On Fri, Apr 26, 2024 at 11:23 AM, Steve T

<srtimm@...> wrote:

I just called Renogy and asked if I could run their 20 amp DC to DC charger in half power mode (10 amp), on 24 ft long 10 awg input wires. ?The guy said yes. ?I also asked what happens if there's too much voltage drop on the input wires due to small wire sizes, and would the charger catch fire or something bad like that. ?He said the charger would just shut off.

In the past when this question was asked the renogy rep said you can't do that, or maybe it wasn't recommended. ?I don't recall the exact wording. ?I was expecting to have to push harder for an answer and maybe get the question referred to the R&D group. ?But nope, the guy just said yes.?

The guy answering my questions was consulting tables for max current on various wire sizes and voltage drop, so he knew a bit about electronics. ?But he didn't consult anyone in the R&D department or already know the answer from prior consultations. ?So I don't know if his answer is completely reliable.

But assuming this answer is valid, that means you can just cut the power wires coming out of the Bargman cable, connect the vehicle side to the inputs to the renogy charger, and the other side to the outputs of the renogy charger, and the problems of battery drain, blown fuse, and alternator damage will not happen. ?You also need to set the jumper wire on the charger to set half power mode.

That's not enough power to run the fridge on DC mode, so you still have to exclude that. ?But at least you don't have to run the extra set of larger gauge wires.

On Sat, Apr 27, 2024 at 7:26 AM, Jean-Francois Larin

<larinjf@...> wrote:

What is that table 'condition of use'?

I mean, normally there are 2 conditions related to wire size. First, there is the maximal ampacity related to the temperature and wire insulation.? If you draw too much current, the wire will become too hot and will melt insulation. That is not related to length and any length that carries too much current will become too hot.

Then there is the voltage drop.? The wire resistance increases with the circuit length and the voltage drop increases with both resistance and current.? So at a specified current, AND FOR A SPECIFIED VOLTAGE DROP you can find a maximal allowed length. From the table capture you provided, I can only guess that the table shows the maximal length for a voltage drop percentage. What would that number be?

In a normal circuit operation, you'll want to limit the voltage drop because the load might not operate correctly if voltage is too low. It may draw too much current, which may overheat internal component, or not operate all all, etc. In a DC-DC converter application however, the circuit is made to be more tolerant. For example if we look at the Renogy DCC1212-xx spec:

It is made to operate down to 8V.? That is 33% voltage drop on a 12V circuit, so much more that usual 3% or 5% drop. So to find if a specific wire size is good, you'd have to take the minimal source voltage from the TV, add all the connection resistance to the destination (I agree that's difficult to find and it changes depending on how well the connector is inserted...), then add the wire resistance for the wire diameter to test and the wire length of the installation? (yes counting both 24 ft from source to load and 24 ft to return the current from load to source) and compute what will be the voltage at the load (the DC-DC converter) under that current.? If that voltage is above 8V, the wire will be good.

?

Of course, the bigger the wire, less drop in voltage there will be.? And since it is a DC-DC converter load, lower is the voltage input, more current will be drawn, so it helps in both ways.

?

That is not as the first ampacity criteria that you must not overheat the wire.? Going over the heat rating is big no-no.

jf

On Sat, Apr 27, 2024 at 7:54 AM, dan smartt via groups.io

<dansm@...> wrote:

Voltage drop of 3% (which is standard allowance in design of circuits)?

Sorry should of included that , hope that clarifies that for you

Happy camping

Dan S

On Sat, Apr 27, 2024 at 7:26 AM, Jean-Francois Larin

<larinjf@...> wrote:

What is that table 'condition of use'?

I mean, normally there are 2 conditions related to wire size. First, there is the maximal ampacity related to the temperature and wire insulation.? If you draw too much current, the wire will become too hot and will melt insulation. That is not related to length and any length that carries too much current will become too hot.

Then there is the voltage drop.? The wire resistance increases with the circuit length and the voltage drop increases with both resistance and current.? So at a specified current, AND FOR A SPECIFIED VOLTAGE DROP you can find a maximal allowed length. From the table capture you provided, I can only guess that the table shows the maximal length for a voltage drop percentage. What would that number be?

