Electrical – Battery terminal voltage as it relates to its proper functioning

Summarised_solution:

• Interuption of load to an alternator can result in a voltage spike of 100 - 200 volts for typically up to about 0.5 seconds. More on a bad day downhill with the wind behind you.

• If you are small and electronic or an LED or a light-bulb you want to be somewhere else if/when this happens.

• The unusual connection means when jump starting can persuade an alternator system to "see" load open circuits due to unexpected interactions. Trying to stop these happening is usually felt to be a good idea.

• Special regulators designed to survive such "events" are available for use in automotive equipment. Some of these are more often used in everyday applications without realisation of their special qualities. (eh LM2940).

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People tend to over react to be on the safe side BUT there are real problems to be avoided.

When an automotive system is connected as normal, energy flows from alternator via rectifier diodes, via or past any controller that is used and into the 12V "bus" with battery attached. When you start introducing alternator level energy input (10's of amps, possibly 50A) and battery to battery connection which will give about 0 amps if batteries are similar in voltage BUT potentially very large if one battery is well charged and the other flat (which is usually exactly the case) then current flows relative to regulator etc are different. If the sending alternator's regulator gets a wee bit confused and misregulates and all that is on the receiving end is an additional car battery and perhaps a starter motor then little harm is likely to occur. A brief transient 20 or 30 or 40 volts will be transient indeed if a car battery is on the receiving end. If a light bulb, radio, MP3 player, cell phone, coputer or other fragile electronic citizen happens to be present when the big boys are partying then they may get hurt.

As a demonstration of what can happen, several decades ago people here sold a device that allowed you to use ag a 230 VAC skilsaw powered by a car alternator still in the vehicle. This may have required a "universal motor" type motor to work - maybe not. The "black box" connected to the alternator connections inside the normal regulator etc and could be switched in when required and output taken from the alterneotr. To use this the engine was set to a very fast idle and the device switched on. Something around 200 Volts AC was provided by th alternator at whatever the rated current was. So eg a 50A alternator was good for about 50 x 200 = 10 kW. !!!. In fact !!!!!!!!!!!! . 10 kW may have been a tad high for what could be achieved in reality but people were running skilsaws from them. That usually needs several KW- depending on the saw.

This works because the alternator is essentially a constant current device with saturation of the "iron" acting as the output limiter. For an eg 30A alterantor, if you clamp it at 12 V you get about 12 x 30 = 360 W. Clamp it at 24 V and you get 720 W. Clamp it at high and you get - WoW!. Don't clamp it at all and !!!. As long as the regulator is happy and stable you never see this. Play with its brain, whether with a magic box or strange connections and it may occasionally impress you.

Note that the above super high voltage is not potentially unlimited (unlike eg inductive ringing) and is not the same in waveform as voltage increases. If you rev the engine so the open circuit voltage peaksat 100 V peak half wave peak to ground then you will get approx a half sinusoid. As you star to clamp this at successively lower voltages uou get a swuarer waveform as you cut the tops off the sine wave. When you get down to 12v you have a near square wave AC output that rises nearly vertically after the zero crossing and then flat-tops at the clamp voltage. This means that the RMS value of the waveform is much higher than a sine wave. Sine wave Vrms = 0.771 Vpeak.This trapezoid is close to Vpeak RMS. As you use higher voltages as for eg the skilsaw example the waveform will be more sinusoidal.

Load dump:

Load dump is what happens when the alternator is running at high output and the load is suddenly reduced or removed OR when something such as a regulator gets confused and makes it appear that this is what has happened (load removed by switching or perhaps short then open or ...).

Wikipedia says re load dump

• In automotive electronics, it refers to the disconnection of the vehicle battery from the alternator while the battery is being charged. Due to such a disconnection of the battery, other loads connected to the alternator see a surge in power line. The peak voltage of this surge may be as high as 120 V and the surge may take up to 400 ms to decay.

Which sounds like fun. And ...

