Amps Draw and Voltage Drop Question 2

[BFL13]

Question 2 -17 Oct
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There are no dumb questions--until now!

If the furnace draws 8 amps and there is a 0.2v drop when it is on, and you change the positive wire to the furnace to a fatter wire, will the furnace draw fewer amps, or will it still draw 8 amps but have less voltage drop at the battery bank?

[Gdetrailer]

BFL13 wrote:


DC-DC in thread about that last Spring it came out that if you used thin wire the draw on the alternator would be 30 amps for a 20 amp output, but if you used fat wire, that draw was much less, but still above 20 of course.

OK, got it!

Yes, feeding a DC to DC charger with a wire ga of sufficient size that will have much less voltage loss because the resistance of the wire is much lower.

Yes, the draw will be more than 20A with a larger wire to a DC-DC charger but will be less than if you feed it with a small wire ga that has higher resistance..

The key here is with high amperage 12V items, use as large of wire and as short of a run as possible. Yes, 12Ga wire will support 20A, but as the run gets longer the resistance of the wire adds up zapping your voltage at the end of the line.

Wire resistance is a big issue when it comes to high current 12V items. Basically in 12V battery world we are talking a operating voltage range of 1V-2V between fully charged to a dead battery.. That means every tenth of a volt reduction in voltage drop means a lot. Comes into play especially when you are dealing with heavily discharged batteries.

[BFL13]

Tried to clarify the readings, but ran into not being able to get a good reading on the running 8531 unless I did! Can't find a for sure good place to get pos and neg contact points.

Knocked surface charge off the battery bank (four good 6s) and waited a bit, meter on the batteries 12.71v steady and Tri says amps 2.1 before starting the furnace.

Furnace on with heat coming out, where heat adds about 0.5a, now

12.64v meter on batts and 9.3 amps on Tri, so furnace amps 9.3 -2.1 = 7.2

I got 10.67v on the furnace outside using the best reading with one of the positive wires showing and neg to a frame ground BUT not confident I used the proper places to measure. Can't get at where it is "grounded" or where maybe the full 12v goes to. (anybody know how?)

So voltage drop was not that 0.2 I reported as seen while camping but was 12.71 - 12.64 = 0.07 I have no idea where the 0.2 I saw on the Tri came from unless it was rounding funny with its one-decimal place readings.

Anyway, that's what I just got. No plan to change the furnace wiring since it has worked this last four years we have owned this RV and for the 31 years since it was built I guess.
----------------

DC-DC in thread about that last Spring it came out that if you used thin wire the draw on the alternator would be 30 amps for a 20 amp output, but if you used fat wire, that draw was much less, but still above 20 of course.

[Gdetrailer]

BFL13 wrote:
Just pondering all this (to no great effect)

The current to the furnace fan rises with more supply voltage but the current to the input of the DC-DC falls with more supply voltage.

The reason is the furnace fan is not a fixed demand like the DC-DC or inverter's output-- it will take more if you give it more.

The 8531 furnace is rated at 8.2 amps at 12v. So that is the expected "load", but it is not the "Demand". The fan runs faster with more amps if the "Supply" voltage is higher, so the demand is a variable depending on the supply.

The inverter load side is fixed by the load's needed input watts so the "demand" is not a variable. The DC supply tries to meet that demand. As the battery voltage falls the amps drawn go up to maintain the demand watts.

The DC-DC charger has an output set at eg, 20 amps at 14.6 volts, so that is the fixed demand, not a variable. The input supply tries to meet that demand.

However, the input is itself a demand on the engine battery and alternator, which has to meet the input demand for the output demand.

Using fatter wire in the DC-DC input to the battery keeps the amps demand lower with the input voltage higher, so the alternator doesn't need to supply so many amps

The engine battery voltage is somewhat fixed as maintained by the regulator. The supply variable is the alternator amps to meet the demand.

The Dc-Dc charger "size" (output setting demand) has to be chosen, limited to match the alternator "size", where using fatter wire will let you have somewhat more Dc-Dc size for the same alternator.

Things are bit complex.

Inverter can have say a 850W 120V load, but if the 12V input voltage drops the inverter must draw more battery amperage to support the 850W 120V load. The reverse happens when the 12V input is higher, the inverter does not need to draw as much amperage.

