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How much Amperage is compromised by a 4mV drop.

Bhupinder

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Nov 19, 2019
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Hello there,
need some advise.

Am using 6mm sq copper single core wires to connect 24V Solar panels of ~300W to SCC placed inhouse.
Since install distance is 20 meters and there is room for future expansions(30A), 6mm was kind of safe to go with.

Later, I planned to place relays on the path to solar panels which could be managed via a separate circuitry.
When checking Voltage Drop across the each wire with 1Amp current input, following readings were noted:
-> Direct Connection - voltage drop of 0.108v (20 meters length)
-> Routed via Relays - voltage drop of 0.112v (20 meters length)

Curious to know from Electrical engineers - How much Amperage capacity is being degraded by adding the relay
which is inducing a 4mV voltage drop?
Because if that's significant - shall put a single big relay or change relay connections with thicker wires.

Thanks for reading through. Please post your valuable comments.
Regards
 
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you just multiply the voltage by the Amperage and you get the Wattage.
so a relay that loose 4mV under 30A is heating about 0.12W, that is actually nothing.
so you loose a lot more in the 20m cabling that is 100mV, under 30A it is 3W
 
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you just multiply the voltage by the Amperage and you get the Wattage.
so a relay that loose 4mV under 30A is heating about 0.12W, that is actually nothing.
so you loose a lot more in the 20m cabling that is 100mV, under 30A it is 3W

Many thanks nosys70!
So, 0.112V is the loss across cable of length 20M X 2 = 6.72W
+
6 Watts for driving the Relay
Total~12W.
That's fine and kind of acceptable for benefit it provides.

However, one question remains - these calculations are based on losses calculated with 12V & 1Amp applied across ends of a single cable(20M).
Can these be safely projected in a linear upscale fashion to calculate losses @ 32V & 30A?
Wouldn't this 4mV loss out of relay degrade overall current carrying capacity of the conductor? and by how much.
(Meant to ask if under these slightly degraded conditions, would conductor behave like a 5sq mm one?
Is there a way to quantize Amperage degradation?)

Regards
 
resistance is not supposed to change a lot in normal condition.
if you stress a wire enough (with heat for example) you could get drastic changes.
the problem with relay is the contact can vary a lot with time.
on the long range (months) you can get oxydation of contacts and in the short range (milisecond) you get the spark problem.
these two phenomenon reinforce each other, that is why relays wear.
usually high current relay use contact under neutral gaz so spark is limited.
nowaday we use transistors.
 
SSR are in fact transistor relay.
if you only use the relay unfrequently, the good old mechanical contact should be ok.
but if you often use it (several time a day for example, it could be good to consider another solution).
 
if you only use the relay unfrequently, the good old mechanical contact should be ok.
but if you often use it (several time a day for example, it could be good to consider another solution).
Idea is to isolate panels when not in use.
Connect them to SCC only during day time to save on accidental surges/ lightening situations.
Additionally the SCC I have is prone to blow fuse when accidentally Panels are connected without connecting the Battery first.

Hence Driving relay with battery would ensure both requirements are met.


SSR are in fact transistor relay.
Would SSR be a better bet or some DIY mosfet based switch (Multiple power mosfets put in parallel)?
 
" Idea is to isolate panels when not in use. "
if you use the relay to switch of panels when light is low enough, then there should be no power to switch, then no problem.
but frankly most of people just keep their panel connected all the time, so it should be ok for you too.
i am not sure a relay would be enough to protect against a lightning.
in that case you can build et manual switch with a long copper busbar clamped.
when there is a lightning risk, you just remove the bar or place it in a grounded position
there are also protection (lightning tubes) that can be mounted permanently in the circuit.
you can probably use those form an old telephone line or build your own.
Basically the idea is to make a near contact that is isolating at low voltage and conductive at high voltage.

1596456914873.png
1596456967366.png
 
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Thanks for your valuable suggestions Nosys70.
Another reason for the Relay schematic was that my usage would be mostly mobile, frequently installations and packing things again means there could be chances of overriding precautions especially in sequence of connecting Battery and SCC - it has happened before which got the SCC roaster ( Had to replace) .

Also while we are at it, am facing a peculiar situation where in am trying to figure how to connect SCC output to the battery. I have a separate port BMS, which requires additional charging source(C-) other than the battery((P+) and drain terminals(P-).
1596474790353.png

In case of common port one may directly attach SCC output to the common - Drain/Charging port(P-,P+) - this way SCC receives battery power to operate its functions and can dump excess charge to the battery when it has it.
1596474826813.png
But for the separate port BMS, how is made to work with SCC? Cause in this case when SCC does not receive reverse charge from the charging port - it would amount to "No-Battery Connected" and when Panels are plugged-in -> its default behavior would be to burn internal fuses.

Have you or someone faced such a situation?
Well I was thinking of attaching a secondary battery for this purpose, have a spare one and should increase total backup capacity - but then its going go complicate things for no big reason.
 
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Thanks @BiduleOhm.
Looks like we need your BMS to be out sooner than planned:LOL:.

