Voltage Drop . . . How Does it Affect Amps?

[MarkJ]

OK I've looked at Gords spinning wheel of volts amps and them other bitties...
But can I get a simple answer to this question?

I am pondering a voltage drop I may have to the fridge compressor. So this is a hypothetical.

If my fridge uses 4 amps per hour.... will I drain my battery faster if the voltage the compressor is getting is 11.7 volts instead of 12.4 volts?


Thanks


Mark

[DeepFrz]

Think power (V x A). The lower the voltage the more current you need to supply the same power.

[senormechanico]

Mark,
There are several variables here.

On many units, the electronics usually put out three phase AC which power the motor.
The electronic module will feed whatever is necessary to run the motor at a proper speed (which can also vary depending on the software).
Basically, I don't think the power comsumption will vary a lot in this scenario.
IOW, the current may actually INCREASE to make up for the lack of voltage.

I'm sure Richard Kollman will have a good take on this as well.

[MarkJ]

Originally Posted by DeepFrz

Think power (V x A). The lower the voltage the more current you need to supply the same power.

So the battery drops faster?

Mark

[DeepFrz]

Yes, you will be drawing more current from the battery in order to supply the same amount of power.

[capt.bobfm]

Yes. The battery will drain faster.

[Cruiser2B]

no, not exactly. your frig draws 4amp at the rated voltage.

Amps= Voltage/resistance(load)


if the load is the same(fridge) and the voltage changes(12.4 to 11.7) then the Amps change proportionately( less current)

[DeepFrz]

Cruiser2B, that is only true on a purely resistive load, which refrigeration is not.

[Cruiser2B]

DEEP,
if it was to work the way you say, the fridge would have to have a regulator ckt of some sort that would maintain the current draw, since I am not a refrigeration tech i cannot say but doubt it. It may seem to draw the battery down fastest because its duty cycle will increase to maintain the temperature, but if we are talking DC current and voltage drop, the way I described is the way it will work. if he decrease voltage to the fridg, the output of the fridge will be less...has to be.

[DeepFrz]

Cruiser2B, take a look at seniormechanico's post above. It explains, briefly, how the electronics module works. However, even if it was a DC motor it would still draw more current. Again, think power. You are still thinking purely resistive circuit.

[David M]

Mark asked if the battery will drain faster if it is at 11.7 vs 12.4.

Well Mark, it is guaranteed it will drain sooner if it is at 11.7.

Lower voltage and more current creates more waste heat, therefore a higher percentage of the power being delivered in the circuit will be converted to waste heat....assuming the circuit tries to draw the same mount of power at 12.4 as it does at 11.7. If not it all depends on how much power it tries to draw at the lower voltage. I think they do because they sure get hot when run at lower voltages.

We have all sen how bulbs dim at lower voltages, they are then also drawing less power. I'm not sure how it works for a DC motor, if the motor tries to make up for the lower voltage by drawing more current. I think they do because they sure get hot when run at lower voltages.

[HopCar]

I think we need to look at how many watts it takes to run the fridge.
12.4 volts X 4 amps = 49.6 watts
to calculate the amperage draw at 11.7 volts
49.6 watts / 11.7 volts = 4.23 amps
So to do the same work that you can do with 12.4 volts and 4 amps you would need 4.23 amps at 11.7 volts.

[fairbank56]

Is this a test? What is the prize for the winner? If this is a simple relay controlled dc compressor refer, the answer is.....the current remains the same.

A dc motor that is running produces a counter electromotive force CEMF also know as a back emf. This is a voltage that opposes the supply voltage. The overall voltage to the motor is the net voltage. When you reduce the supply voltage, the motor speed slows which reduces the back emf so the overall voltage remains the same as does the current. The reason a motor runs hotter when running at slower than it's design speed is because there is less cooling effect of the rotating armature.

Keep reducing the voltage and you will reach a point where the overall voltage is higher than design net voltage and the current will increase somewhat. However, you can vary the supply voltage to your refer from around 9 volts to 14 volts and the current remains the same.

The fact that the motor "needs" more current to get the same power is irrelevant. Reduce the supply voltage, you have reduced the power it's getting. It doesn't magically produce it's own power.

So, reduce the power supply to your refer by about a volt, the drain on your battery remains the same and your compressor runs a tad slower.

BTW, I have confirmed all this with a real world bench test. That's the real proof.

Eric

[Bash]

not to bypass all the theoretical discussions, but remember that these compressors draw more power at startup, and most power heads are programmed to turn themselves off if they're getting a low voltage reading the moment the unit switches on.

so if you're trying to talk yourself out of replacing wiring where you've HYPOTHETICALLY got too much of a voltage drop, don't.

[Cdesousa43]

Actually the voltage is proportional to the square of applied voltage to rated voltage of the motor. So you would think it will be less. But what will happen is you will not cycle as often so the motor will run longer. You may actually see a higher drain. You should be looking at amp-hours not just amps and volts. Figure on a 50% duty cycle. For 4amps of draw it should be 12v x 4a x .5hr or 24amp hours. Your battery might be rated for 110 amp hours. So it would take about 5 hours of operation to drain the battery.