Capacitor Voltage Hold up (During Input Power Interruption)

I want to understand how long a capacitor can provide the stable voltage for a given load current, in the event of an input voltage interruption?

Suppose, I have a single 100nF input capacitor at my 12V. Across the capacitor, I have a load of 50mA. My stable input voltage is 12V.

Suppose, the 12V drops to 0V for some, say 100us, and then comes back up to stable 12V. Will that 100nF capacitor be able to provide the stable 12V with 50mA to the load during the momentary power interruption? If not, how long will it provide the stable 12V maintaining the same load current?

Is there any formula or a way to figure out the minimum capacitance needed to support this requirement? Or how long the capacitor can provide the stable 12V with the same 50mA Load?

• Before even thinking of voltage (which will start dropping immediately, regardless of the value of C), think of energy. $E_\mathrm{device} = t * I*V$ and $E_\mathrm{cap} = \frac{1}{2}CV^2$ - how do the values work out? – awjlogan Mar 18 '20 at 15:17
• Why would the voltage drop immediately regardless of the Capacitor value? – Newbie Mar 18 '20 at 15:42
• $Q=CV$, so if you start drawing charge off the capacitor, $V$ must go down. This is true for any value of $C$. The limit to which your circuit tolerates that drop is one part of calculating the required $C$. – awjlogan Mar 18 '20 at 16:43

Suppose, I have a single 100nF input capacitor at my 12V

And

Suppose, the 12V drops to 0V for some, say 100us, and then comes back up to stable 12V. Will that 100nF capacitor be able to provide the stable 12V

No, the 12 volt dropping to zero volts will drag the capacitor voltage along for the ride and you get no benefit. You need to put a diode between the raw 12 volt supply and the capacitor to stand any chance of holding up the voltage.

$$I = C\dfrac{dV}{dt}$$

That formula will allow you to plug 50 mA into the "I" part and examine how quickly (or slowly) voltage changes against time (dv/dt).

Bear also in mind that the diode will have a forward volt drop of circa 0.6 volts so your actual supply to the thing that draws 50 mA will be more like 11.4 volts. That may be a show-stopper for you.

• Ok, If I put a diode between the 12V raw supply and the 100nF capacitor, the capacitor would hold the voltage of (12-0.6)V at the load whilst providing 50mA of current for 100us of Raw power supply interruption? – Newbie Mar 18 '20 at 15:45
• Have you calculated dv/dt using the formula? – Andy aka Mar 18 '20 at 15:51
• Please check whether my calculation is correct. dv = (ICdt). So, the change in voltage at the output capacitor for 100us would be = (50mA * 100nF * 100us) = (5 * 10^-13)V. So, The voltage drop during the 100us at the output would be (12-0.6- (5 * 10^-13))V . Am I correct? – Newbie Mar 18 '20 at 15:51
• No, it’s 50 mA divided by 100 nF then multiplied by 100 us. And that gives you 50 volts. In other words the capacitor voltage would try and drop 50 volts in 100 us. If you used a 10 uF capacitor, the voltage would drop 0.5 volts and is more acceptable. – Andy aka Mar 18 '20 at 15:55
• That’s what I put in my answer and I have no reason to change it. – Andy aka Mar 18 '20 at 15:59