In a normal circuit operation, you'll want to limit the voltage drop because the load might not operate correctly if voltage is too low. It may draw too much current, which may overheat internal component, or not operate all all, etc. In a DC-DC converter application however, the circuit is made to be more tolerant. For example if we look at the Renogy DCC1212-xx spec:

It is made to operate down to 8V.? That is 33% voltage drop on a 12V circuit, so much more that usual 3% or 5% drop. So to find if a specific wire size is good, you'd have to take the minimal source voltage from the TV, add all the connection resistance to the destination (I agree that's difficult to find and it changes depending on how well the connector is inserted...), then add the wire resistance for the wire diameter to test and the wire length of the installation? (yes counting both 24 ft from source to load and 24 ft to return the current from load to source) and compute what will be the voltage at the load (the DC-DC converter) under that current.? If that voltage is above 8V, the wire will be good.

?

Of course, the bigger the wire, less drop in voltage there will be.? And since it is a DC-DC converter load, lower is the voltage input, more current will be drawn, so it helps in both ways.

?

That is not as the first ampacity criteria that you must not overheat the wire.? Going over the heat rating is big no-no.

jf

Is your 24ft total length of 10 gauge wire or is it 2 runs of 24ft,? 24 for red & 24 for the ground?

?

?If that is the case then your total run is 48ft.

?

Here is a table for voltage drops for automotive DC circuits.

?

If it's a total of 48' you might be on the small size for 10 amps?

?

?

On Fri, Apr 26, 2024 at 11:23 AM, Steve T

<srtimm@...> wrote:

I just called Renogy and asked if I could run their 20 amp DC to DC charger in half power mode (10 amp), on 24 ft long 10 awg input wires. ?The guy said yes. ?I also asked what happens if there's too much voltage drop on the input wires due to small wire sizes, and would the charger catch fire or something bad like that. ?He said the charger would just shut off.

In the past when this question was asked the renogy rep said you can't do that, or maybe it wasn't recommended. ?I don't recall the exact wording. ?I was expecting to have to push harder for an answer and maybe get the question referred to the R&D group. ?But nope, the guy just said yes.?

The guy answering my questions was consulting tables for max current on various wire sizes and voltage drop, so he knew a bit about electronics. ?But he didn't consult anyone in the R&D department or already know the answer from prior consultations. ?So I don't know if his answer is completely reliable.

But assuming this answer is valid, that means you can just cut the power wires coming out of the Bargman cable, connect the vehicle side to the inputs to the renogy charger, and the other side to the outputs of the renogy charger, and the problems of battery drain, blown fuse, and alternator damage will not happen. ?You also need to set the jumper wire on the charger to set half power mode.

That's not enough power to run the fridge on DC mode, so you still have to exclude that. ?But at least you don't have to run the extra set of larger gauge wires.

So, taking the number provided:

48 feet round-trip, 120W load, AWG 10, 12V source input (at up to 15A) and I put that at a first-found voltage drop calculator:

I get the following results:

Voltage Drop

?

12.33%

?

?

Voltage drop will be limited to 12.33% or less with a 10 AWG copper conductor delivering a total of 15 amps on a 12 volt single phase system for a distance of 48 feet.

Engineering Information

30

Amps Rated ampacity of selected conductor

1.1417

Ohms Resistance (Ohms per 1000 feet)

If we take the source 12V, remove 12.33%, we get a voltage at destination of at least 10.52V, which is within the operating voltage of the load.

?

Another consideration is the wasted energy along the wire. That would go up to 12.33W (along the wire). So it would 'work'. Current would be less 15A (since voltage is more than 8V), so voltage drop would be less 12.33%.

?

Still, there is one missing piece: the efficiency. Although the load AFTER the DC-DC converter is 120W, it does not mean that the DC-DC converter is a load of 120W. First, charging voltage is higher than 12V:

Second, the efficiency is UP TO 90%. That is not a spec you can work with!? To do proper engineering computation, we'd need minimal efficiency at maximal load (10A when configured that way).? Then, from voltage and current at that specific point, we get figure and output power, divide it for the efficiency to get the DC-DC converter power input. And redo the computation.

?

But, let's ignore that fact and check the effect of the wire size with a AWG 8.

