I have a 9V 500 mAh battery that has an open circuit voltage of 10.54V. As soon as I connect it to my load, the battery terminals measure just 5.5V. The current drawn is just 1.5 mA. Why does this voltage drop occurs at the supply?
Hi,
Given that the current draw is so low, the most likely reason for the huge drop in voltage is that the battery is exhausted. While a battery is not a capacitor, it accumulates charge in the same way. If it is connected to a very high impedance load (such as a voltmeter) then this accumulated charge remains virtually intact and the voltmeter gives a high voltage reading. But as soon as a low impedance load is connected to the battery, this accumulated charge, which takes time to accumulate, very quickly discharges and you read the lower voltage reading.
Batteries, capacitors, and power supplies all have a feature called an Equivalent Series Resistance (ESR). Conceptually, a real world battery is a theoretically perfect battery in series with a resistor. That resistor forms a voltage divider with the equivalent resistance of whatever is connected to the battery as a load. For a normal battery the ESR is less than 1 ohm. But as a battery discharges, the ESR increases. In your case the ESR is over 3.6K. So either the battery is discharged or it is defective.