How do I calculate capacitance for a specific voltage drop, over a specific time, drawn by a specific load?

Thread Starter

Mark Flint

Joined Jun 11, 2017
136
So for example, if I have a capacitor at 10v and I want a voltage drop of no more than 0.1v over a time period of 10 microsecond, powering a 1w load, what must the capacitance of the capacitor be to start with?
 

MrChips

Joined Oct 2, 2009
27,727
So for example, if I have a capacitor at 10v and I want a voltage drop of no more than 0.1v over a time period of 10 microsecond, powering a 1w load, what must the capacitance of the capacitor be to start with?
Power P = V x V /R
Load R = V x V / P = 10V x 10V /1W = 100Ω
or
Current I = Power/Voltage = Voltage/Resistance = 0.1A

Using the formula provided by eric
C = t x I / V = 10 x 0.1 / 0.1 μF = 10μF
 

Ian0

Joined Aug 7, 2020
6,737
Or alternatively. . .
E=½CV^2
Energy lost = 1W*10us =10uJ
½C(V0^2-V1^2) = 10uJ

½C(10^2-9.9^2)=10uJ
C=10.05uF

because the current at the end of the 10us is no longer 0.1A
 

Thread Starter

Mark Flint

Joined Jun 11, 2017
136
In the equation, why is time (t) not subject to a unit, like uS, mS, S, but is just 10? If I wanted to calculate this and change it from 10 microseconds to 10 milliseconds would t = 10000 instead of 10?

Using the formula provided by eric
C = t x I / V = 10 x 0.1 / 0.1 μF = 10μF
 

MrChips

Joined Oct 2, 2009
27,727
In the equation, why is time (t) not subject to a unit, like uS, mS, S, but is just 10? If I wanted to calculate this and change it from 10 microseconds to 10 milliseconds would t = 10000 instead of 10?

Using the formula provided by eric
C = t x I / V = 10 x 0.1 / 0.1 μF = 10μF
Sorry. In order to keep the equation simple, I omitted the units.
t = 10μs = 10 x 10^-6 seconds.

The 10^-6 factor appears in the result as μF.
 

Ian0

Joined Aug 7, 2020
6,737
Always calculate in the base units (Volts, Amps, Seconds, Farads) then you never get the prefixes wrong - unless you're doing it on a slide-rule.
 

Thread Starter

Mark Flint

Joined Jun 11, 2017
136
The 10^-6 factor appears in the result as μF.
Oh, yes that makes sense.
So, to see if I really understand this I'll take another problem: I have a capacitor at 12v and I want a voltage drop of no more than 0.1v over a time period of 2 milliseconds, powering a 500w load.
500w/12v = 41.667A
C = 2mS x 41.667A / 0.1 volt drop = 833.34F / 1000 = 0.833F
 

WBahn

Joined Mar 31, 2012
27,942
In the equation, why is time (t) not subject to a unit, like uS, mS, S, but is just 10?
Because of sloppiness and a preference for refusing to properly track units and, instead, tacking on the units that they hope, want, and expect the answer to have.

The correct formula, assuming constant current, is

Q = CV
dQ = C·dV // Here, dQ is the change in charge stored on the cap and dV is the change in voltage.
C = dQ/dV

dQ = I·T

C = I·T/dV

To put it in terms of power, as per your original post, and assuming that the voltage doesn't drop enough to appreciably change the power, you have

P = I·V // Note that THIS V is the total voltage on the cap, and NOT dV, which is the slight change in that voltage.

I = P/V

So this gives us

C = (P·T)/(V·dV)

From your first post:

V = 10 V
dV = 0.1V
T = 10 µs
P = 1 W

C = (1 W · 10 µs) / (10 V · 0.1V) = 10 µWs/V²

Now it's just a matter of converting the units.

C = (10 µWs/V²) (1 A·V / 1 W) (1 C / 1 A·s) (1 F / (1 C/V)) = 10 µF

With a bit of practice, this last part can be done in your head in a matter of seconds -- and if you did something wrong in the math, it will probably yell at you kicking and screaming because the units won't work out.

If I wanted to calculate this and change it from 10 microseconds to 10 milliseconds would t = 10000 instead of 10?
Just use 10 ms.

Want to change dV from 0.1 V to 50 mV, just use 50 mV.

Want to change P from 1 W to 500 mW, just use 500 mW.

ASIDE:

You seem to have a habit of not using the correct case with your units.

s : seconds
S : siemens
V : volts
v : ????
A : amperes
a : ????
W: watts
w: ????
 

WBahn

Joined Mar 31, 2012
27,942
Capital letter if it's called after a person, lower case if it isn't.
In general, yes. I seem to recall that there's an exception, but I can't recall what it is.

With scaling prefixes, if the prefix is larger than 1 it is capital and if it is smaller than 1 it is lowercase. The exception is k for x1000 due to K being the unit for kelvins.
 
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