# 12v/54v Dual Voltage Question

So I have a Peplink router I’m building a custom case for. Here’s my dilemma, I have one power cord and one switch and two separate power sources. When the unit runs on mains power it will use the units supplied power supply which outputs 54v @ 3.34amps which will allow the Peplink to provide PoE. When no mains power is available, it will run on cinema batteries which output 14.4v. So I need to work out a way to run both voltages through one power cord.

My original thought was to use schottkey diodes to prevent reverse voltage back into the power supply and battery. I just want to make sure this is a good option or if there is a better way.

Is there a better way to do this or is my method the most simple?

See this recent thread:

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I do not…I just wanted to prevent voltage from being pushed back into the battery or power supply.

Ill check out that post.

You’re spot on, it’s exactly what I need. A quick google search seems that most turn-key solutions are either way under my needs or way over. Given my lack on knowledge on these, what would you recommend?

To ensure we understand what you want to do:

• When on mains the internal DC supply provides 54V [very weird voltage] to the internal board.
• When on battery you want to use 14.4V to power the internal board?
• How exactly do you plan to wire this?
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Yes, it is a very weird voltage, I’ll attach a photo of the power supply.

So yes, on mains AC power runs through the DC power supply to the unit.

When on battery, the unit will receive 14.4v.

All power will run through one power cord, and one switch.

In terms of wiring, I’ll come up with a plan once I know the package of the diode you suggest. If I need to build a small PCB and surface mount it I can do that.

What I am puzzled by is what voltage the circuitry on the router needs, 54V?
If so how do you plan to power it with 14.4V battery?

It’s the Power Over Ethernet that requires the 54v. I think the minimum for the PoE to be enabled is 49 or 50 volts.

The battery’s are standard gold mount cinema batteries. They have a specific mount with a built in fuse and output power leads. The batteries themselves have built in BMS as well.

EEE 802.3at (PoE+): is 50-57V

What I do not get is how you are going to get to 50+ volts from 14.4V when the battery is powering the system.

I am assuming you want to connect the two DC outputs together and then to the router; i.e. the output of the brick (57V) and the output of the batteries (14.4V) together through a blocking diode.

Ahh…okay, now I understand your question. The 50+ volts will only be available when powered by mains AC power. When powered by battery, the unit will function normally but without PoE. The power sources are a this-or-that scenario, there SHOULD NEVER be a time when battery and AC will be powered on at the same time, so the blocking diode is simply a safety measure to prevent anything from happening IF someone was to have a battery and AC power on at the same time.

Does that clarify things a bit?

I know I should have figured it out sooner when “POE” was mentioned, but I still made it to the end of this thread before realizing that we were talking about an internet router, not the spinny wood carvey kind.

When plugged into mains power what voltage is the router receiving and from where?

The router receives 54volts from the external power supply, which has been installed inside the case and wired to an exterior facing power entry module.

A lot … thanks.

When the battery is the source there are 14.4 volts at the point where normally there are 50+ volts. The router can run at a lower voltage?

What is the lowest voltage the router will run at during battery operation?

Pick a diode whose forward voltage drop subtracted from the terminal voltage will keep you above that level.

12v is the minimum required.

How much heat do you think these diodes will dissipate?
Is there any disadvantage to going with a 100v / 5amp diode vs a 60v / 5amp?

If you use schottky diodes, they have a typical Vf of 0.2V. They basically act like whatever resistor would give you their Vf so you can use Ohm’s law to calculate power dissipation for each of the power sources.

If you want, you could use a schottky for the mains and an ideal diode for the battery, because the RDSon for the MOSFET in the ideal diode will have lower resistive losses than the schottky. You might choose this to preserve battery life and optimize for cost.

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Your explanation of what you want to do is a bit confusing. But I think you’re trying to provide POE power as well as charge your batteries to provide power to another circuit. A buck converter will drop your 54V to the 12V for your batteries via BMS while also providing power for your other circuit. Hence, one power source providing power to two circuits. The only thing is if your source can provide enough current.

This is a router which provides PoE if the supply voltage is 54V, but will run an a router without providing PoE if the supply voltage is 12V.

@Dustin_B wants to provide 54V in normal operation, but seamlessly fall back to 12V from battery during outages, without exploding his 12V nominal battery by charging it from the 54V mains-derived source.

This is a classic hot fail situation, though with a much larger disparity than usual.

Schottky diodes have been the classic answer, or active switching with discrete components, or more recently active switching with integrated “ideal diodes”

So basically any PoE device is dropped from the network if main power goes out. I would think then he builds a 48v power pack, and nothing goes down.

PoE is a bonus and would only be in play when the production crew can guarantee uninterrupted power.

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Redundancy and keeping all things running has been hardwired into my brain throughout my career as an IT person.