clintc avatar image
clintc asked

VE Direct common negative with battery?

I have a 100/50 SmartSolar MPPT charge controller that I am using to charge 6 AGM's to power a small 240v inverter.

This model has VE Direct (no separate load outputs).

I intend to use the VE Direct to drive a logic level MOSFET or transistor, which will then actuate the solenoid in a relay that will be able to cut power to the inverter if the battery voltage drops too low.

To do this, I will need to common the VE Direct negative lead to the system battery negative line, in order that my transistor has the correct voltage reference wrt system ground and the VE Direct power pin (+5V).

Are the VE Direct negative pin and the battery negative terminal isolated from each other inside the controller? If so, will commoning them cause issues / be dangerous?

Thanks, Clint

2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

VE.Direct is a serial connection I don't think you can drive a transistor, to switch a relay, with that directly, whatever the ground arrangement.
That said I may be missing something and will keep an eye on this thread out of curiosity.

You might be better buying a relay/controller designed for that, Victron have a
'Battery Protect' product, several in fact, which do exactly what you want.

1 Like 1 ·
clintc avatar image clintc Alistair Warburton ·
The documentation for my 100/50 says it can be used as a "virtual load output" (that obviously doesn't then utilise any communication protocols). The documentation for the VE Direct TX digital output cable says that it can be used to direct drive SSRs; I bought a chinese SSR which was much cheaper than the Omron one recommended, but this required more than the 10mA that the VE Direct port could supply. Then I realised that I had enough transistors/mosfets lying around to build my own solution, the only functional difference being that in SSRs, the inputs are completely isolated from the outputs; using a transistor, they are not.

But I have proven that it's referenced to battery ground anyway so it should work ok in this regard, unless you can think of some other reason (and I've probably overlooked something!).

Thanks for your help though, much appreciated.


0 Likes 0 ·
That is interesting, glad I commented as it is always good to learn something. How are the 'outputs' controlled?

Perhaps it is only for modles that dont have a dedicated load output. it would be pretty cool if all the VE.Direct ports on a Cerbo GX ould be used as outputs.

(I am going to have to look now but I will be gobsmacked if they can.)

Cheap SSR's, not sure I would be trusting those long tern. If the gate shoves the TX pin high due to a fault you could do serious dammage. Have you considdered driving your relays with an opto isolator?
You should be able to find an opto FET with a low current driver LED that will handle enough DC to drive any SSR, no matter the quality, just make sure you calculate the current limit resistor value in the input circuit carefully.

0 Likes 0 ·
clintc avatar image clintc Alistair Warburton ·
Outputs are controlled via the Victron app, but it's only a logic level 5v on/off for the "virtual load output option".

An opto-isolator is a good idea, I'll look into that.

I've just done some more testing, including shorting the +ve and -ve pins on the output and could only get a maximum of 2.6mA! So it can't even drive the SSR they recommend (which requires 7mA).

0 Likes 0 ·

Have a look for a line driver IC, or a totem pole transistoe driver circuit.

If uou use a ULN2003a, a 7 channel NPN darlinton array you can drive up to 100mA/channel and paralell channels if necessarry.


Using a couple of BJT's NPN/PNP in a totempole arrangement will produce a low impedence output that follows your high imedance voltage from the port. A single vurrent limit resistor to the commoned base pins is the only other component you need. 2N3904 / 2N3906 are a good general purpos complimentarry pare and available - almost free!

You could also use logic level shifters... Many are a clever arrangement of two FET's and can support significant voltage and current on the higher voltage side allowing them to drive relays directly.

Explanation here.

0 Likes 0 ·
clintc avatar image clintc Alistair Warburton ·
The simplest solution IMO (and that I've decided to go with) is to use an IRLZ44 logic level power mosfet. I'll put a diode on the gate to protect the VE direct output... That should be it.

All interesting info you've provided though, thank you.


0 Likes 0 ·
2 Answers
wkirby avatar image
wkirby answered ·

Yes, the VE.Direct negative is referenced to the battery negative.
Driving a MOSFET gate should be fine. The VE.Direct Tx line is pretty high impedance, so switching time may be slow as the gate charges up, maybe a few mS waiting time.

1 comment
2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.

I was such an idiot, I had tested it with a multimeter and already proven that it was referenced to battery ground (by checking voltage from the VE Direct +ve to battery negative, and it was stable), so in my arrangement I don't actually need to connect the VE Direct -ve pin to anything.

But thanks for your help confirming that anyway!


0 Likes 0 ·
Alistair Warburton avatar image
Alistair Warburton answered ·

That may not work well, it depends on the way the output pin is driven, if the pin can both source and sink you are good.

If the output can only source, or you add a diode to protect it so it can only source then you have a problem to address...

A FET gate is essentially a capacitor, you charge it up to switch the thing on at which point almost no current flows into it, practically, 0 for circuit design and conceptual purposes.

The issue is that, unlike a transistor which is a current input for want of a better term, a FET needs to be discharged, if you simply stop driving the gate it will stay on, until it discharges through parasitic coupling, if it turns off at all. Additionally when the gate is partially charged the FET is partially on and will have a significant RDS, causing it to dissipate heat across its channel.

If you try to drive the FET directly you should add a current limit resistor to ensure that the output stays within its spec. You will also need a drain resistor, gate to source, so that the FET can discharge.

to calculate to limit resistor subtract Vf of the diode from V Out and pic a resistor value that gives you a shade under I Out (Max).

Now pic the smallest, drain resistor you can that gives you a VGS about 10% above VGS (On).

You are creating a voltage divider and connecting it mid point to the gate of the FET
The resistor to ground will end up being far larger than the one above to the output, to get a high enough VGS, which will impact turn off time, significantly probably.

The end game is to charge and discharge the gate as quickly as possible, without consuming more current than the output pin can supply. charging enough to achieve the specified RDS (On) but no more is the key.

Depending on the specific FET you may find that the current you have to work with is too small to achieve a reasonable switching time if you only use a passive driver.

Is your FET N Channel, it will need to be so that the Source can be grounded as the output is relative to that and not B+ (The configuration is called a low side switch) the relay coil is between B+ and the FET Drain. Add a freewheel diode on the relay for good measure.

Hope that helps.


2 |3000

Up to 8 attachments (including images) can be used with a maximum of 190.8 MiB each and 286.6 MiB total.