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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

VE.Direct
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Alistair Warburton avatar image Alistair Warburton commented ·

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.

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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.

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clintc avatar image clintc commented ·
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!

Clint

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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.

Al


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Related Resources

VE.Direct protocol FAQ