# Safe discharge for capacitor

If I have a source voltage of 5 V going to a 10000 µF capacitor will the voltage/amps remain the same (obviously it will discharge) when I disconnect the source?

Example:

5 V 1 A Src ----- 10000 µF --- (5 V 60 mA electronic drain) -- (back to the 5 V Src) simulate this circuit – Schematic created using CircuitLab

Will the capacitor dump all its voltage/amps at once? or will it only be drained what the electronic drain is meant to use(5 V 60 mA until capacitor is fully discharged)?

• Capacitors retain their charge (other than loss due to self-discharge) if there is nothing to drain it. The capacitor will only supply what is drawn from it, just like a battery or power supply. – DKNguyen Jul 17 '19 at 19:21
• Welcome to EE.SE. A schematic is better than words. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. – Transistor Jul 17 '19 at 19:23
• So if I understand correctly it would be a 5v battery essentially? – James Jul 17 '19 at 19:24
• @James A really crappy one that doesn't hold it's voltage as it is discharged. – DKNguyen Jul 17 '19 at 19:31
• Electronic drain is most likely non-linear. After the +5V source is disconnected, the capacitor will start to discharge with 60mA, but that discharge current will likely decrease as the capacitor voltage slowly drains away. Eventually, the capacitor will self-discharge because it is "leaky". There's no way to tell how your electronic load reacts to a slowly-decreasing +5v supply. – glen_geek Jul 17 '19 at 19:45

If I have a source voltage of 5v going to a 10000uF capacitor will the voltage/amps remain the same (obviously it will discharge) when I disconnect the source?

Capacitors have a leakage current (no material is a perfect insulator) which can be modeled as parallel resistance internally. Many times leakage current is listed in the datasheet. The capacitor will slowly drain according to the leakage.

Capacitors can be discharged with a resistor across the terminal, the problem is the current from the capacitor can heat the resistor beyond it's temperature rating. A resistor with the appropriate wattage rating must be used, or a larger resistor can be used. The time to discharge is the RC time constant, and the discharge follows an exponential curve.

For example to drain to ~60% of the charge, a 10Ohm resistor would take:

$$\R C = 10000*1e-6*10 = 0.1 s\$$

(A 10Ohm resistor would be about 50mA. A 60mA load would be about 8Ω or 0.83s to drain to ~60% or 3.15V)

If you found the leakage to be say, 1nA, this would be similar to having a 5GΩ resistor.

$$\R C = 10000*1e-6*5e9 = 5e7 s\$$

or over 1 year to drain to 60% of it's value.

If you want to find different values of discharge (like 1V which would be 20% of the initial value) this chart can be used. Look up the percent discharge, and then find out how many time constants needed.

For a 60mA load (and a time constant of 0.083), it would take 4 time constants to drain to 1.8% of 5V or 90mV. 4 time constants is 0.333 seconds.

Clarification: The capacitor will drain by the load placed on it so in the example above when 5V is removed from the capacitor and the voltage regulator, the voltage will fall exponentially (assuming that the current is constant, which it won't be if the device is a voltage regulator or microprocessor). 