In both cases the wattage will be the same..

DC-DC chargers and 12V to 120V inverters employ the same switching power supply technology.

Basic switching power supply consists of a chopper circuit that operates at high frequency just above your hearing range.

The chopper output feeds a small high frequency transformer. The transformer will step up the high frequency chopped voltage to a higher voltage.

The output of the high frequency transformer is rectified and filtered to a higher voltage DC.

For DC to DC charger, it stops there.

For 12 to 120V AC inverter the higher voltage DC is now switched on/off via a PWM signal that mimics 60hz sinewaves.

The key here is switching power supplies have a very large input voltage range..

Old school inverters used a big heavy 60hz transformer which did the voltage boost, not as efficient and less tolerant to 12V input voltage sags.

DC to DC charger will not reduce the amps required from the alternator. Basically a DC to DC charger is what is known as a "Boost Converter".

For some technical reading you can take a look at some basics of Boost converters HERE

From Page 4..

The simplest way to calculate the input current of a boost regulatoris to use the power balance equation, shown in Equation
1. For a DC/DC converter, the input and output powers
are just the product of their respective currents and voltages.
Adding the triangular ripple current, we arrive at Equation 2


This equation highlights the biggest stumbling block
when working with boost converters: the input current will always
be larger than the load current (IOUT). Since the output voltage
of a boost is always greater than the input voltage, the input current
must be greater than the load current. This is a simple consequence of
conservation of energy: the input power will be equal to the output
power plus the losses. In this case the losses are taken care
of by the efficiency factor, ?. Equation 2 also applies to a buck converter. And since the output
voltage of a buck is less than the input voltage, the input current
will be less than the load current, for any reasonable efficiency.

As an example, suppose we wish to convert 6V to 12V (IE BOOST CONVERTER) at a load
current of 2A. If the efficiency is 90%, and the peak-to-peak
ripple current is 30% of the load, then Equation2 gives 4.74A


I couldn't copy and paste the equations but they are there on page 4 of the PDF in the link above if you want to check their math..

In reality, to get 20A on the output of a DC-DC charger, the input current will be much higher than 20A on the alternator side..

Not really exactly sure where you are going with this, saving your alternator or just trying to get you vehicle to charge house batteries a bit faster.. If you are trying to save the alternator with a lower amperage load, that is the wrong assumption..

[Gdetrailer]

MNRon wrote:
Unless I’m missing something, most of responses are missing the point of the issue (as the OP might have with how he’s thinking about this).

I believe the 0.2V drop is not IR from the 8A flowing through wiring to the furnace, but instead is internal voltage drop in the batteries (assuming normal battery connection wires). OP is not measuring 0.2V drop across the wire to the furnace, but sees that drop in battery voltage. Doesn’t sound like a big battery bank, or strong (low internal resistance) batteries like Lithium or Lifeline AGMs. There is a reason that SOC is measured with resting batteries and not under load, any battery will have some voltage drop when sourcing significant current.

To answer the OP question - no, a larger wire to the furnace will not help the situation. Any small IR change by beefing up this wire would have 2nd or 3rd order trivial change in motor draw, fan speed, etc. Better batteries and/or more capacity is the issue you’re dancing around.

OPs voltage measurement is highly dependent as to exactly where the voltage meter is connected.

If not directly connected to the battery post the measurement is taking into account the IR of the wires to the connection point of the voltage meter.

I would however concur, part of the voltage drop is most likely from the internal resistance of the battery bank which goes up as the batteries are depleted. That can't be changed other than adding extra battery capacity.

However, what one can do is minimize as much as possible the voltage drop to the high amperage loads like an big inverter (OP is using the inverter to run a toaster if I remember correctly, typical two slice toasters are 850W). That IS done by using as large as possible wire ga that will fit the inverter terminals and keep both the pos and neg wire runs as short as possible. With 12V items, even saving .1V drop under a heavy load can make a difference when dealing with depleted batteries.

[MNRon]

Unless I’m missing something, most of responses are missing the point of the issue (as the OP might have with how he’s thinking about this).

