop when they are conducting. Typical power diodes will drop about 0.7 volts. Schottky diodes have a lower forward voltage drop: 0.15-0.45 volts. So you drop less voltage and need less heat sinking.

The new kid on the block is often called an “ideal diode” which has even less forward drop. It’s not really a “diode,” but rather a very-low-resistance FET with built-in driver circuitry. I just saw that www.mpja.com had these on sale for under $5. Rated at 10a with a maximum of 15v. Voltage drop at 8 amps is 0.036 volts.

[moctrams]

This how I fed my nav lights from one breaker. I'm sure you can wire up something similar.

[Cat36Mahalo]

Look at yacht devices smart relay. Turning it on twice in 2 seconds activates second circuit, no parasitic draw.

[iduff]

Originally Posted by smac999

10a is a big load for normal diodes.

I’m gurssing that’s the breaker size and not the inst light draw

Right you are. I'd have to put a meter on the actual light loads, but guessing they are indeed quite small.

[iduff]

Originally Posted by Cat36Mahalo

Look at yacht devices smart relay. Turning it on twice in 2 seconds activates second circuit, no parasitic draw.

That's pretty cool, but the inverse of what I'm seeking. I have one load, two switches.

[Paul Elliott]

Something like this?

The switches are either plain switches or your circuit breakers. The diodes might be something like these: https://www.digikey.com/product-deta...3TR-ND/5965303 (Schottky diodes, 20A, 45V). It would be good if the diodes have a higher current-ration than your breakers. That way a short-circuit pops the breaker and probably not the diode. Everything shares a common ground (-) connection.

One problem with this is that you will see a few tents of a volt drop in the diodes. This may pose a problem with any electronics you are running off the "Instruments" circuit. In that case, some of the lower voltage-drop devices mentioned above may be more suitable. I have a similar circuit powering my binnacle compass light.

[iduff]

Originally Posted by Bycrick

<<good stuff snipped>>
The new kid on the block is often called an “ideal diode” which has even less forward drop. It’s not really a “diode,” but rather a very-low-resistance FET with built-in driver circuitry. I just saw that www.mpja.com had these on sale for under $5. Rated at 10a with a maximum of 15v. Voltage drop at 8 amps is 0.036 volts.

Just found this on their site, bought three, thanks!

[AndyEss]

You can also wire multiple diodes in parallel to increase current capacity.
Marlin P. Jones is a great source, but they are jobbers, and some of their products are over runs, out of spec (below) and not infrequently poorly documented.
Order extra pieces.

[iduff]

Here's a quick picture of what I'm trying to do. Hopefully I've figured out the Cruiser's Forum file attachment bit.

[iduff]

Originally Posted by Paul Elliott

Something like this?

The switches are either plain switches or your circuit breakers. The diodes might be something like these: https://www.digikey.com/product-deta...3TR-ND/5965303 (Schottky diodes, 20A, 45V). It would be good if the diodes have a higher current-ration than your breakers. That way a short-circuit pops the breaker and probably not the diode. Everything shares a common ground (-) connection.

One problem with this is that you will see a few tents of a volt drop in the diodes. This may pose a problem with any electronics you are running off the "Instruments" circuit. In that case, some of the lower voltage-drop devices mentioned above may be more suitable. I have a similar circuit powering my binnacle compass light.

Bingo! Thanks for the steer to a possibility on the Digi-key site, I just bought some.

I'm only powering the instrument lights, not the instruments themselves, so I think any variance in volts might be OK. Hopefully.

[mbartosch]

Originally Posted by iduff

Bingo! Thanks for the steer to a possibility on the Digi-key site, I just bought some.

I'm only powering the instrument lights, not the instruments themselves, so I think any variance in volts might be OK. Hopefully.

The diagram Paul Elliott posted is spot on. I am using Schottky diodes myself on the boat for the same purpose a lot, as they are essential for such applications.

However, there is a caveat.

Note that the maximum current rating does not imply that this current may be sustained. In fact, higher rated diodes may be able to withstand the rated current only briefly unless they are properly cooled.

Due to and depending on the voltage drop on the diode some electrical power will be dissipated as heat. The generated heat leaves the diode via via convection and heat radiation, resulting in an equilibrium temperature. This equilibrium temperature must be lower than the rated junction temperature of the diode (150 °C), otherwise the diode will fail. It is advisable to keep the temperature significantly lower for longevity and other obvious reasons.

This means: either limit the current through the diode to sensible values or add cooling to the diode. For high current application every millivolt less voltage drop of the diode helps reducing heat dissipation.

You can calculate the power dissipation by multiplying the voltage drop of the diode (the referenced Schottky diode has a voltage drop of max 0.58 V) with the current flowing through the diode. E. g. if you instrument panel lighting uses 2 W of power then the current is 2 W / 12 V = (roughly) 0.17 A.

0.17 A * 0.58 V (voltage drop) = roughly 100 mW

100 mW heat dissipation in a diode is usually not a problem, you should barely be able to feel that the diode gets warmer.

Just don't expect a diode to properly dissipate e. g. 10 A * 0.58 V = 5.8 W without dealing with the excess heat via a cooling element.

[AndyEss]

Yep, diodes become somewhat expensive fuses if too much current flows through them. If you have an accidental short, they will often pop faster than a fuse or breaker.
Carry spares and look into fast blow fuses to protect the diodes.
As in my first post, multiple diodes in parallel greatly decrease the current flowing across any one diode. Two diodes in parallel carry 1/2 the current, etc.

[AKA-None]

Originally Posted by AndyEss

Yep, diodes become somewhat expensive fuses if too much current flows through them. If you have an accidental short, they will often pop faster than a fuse or breaker.
Carry spares and look into fast blow fuses to protect the diodes.
As in my first post, multiple diodes in parallel greatly decrease the current flowing across any one diode. Two diodes in parallel carry 1/2 the current, etc.

If they are matched diodes then yes. Unmatched then no as one will draw more than the other. You can buy matched diodes.