@BiduleOhm : Could you share your opinion on what would be better suited to isolate PV Panels in the Night time?
1) Mechanical/Automotive type Relays
2) Mosfet based Switch (Combining multiple mosfets in parallel) Do you have a DIY circuit for this as you generally have for other things.

Requirement is to ensure Battery is in place before PV Panels + Lightening protection (Nosys70 has suggested to use "lightening protection tubes" or use a mechanical contact point for this use case).
 
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20 meters of 6 sq mm of copper (between #9 and #10 gauge solid) with 1 amp should have 0.06 ohms with 0.06v drop
20 meters of 6 mm X 6 mm square of copper (36 mm sq, bit larger then #2 gauge solid) with 1 amp should have 0.01 ohms with 0.01v drop.

Should measure wire with four point method, Force known current at ends of wire and measure inside ends directly on wire with voltmeter contacting wires at specific distance separation. Do not measure voltage drop at current connection terminals as measurement will be impacted by current contact point resistance. This is exactly the same way a current shunt works.

Relay contacts have resistance (that will likely degrade over use life). Not specified as voltage drop unless stated at a particular current.
 
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20 meters of 6 sq mm of copper (between #9 and #10 gauge solid) with 1 amp should have 0.06 ohms with 0.06v drop
20 meters of 6 mm X 6 mm square of copper (36 mm sq, bit larger then #2 gauge solid) with 1 amp should have 0.01 ohms with 0.01v drop.

Should measure wire with four point method, Force known current at ends of wire and measure inside ends directly on wire with voltmeter contacting wires at specific distance separation. Do not measure voltage drop at current connection terminals as measurement will be impacted by current contact point resistance. This is exactly the same way a current shunt works.

Relay contacts have resistance (that will likely degrade over use life). Not specified as voltage drop unless stated at a particular current.

Thats insifghtful @RCinFLA .
Could you please clarify on "measure inside ends directly on wire with voltmeter contacting wires at specific distance separation"?
What you say is the correct way and probably how its done in lab experiments with bare wires.
But this one is a 20 meters insulated conductor - how do we measure voltage/current at different distance points with insulation?
 
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Thanks @BiduleOhm.
Looks like we need your BMS to be out sooner than planned:LOL:.

@BiduleOhm : Could you share your opinion on what would be better suited to isolate PV Panels in the Night time?
1) Mechanical/Automotive type Relays
2) Mosfet based Switch (Combining multiple mosfets in parallel) Do you have a DIY circuit for this as you generally have for other things.

Requirement is to ensure Battery is in place before PV Panels + Lightening protection (Nosys70 has suggested to use "lightening protection tubes" or use a mechanical contact point for this use case).

I go as fast as I can ^^

It depends on a lot of factors but if protection against lightning is the priority then a physical contact is the best solution.

Yea, spark gaps are one of the best solution but they allow quite a high voltage to develop before they are active and are best used with some other tech at the same time like MOVs for example.

I have some ideas for a DIY lightning protection device which involves bare wire inside a grounded copper tube on some part of it's length (total wire length of maybe 1 m, copper tube maybe 10 cm) and the whole thing inside sand to absorb the energy (basically it's a giant fuse with a spark gap at one end but cost only a few dozens of $ to make), then breakers plus of course some classic arrestors after that, and then a bunch of MOVs. And even with all that you can't be sure to stop lightning... Distance and grounding is your best bet with lightning.
 
You basically want a SSR. What voltage will be the PV string at most?
Open Circuit Voltage would be ~40V and under load conditions it would be ~32V.
Don't SSR waste a lot of power? I mean aren't mosfets more efficient?
 
SSRs use mosfets (or something else for the AC ones, but we don't care about those ones here) :)

Ok, so you can buy a SSR (must be a DC one, at least 50-60 V and at least 35-40 A) or build one.
 
Thats insifghtful @RCinFLA .
Could you please clarify on "measure inside ends directly on wire with voltmeter contacting wires at specific distance separation"?
Picture worth 1000 words Set power supply with desired current limit. Alligator clips on ends. If you don't have an adjustable current limiting power supply then use a voltage power supply with resistor in series of few hundred times higher then expected wire resistance. Just be aware of resistor power dissipation based on I^2 x R

Four point measurement.png
 
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Okey, I think that's exactly how the Initial Internal Resistance readings were made.
Current limited to 1A@ 12V was induced through both ends of 20M 6mm sq cable and
Voltage difference between the same two ends was noted with a Voltmeter to be showing as 0.108 Volts.

Since current was 1 Amp so "Voltage Drop" equals Resistance hence 0.108 Ohm.
Test was repeated 2/3 times and reading was consistent and as expected with the relay put in series it increased by 4mV to 0.112Volts.

But then question arises - why did the voltmeter showed 108mV when it should be something in the range of 6mV??
Could it has to do with the Voltage across wire ends -> 12V - I think "YES".
May be we need to bring down voltage to 1v for the power equation to be symmetrically balanced or
normalize the 12V results with a factor of 12 = 108/12 ~9mV kind of close to standard 6.

Not sure, not an expert in electrical derivations - may be you could suggest what went wrong???
 
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