• The windings of an alternator have a large inductance. When the vehicle battery is being charged, the alternator supplies it with a large current. If the battery gets disconnected while it is being charged, the alternator current drops sharply and suddenly. This causes a high voltage across the alternator due to the inductance of the field winding. The field current cannot drop quickly, so the magnetic field remains large, so the rotation of the alternator continues to generate a large voltage. All the loads connected to the alternator see this high voltage spike. The strength of the spike depends on many factors including the speed at which the alternator is rotating and the current which was being supplied to the battery before it was disconnected. These spike may peak at as high as 120 V and may take up to 400 ms to decay. This kind of a spike would damage any semiconductor device, e.g. ECUs, that may be connected to the alternator. Special protection devices, such as TVS diodes, varistors which can withstand and ground these spikes may be added to protect such semiconductor devices.

  Different automotive standards such as ISO 7637-2 and SAE J1113-11 specify the standard shape of the load dump pulse against which automotive electronic components may be designed.

  There can also be an inductive spike due to the inductance of the field winding. That may have a larger voltage, but it will be for a much shorter duration, as little energy can be stored in the inductance of a field winding. Load dump can be more damaging because the alternator continues to generate power from the rotation of the engine, so much more energy can be released.

  The inductance of the current-carrying windings has no direct effect on the load dump, but it has a large indirect effect. In normal running, current flowing in the inductance of the windings causes a large voltage drop. To keep the correct terminal voltage, the magnetic field from the field winding has to increase a lot at large loads. When the load is disconnected, there is no current, so there is no voltage drop in the inductance of the windings. The full voltage appears on the alternator terminal until the field current falls. .

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Note that electronics regulators for electronic devices made to operate in an automotive environent have a very rugged and specialsesd front end. Wprst case they are designed to turn off on high peaks rather than try to ride the wave. You may drop-out your output but your circuit survives. Hopefully.
The NatSemi LM294x series is a classic example of this approach.
LM2940 data sheet here
They say:

• Designed also for vehicular applications,
  the LM2940/LM2940C
  and all regulated circuitry
  are protected from

  • reverse battery installations or
  • 2-battery jumps.

• During line transients,
  such as load dump
  when the input voltage can momentarily exceed the specified maximum operating voltage,
  the regulator will automatically shut down to protect both the internal circuits and the load.

  The LM2940/LM2940C cannot be harmed by temporary mirror-image insertion.

  Familiar regulator features such as short circuit and thermal overload protection are also provided

Also the standard but good stuff

• The LM2940/LM2940C positive voltage regulator features the ability to source 1A of output current with a dropout voltage of typically 0.5V and a maximum of 1V over the entire temperature range. Furthermore, a quiescent current reduction circuit has been included which reduces the ground current when the differential between the input voltage and the output voltage exceeds approximately 3V. The quiescent current with 1A of output current and an input-output differential of 5V is therefore only 30 mA. Higher quiescent currents only exist when the regulator is in the dropout mode (VIN − VOUT ≤ 3V).

Battery heating can be roughly defined using its internal resistance. We can get a rough estimate of a battery's internal resistance using it's cold cranking amp rating. This is rated at -18C, which is close to -20C.

For a nominal 12V battery: $$ R = \frac{12V - 7.2V}{CCA} $$

Let's assume this is constant vs. current draw.

Let's assume at time t=0, the battery is completely insulated from the world and has a heat capacity similar to pure water.

The amount of energy required to heat 17 kg of water from say even -20C to -18C is: $$ E = 4.2 \frac{J}{g K} \cdot 17 kg \cdot (20^oC - 18^oC) = 142.8 kJ $$

For a 9A draw, this would take:

$$ t = \frac{E}{(9A)^2 \cdot R} $$

Suppose you have a 200 CCA battery. Then we can find t = 73 seconds.

Problems with this analysis: Obviously the full weight of the battery is not just the liquid inside. Thus the time required to heat the battery 2 degrees C would be less. However, a larger problem is that you probably don't have a perfectly insulated battery. The 9A draw only produces ~2W, which could easily be lost from the battery (the cables are extremely good conductors away from the battery).

At temperatures much colder you're going to have to start worrying about fluids freezing, thus many things could just stop working. You're much better off following proper car winterization guide.