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One 12v car battery powers 2 loads. One load is 3.5A and the other is 100mA. The 100mA load is always on. When the 3.5A load gets switched on, scope readings show the amp draw momentarily dropping to half or less and voltage dropping from 12v to about 9.5v on the 100mA load. How can I maintain steady, smooth voltage and current to the low amp load during the brief time that the much larger load is being energized? Would I need to use a large inductor or large cap to help momentarily sustain the 100mA current and voltage during this time? I think I need an "isolator". If so, can it be explained in more detail?

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  • \$\begingroup\$ What do you mean by "scope readings show the amp draw momentarily dropping to half"? Half of what? \$\endgroup\$ – Phil Frost Jan 16 '13 at 3:34
  • \$\begingroup\$ @PhilFrost Presumably half the 100 mA that the always on part uses. \$\endgroup\$ – Anindo Ghosh Jan 16 '13 at 3:43
  • \$\begingroup\$ half of it's normal amp draw. It's normal amp draw is 100mA. When the high amp load is energized, current is sucked away from the low amp load, momentarily depriving it of the current and voltage it needs to run. \$\endgroup\$ – Arewfawe Waefadffawe Jan 16 '13 at 3:43
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What's likely happening here is when you switch on your 3.5A load, it is as first drawing much more than that. This is called inrush current and is often caused by capacitors being charged up to the supply voltage. Since you mention a car battery, it's also possible the initial current is high because you are starting a motor. A stopped motor provides no back-EMF and so will draw more current until it's up to speed.

This isn't a problem in itself, but all batteries have an internal resistance which limits the maximum current they can drive. Ohm's law states that as the current through a fixed resistance increases, the voltage across that resistor also increases. The practical consequence of this is that as the current draw on any battery increases, its voltage decreases.

Most likely, you see the current draw on your 100mA load decrease because the battery voltage has decreased so significantly that there is no longer sufficient voltage to drive 100mA through it. If your load were simply a resistor, you could use Ohm's law to know the relationship between voltage and current.

You could, in theory, solve this by putting a capacitor across the battery terminals, which could provide some stored energy to fill in transient demands such as this. Ostensibly, the capacitior would have a lower internal resistance than the battery, and it would temporarilly be able to drive the needed current, as long as it isn't needed for too long.

However, the internal resistance of a car battery is pretty small, and a capacitor large enough to store enough energy to supply this transient load is probably impractically large and expensive.

One solution is to limit the inrush current to your 3.5A load. Look for "soft" or "slow start" circuits as a starting point for your research. The other solution, if you only care about the 100mA load, is what Eric Gunnerson's answer suggests.

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    \$\begingroup\$ you are a dude!! \$\endgroup\$ – sheetansh Jan 16 '13 at 9:28
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If the 100mA load can tolerate a slightly lower voltage, you can put a diode in series with the positive terminal, and then put a capacitor across the 100mA load. The capacitor will charge up, and it will help keep the voltage more constant when the other load kicks in.

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  • \$\begingroup\$ This was the answer that I was going to submit. A simple Shottky diode in series with the 100mA load would isolate the added storage capacitor at the 100mA load so that it's voltage would not be affected by the drag down of the battery when the high current load is engaged. \$\endgroup\$ – Michael Karas Jan 16 '13 at 7:53

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