I believe the 0.2V drop is not IR from the 8A flowing through wiring to the furnace, but instead is internal voltage drop in the batteries (assuming normal battery connection wires). OP is not measuring 0.2V drop across the wire to the furnace, but sees that drop in battery voltage. Doesn’t sound like a big battery bank, or strong (low internal resistance) batteries like Lithium or Lifeline AGMs. There is a reason that SOC is measured with resting batteries and not under load, any battery will have some voltage drop when sourcing significant current.

To answer the OP question - no, a larger wire to the furnace will not help the situation. Any small IR change by beefing up this wire would have 2nd or 3rd order trivial change in motor draw, fan speed, etc. Better batteries and/or more capacity is the issue you’re dancing around.

[BFL13]

Just pondering all this (to no great effect)

The current to the furnace fan rises with more supply voltage but the current to the input of the DC-DC falls with more supply voltage.

The reason is the furnace fan is not a fixed demand like the DC-DC or inverter's output-- it will take more if you give it more.

The 8531 furnace is rated at 8.2 amps at 12v. So that is the expected "load", but it is not the "Demand". The fan runs faster with more amps if the "Supply" voltage is higher, so the demand is a variable depending on the supply.

The inverter load side is fixed by the load's needed input watts so the "demand" is not a variable. The DC supply tries to meet that demand. As the battery voltage falls the amps drawn go up to maintain the demand watts.

The DC-DC charger has an output set at eg, 20 amps at 14.6 volts, so that is the fixed demand, not a variable. The input supply tries to meet that demand.

However, the input is itself a demand on the engine battery and alternator, which has to meet the input demand for the output demand.

Using fatter wire in the DC-DC input to the battery keeps the amps demand lower with the input voltage higher, so the alternator doesn't need to supply so many amps

The engine battery voltage is somewhat fixed as maintained by the regulator. The supply variable is the alternator amps to meet the demand.

The Dc-Dc charger "size" (output setting demand) has to be chosen, limited to match the alternator "size", where using fatter wire will let you have somewhat more Dc-Dc size for the same alternator.

[Gdetrailer]

MT BOB wrote:
OK,do you have an issue with your furnace? A little background-- yes the motor will slow down when voltage drops, all is normal.You are protected by the sail switch. Manufacturers build them to be in a RV, so they anticipate lower voltages.
They are rated to operate fine and safely down to 10.5 volts. That is why they have 10 volt motors in them.I have seen older Hydroflames run until voltage dropped below 10.0. If you suspect an issue, you must measure the voltage AT the furnace, not on the RV's panel,which reads the battery itself.

I think the OP is more concerned about when the furnace is running plus the inverter is heavily loaded the combined load is enough to send the inverter into low voltage alarm or shut down when the batteries have been discharged to 50%..

Op may have though reducing the voltage drop at the furnace would reduce the current drawn by furnace, in the case of DC motors, nope..

OP could try much bigger wires to the inverter, even reducing the voltage drop by .1V at the inverter terminals can make a big difference.. Reducing the voltage drop between inverter and batteries can have a pretty positive effect, as the DC input voltage goes up to the inverter, the lower the DC current will be required to maintain the 120V load.. Watts is Watts..

[MT_BOB]

OK,do you have an issue with your furnace? A little background-- yes the motor will slow down when voltage drops, all is normal.You are protected by the sail switch. Manufacturers build them to be in a RV, so they anticipate lower voltages.
They are rated to operate fine and safely down to 10.5 volts. That is why they have 10 volt motors in them.I have seen older Hydroflames run until voltage dropped below 10.0. If you suspect an issue, you must measure the voltage AT the furnace, not on the RV's panel,which reads the battery itself.

[Gdetrailer]

BFL13 wrote:


Lights dim with lower voltage and also draw fewer amps when dimmed, so that fits.

Slower fan speed in the furnace with lower voltage would get to the point the sail switch would open, so that is the safety there.

My 8531 older furnace says the 12v to it should not be above 13.5v. That is in the manual's section about the flame sensor's microamps, but I can't find a minimum voltage specified for when the blower fan gets too slow.

The furnace does work ok when battery charging is happening and the battery voltage is above 13.5v. I have not measured for voltage at the furnace then. Doug R and the other techs here would know about all that.
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Part 2--

Not sure it is the same thing for how amps work--I am now looking at the DC-DC charger

With the input of that thing, you must use fatter wire to the battery to keep the amps down and the voltage up.

