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

#### 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?

#### ericgibbs

Joined Jan 29, 2010
16,835
hi Mark,
Q=C * V
Q= I * time
so C= (I*time)/V
W=V * I
So I= 1W/10v =0.1A

Can you calculate from that.?

E

• Mark Flint

#### BobTPH

Joined Jun 5, 2013
6,095
dV / dt = i / C

0.1 / 10e-6 = 0.1 / C

C = 10e-6 = 10 uF

• Mark Flint

#### 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

• Mark Flint

#### Mark Flint

Joined Jun 11, 2017
136
Using the formula provided by eric
C = t x I / V = 10 x 0.1 / 0.1 μF = 10μF
So is 10uF the amount of drop of capacitance that a 0.1 drop of voltage cause in that time period (10 microseconds)?

#### 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

• Mark Flint and BobTPH

#### BobTPH

Joined Jun 5, 2013
6,095
The capacitance does not drop.

That is the capacitance required to provide 0.1A of current for 20 microseconds while dropping the voltage by 0.1V

• shripatil82 and Mark Flint

#### ericgibbs

Joined Jan 29, 2010
16,835
hi Mark,
Graphic plot of the circuit.
E

#### crutschow

Joined Mar 14, 2008
31,151
That is the capacitance required to provide 0.1A of current for 20 microseconds while dropping the voltage by 0.1V
I think you mean 10μs.

#### Mark Flint

Joined Jun 11, 2017
136
hi Mark,
Graphic plot of the circuit.
E
Great - much appreciated Eric. #### 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

#### ericgibbs

Joined Jan 29, 2010
16,835
hi Mark,
Consider:
C = (10^-5 * 0.1A)/0.1V = 10^-6/ 0.1A = 10^-5 Farads.

E

#### crutschow

Joined Mar 14, 2008
31,151
If I wanted to calculate this and change it from 10 microseconds to 10 milliseconds would t = 10000 instead of 10?
Yes, which means the capacitor would need to be 1000 times as large, or 10mF for the same voltage drop.

#### 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.

• Mark Flint

#### 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)

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: ????

• Mark Flint

#### Ian0

Joined Aug 7, 2020
6,737
Capital letter if it's called after a person, lower case if it isn't.

• Mark Flint

#### 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.

#### Ian0

Joined Aug 7, 2020
6,737
In general, yes. I seem to recall that there's an exception, but I can't recall what it is.
Maybe you're thinking of the rule that says it is not capitalised if written out in full.