?

Voltage Drop

?

8.01%

?

?

Voltage drop will be limited to 8.01% or less with a 8 AWG copper conductor delivering a total of 15 amps on a 12 volt single phase system for a distance of 48 feet.

Engineering Information

40

Amps Rated ampacity of selected conductor

0.7421

Ohms Resistance (Ohms per 1000 feet)

12V - 8% gives 11.04V.? Power loss in the wire would be (up to) 8 W.? Of course that would be nicer and safer!
jf

On Sun, Apr 28, 2024 at 10:34 AM, Steve T

<srtimm@...> wrote:

Go to calculator.net and they describe how their calculator works:

Voltage drop calculation

Ohm's Law is a very basic law for calculating voltage drop:

V_drop?= I·R

where:

I: the current through the wire, measured in amperes

R: the resistance of the wires, measured in ohms

?

The resistance of the wires is often measured and given as length-specific resistance, normally in the unit of ohms per kilometer or ohms per 1000 feet. Also, the wire is round-tripped. Therefore, the formula for a single-phase or direct current circuit becomes:

V_drop?= 2·I·R·L

Note the 2 in the formula.? That accounts for the round trip.? So you don't need to plug in 48.? The number you plug in is 24.? Calculator.net gives essentially the same results as southwind, although the number do differ by a few hundredths.

The point if this whole query, though, was not to find out if this satisfied some engineering standard for voltage drop.? It was to see if it worked at all, or if it caught fire or something really bad happened.? I don't care if its not efficient.? If the battery doesn't drain while traveling, and even charges a little, that's good enough.? Some guys have reported their lithium batteries draining at the rate of 7 amps while towing. If the battery just charged at 5 amps, 1/2 the rated output of the Renogy charger in low power mode, that would be seen as problem solved.

The other point of this exercise is I don't want to route brand new thicker wires.? The hassle factor of doing that and the hassle of connecting those wires each time I hitch and unhitch is something I want to avoid.


On Sun, Apr 28, 2024 at 10:01 AM, Jean-Francois Larin wrote:

So, taking the number provided:

48 feet round-trip, 120W load, AWG 10, 12V source input (at up to 15A) and I put that at a first-found voltage drop calculator:

I get the following results:

Voltage Drop

?

12.33%

?

?

Voltage drop will be limited to 12.33% or less with a 10 AWG copper conductor delivering a total of 15 amps on a 12 volt single phase system for a distance of 48 feet.

Engineering Information

30

Amps Rated ampacity of selected conductor

1.1417

Ohms Resistance (Ohms per 1000 feet)

If we take the source 12V, remove 12.33%, we get a voltage at destination of at least 10.52V, which is within the operating voltage of the load.

?

Another consideration is the wasted energy along the wire. That would go up to 12.33W (along the wire). So it would 'work'. Current would be less 15A (since voltage is more than 8V), so voltage drop would be less 12.33%.

?

Still, there is one missing piece: the efficiency. Although the load AFTER the DC-DC converter is 120W, it does not mean that the DC-DC converter is a load of 120W. First, charging voltage is higher than 12V:

Second, the efficiency is UP TO 90%. That is not a spec you can work with!? To do proper engineering computation, we'd need minimal efficiency at maximal load (10A when configured that way).? Then, from voltage and current at that specific point, we get figure and output power, divide it for the efficiency to get the DC-DC converter power input. And redo the computation.

?

But, let's ignore that fact and check the effect of the wire size with a AWG 8.

?

Voltage Drop

?

8.01%

?

?

Voltage drop will be limited to 8.01% or less with a 8 AWG copper conductor delivering a total of 15 amps on a 12 volt single phase system for a distance of 48 feet.

Engineering Information

40

Amps Rated ampacity of selected conductor

0.7421

Ohms Resistance (Ohms per 1000 feet)

12V - 8% gives 11.04V.? Power loss in the wire would be (up to) 8 W.? Of course that would be nicer and safer!
jf

Hi Gang:

Two notes I would like to add:

1) 10GA plus a parallel 12GA wire would be better than just the 12GA alone.

2) Even if the plug/jack has only a smaller wires, the heavier wire on each side still help reduce the voltage drop.

Carl.