Could relate to the point made above re wasted amps making heat with voltage drop? Fatter wire means the alternator does not have to do extra amps to make that heat?

Flame sensor could be an issue with a too low of a battery voltage if the control board doesn't have good voltage regulation built in. Flame sensor works on the idea that the flame conducts electricity. So control board places a voltage on the sensor rod. No flame, no current flows. When there is a good flame, the flame will conduct a small amount of electricity which creates a small measurable one way current flow. Controller will monitor the current flow, if not correct current will shut down the furnace..

Potentially you could reduce the voltage to the fan motor to reduce the current drawn but I believe it is in your best interest overall to leave well enough alone as slowing down the fan affects not only your inside air flow but the burner side airflow.

As far as DC-DC setups goes, they work in the same principle as a autoformer does for AC but with DC and perhaps a bit more efficiency.

Lower input voltage will require more current to get a higher voltage at a certain amperage. So say you have a 20A DC to DC power supply and you are taking 13V and changing it to 14V you would draw say 20 amp (260W) on the input but get 18.5A (260W) on the output.. There is no free lunch.. Wattage is the same, but amperage drops..

Using a DC to DC charger, you are simply trying to boost the voltage high enough to get over the voltage lost due to higher resistance of small wiring.

You are also trying to get past the vehicle alternator's regulator which typically tapers down the voltage from the alternator from initial charging voltage after starting of around 14.4V down to 13.8V which basically means your RV battery only gets a tiny trickle charge after a minute or two of engine running..

Keeping in mind, in my example that is assuming 100% efficiency of the DC to DC charger.. Nothing is 100% efficiency in this world and DC to DC chargers are no exception and will run 80%-94% efficient.

[BFL13]

time2roll wrote:
Yes the LFP does seem to move a bit more air at 13.4 volts. Amp draw is significantly lower with steady 13+ volts vs. 11.x with the old battery running the kitchen.

You mean the inverter's amp draw is less without so much voltage drop? It does not have do so much to maintain the watts? That is different from the furnace draw, but might be like the DC-DC input draw, not sure.

[time2roll]

Yes the LFP does seem to move a bit more air at 13.4 volts. Amp draw is significantly lower with steady 13+ volts vs. 11.x with the old battery running the kitchen.

[BFL13]

Gdetrailer wrote:
Will draw slightly MORE current due to less voltage drop, results in motor spinning a bit faster.

RV furnace motor is a DC motor, DC for practical intents can be considered like a resistor since you are not dealing with impedance/inductance like you do with AC devices.

If you want the furnace motor to draw less current, you would need to REDUCE the voltage going to the motor. However, that results in less RPMs for the fan which results in less air flow which results in possibility of overheating the furnace and/or creating a fire hazard..

Lights dim with lower voltage and also draw fewer amps when dimmed, so that fits.

Slower fan speed in the furnace with lower voltage would get to the point the sail switch would open, so that is the safety there.

My 8531 older furnace says the 12v to it should not be above 13.5v. That is in the manual's section about the flame sensor's microamps, but I can't find a minimum voltage specified for when the blower fan gets too slow.

The furnace does work ok when battery charging is happening and the battery voltage is above 13.5v. I have not measured for voltage at the furnace then. Doug R and the other techs here would know about all that.
---------
Part 2--

Not sure it is the same thing for how amps work--I am now looking at the DC-DC charger

With the input of that thing, you must use fatter wire to the battery to keep the amps down and the voltage up.

Could relate to the point made above re wasted amps making heat with voltage drop? Fatter wire means the alternator does not have to do extra amps to make that heat?

[wa8yxm]

The question on how much power (Amps) the furnace will draw is actually rather complex.. But in general I suspect the higher the voltage the greater the amps but it's not a 1:1 relationship also the faster the blower will blow which might result in greater comfort. But you cited a 0.2 volt drop.. NOT WORTH the cost of the wire to upgrade it for 0.2 volts. for 2 volts yes but 0.2 no.

[Matt_Colie]

This is amazing, in the entire thread there is only one line - ONE!
That I do not agree with....

That was the OPs line that there were no stupid questions - UNTIL NOW!!

This does not qualify as a stupid question at all. It was a well thought and interesting question and you got good and clear answers.

Matt