Grounding confusion

The explaination you gave would surely be for the output current, seeing as you are using the output votlage and the load resistance to calculate the current? I am talking about the input current draw. The diagram below shows the grounding resistance modelled as a resistor. Seeing as the amp is in series with the 'grounding resistor', the maximum current the amp can draw is determined by the sum of the resistance of the ground and and the amp's own impedance. Hopefully this helps to explain the problem more clearly.

Yeah, that's why I try to tackle these kinds of problems.  It confuses me, so I have to sort it all out every time.  Some day maybe it will come quick and easy. 

I think I would have had to first determine power output of the amplifier, then use the voltage at the battery as E to find current draw.  The efficiency of the amplifier would have had to be calculated, as well.  And it seems reasonable, in a sense, to add the resistance of the ground circuit to the resistance of the device (as you displayed above) to find an answer, but that resistance is not part of the load resistance.  It is resistance in a conduit back to the power source.

In all reality, the ground circuit is acting just like a wire...in fact, a wire could replace it altogether.  You could have 0.5 ohms resistance in a ground wire run all the way to battery ground just as much as you could have 0.5 ohms of resistance in the chassis itself.  Or you could have more resistance than 0.5 ohms.  That resistance figure still cannot be used, by itself, as the number for R in an Ohm's Law formula (as you used it in the original post).  When something looks reasonable in theory, but obviously makes no sense when you apply logic, you can presume that the theory may not be correct.

If you look at ground resistance as nothing more than "a wire", then think of it as you would think of a wire when determining its responsiblity in affecting power.  Here is a quote from DYohn that I saved to file that was written about wire:

Larger wire does not "produce more current flow" nor does it rob anything from anything to use multiple sizes.  The LOAD (in this case the amp or the speakers) creates the demand for current flow.  Larger gage wires ALLOW more current flow than smaller ones with less HEAT and losses.  ANY wire will try and pass whatever current demand is required by the load.  The wire does not care.  A 36 gage wire will try to pass all 200 amps demanded by a large amplifier if you hook it up that way.  It's just that smaller wire will add enough resistance to burn up faster.  It is impossible for a wire, any wire, to "produce" current all by itself.  Using smaller wire does not "rob your sub of its full potential," this is a myth.  Smaller wire is simply a fire hazard.

(edited to add:  you smarter ones on this forum, feel free to correct me if I'm off-base with this.  it's always a learning experience... ;)

Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.

First off, I suspect most of us don´t have a meter capable of accurately measuring ground resistance. Anyone want to argue that, please post what kind of meter you have. I´ve got a Fluke 88, and well as an 8060 and many others, and they aren´t good enough to directly measure ground resistance.

Besides, circuit resistance itself isn´t our concern.. what we´re concerned with is voltage drop. We can measure voltage drop easily. Thinking about Ohm´s law, voltage drop is determined not only by resistance, but also by current flow. Using the 0.5 ohm circuit resistance mentioned earlier, if this was measured on a circuit powering an LED using 0.020 amps, the circuit drop would be 0.01 volts. Nothing to worry about here. If, however, that same circuit resistance was measured in a circuit powering a headlight using 8 amps, then the voltage drop on that circuit would be 4 volts.. meaning the headlight would only be seeing around 10 volts instead of the full 14 volts that the alternator is pulling out. This is why relays are a good idea for aux lighting as well as upgraded headlights.

Wire size will also determine voltage drop.. smaller wires have a greater resistance per foot than larger wires. Undersized wires will restrict current flow through them, this translates into voltage drop. Does your amp work better at 14v or at 10v.

I don´t like to see more than 1.0v drop on the starter wires.. that means that circuit resistance for those wires at 200 amps can´t be more than  0.005 ohms.

If you have a question or aren´t clear on something, please post.

Jim

Steve, in your example where you replace the chassis with a wire (essentialy the same as where i replaced it with a resisitor), would you agree that if you have a 14V power supply (the alternator) and a wire/grounding resistance of 0.5ohms the maximum current that can flow through that wire is 28 amps. I would like to have that confirmed, or i definately do not understand what is going on! If i am correct, then seeing as the ground is connected in series with the amp then the maximum current that the amp could recieve would be 28 amps (which realistically would not happen as the amp itself would have to have zero resistance).

Jim, I agree that higher resisitance ground will cause a greater voltage drop in the chassis so that there is less voltage available for the amp, but you havn't mentioned the fact that the higher resistance will also reduce the current flow through both the chassis and amp (seeing as they are connected in series). Another way of phrasing my question is 'can an amp that draws 100Amps at full power run in a car that has a grounding resistance of 0.5 ohms?' if the answer is no then no further explaination is required, but if it is yes then further explaination is definately required! Thanks again for trying to answer this difficult-to-explain problem

"can an amp that draws 100Amps at full power run in a car that has a grounding resistance of 0.5 ohms?"

No.. you are correct; with 14v and a resistance of 0.5 ohms, the circuit current would be 28 amps. Ohm's Law, plain and simple. If you measured 3.2 ohms on your amp circuit but everything worked correctly, then you almost certainly got a bad reading somehow. Measuring voltage drop is the best way to determine if your wiring is ok.

Jim

A half ohm of resistance in the chasis ground-to-battery is not uncommon...in fact, it's good.  Isn't it?  Let's double-check your reckoning by employing some Ohm's Law and logic to a common scenario:

Guy has three amps.  Two for subwoofers and one for the mids/highs.  Sub amps have loads wired to 2 ohms, and the third amp is working at a 4 ohm load.

There is 1/2 ohm of resistance between the amplifier common ground point and the battery negative terminal.  You are saying that with a resistance of that much in the chassis to ground, his system can draw no more than 28 amps of current.

Guy measures voltage of each one of the amps after gains have all been set.  The system is set to full power.  He wants to find power so that he can determine the draw of current from the car.

• mono amp 1:  output voltage is measured at 32 volts. P = E^2 / R says that this amp is producing 512 watts (it's a 2 ohm load).  Now, the amp is a class D and is about 85% efficient.  To produce 512 watts, the amplifer has to use enough current to make up for its waste factor.  Figuring in 15% for heat waste, this amplifier needs to consume enough for 602 watts of power to make 512 watts.  With battery voltage at 14, I = P/E tells us that this amp will demand 43 amps of current.
• mono amp 2 is the same, using 43 amps of current.
• 4 channel amp is measured at 38 volts.  It is 60% efficient.  P = E^2 / R says that with this amp's 4 ohm load  its power output is 361 watts.  With a waste of 40%, it needs enough current to make 505 watts.  Battery voltage is 14 volts, so the current needed is 36 amps from the battery ( I = P/E ).

This is not an unreal situation.  This is a big, loud system but not all that uncommon.  And assume that these power numbers are real because the HO alternator keeps up with the demand.  512 + 512 + 361 = 1385 total watts for this system at full power.

To get this power, there is a draw of 122 amps from the car (43 + 43 + 36).

How do you get this power if a chassis ground of 0.5 ohms is limiting the power output that much?  At a maximum limit of 28 amps, there would not be enough power for just the 4 channel amp alone.  28 X 14 = 392 (P = I X E).  You are saying that this system cannot produce any more than 392 watts (and actually quite a bit less than that as we figure the amps' heat waste factor).

The quote above says that a wire (or the chassis, if you will) will try to pass whatever current is demanded by the load.  The loss due to resistance does have an effect, but not to the extent you are saying it does.  Here is a calculator that will help determine losses of resistance in the circuit:  http://www.h-o-alternators.com/TechCalcs/voltdropthroughsystemhoalt.swf