I need a capacitor with X capacitance that can supply 50mA for 10 sec, max voltage 10V. I must buy one like this for my project. Would somebody please calculate the required capacitance? As I said, 10V is the max voltage as I will probably work with 7-9V. Thank you.
C=dV/dT
A capacitor will drop in voltage very quickly. you need to specify a start and an a minimal voltage to get any useful answers
aarg:
C=dV/dT
?
Wouldn't that T, whatever it means, have the unit kg^2 m^4/(s^7 A^3)?
ElCaron:
A capacitor will drop in voltage very quickly. you need to specify a start and an a minimal voltage to get any useful answers
?
Wouldn't that T, whatever it means, have the unit kg^2 m^4/(s^7 A^3)?
No, T is just time in seconds. Also I screwed up. It's 1/C=dV/dT
aarg:
No, T is just time in seconds. Also I screwed up. It's 1/C=dV/dT
So that is a "yes".
After the correction, T has the unit of C * V (which I guess is voltage U, not volume?). Volt*Farad is Coulomb, not seconds.
I think more relevant here would be
U(t) = U_0 * exp(-t/(R*C)) (which I guess is the solution of the correct differential equation)
If you allow a voltage drop of 20% and assume that the 50mA drop accordingly on an Ohmic resistor, this would require around 280mF.
But I am pretty sure if the OP told us what he plans to do, we would have a more sensible approach. Else, he would be quite fine with a Farad range supercap.
Above equation, btw, leads to
dU(t)/dt = - U_0/(RC) exp(-t/(RC))
= U(t)/(R*C)
= I(t)/C
Well you can get picky about whether you need the exponential or linear equation, but the outcome is very similar and for ball park calculations of ripple, the fundamental equation is very close. Not all loads are resistive, the OP didn't specify what it is.
Also the choice of nomenclature is academic. A beginner is not going to understand that U represents voltage.
aarg:
Well you can get picky about whether you need the exponential or linear equation, but the outcome is very similar and for ball park calculations of ripple, the fundamental equation is very close. Not all loads are resistive, the OP didn't specify what it is.Also the choice of nomenclature is academic. A beginner is not going to understand that U represents voltage.
As far as I can see, your equation is just wrong. Not even the units check out. From the exponential equation, I showed how the differential equation has to look. You are missing an I. (Misread I/C for 1/C?)
Can you demonstrate how to calculate anythting with the version that you stated?
You have showed the differential equation with an R. I already pointed out that for the case in point, it is not necessary to achieve the desired precision. In fact, not every circuit has an R in it. Yes, it is I/C = dV/dt not 1/C = dV/dt. The OP wants to start at 10V and end up no lower than 7. That is a delta of 3V. The current is 50mA. The time delta is 10S. Therefore the required cap value is C = (dV/dt)/I = (3/10)/0.05 = 6F.
For Christ's sake, stop trying to do math 
C = (dU/dt)/I is very obviously wrong because it would imply that a higher voltage drop would lead to a higher capacity.
It is C = I/(dU/dt) which not only has the right dependency on he quantities and the right unit (and this, kids, is why your physics teacher wanted you o do the units calculation), but is also consistent with my value with 167mF.
aarg:
C=dV/dT
T is temperature, not time. T is the dimension of time, but this is a differential equation, not a dimensional
equation. And no, C != dV/dt
However C = dQ/dV, but more than that C = Q/V directly. Everything else follows, capacitance is charge per unit
voltage, differentiate once for the rates of change:
C V = Q therefore C dV/dt = dQ/dt = I therefore dV/dt = I/C
All assuming C is constant, which it is for most capacitors. Otherwise there would be a V dC/dt term.
MarkT:
C V = Q therefore C dV/dt = dQ/dt = I therefore dV/dt = I/C
This is what I eventually posted in reply #7. It would result in C = 1/6F = 0.167F. It makes a gross simplification that the current is constant. What I find strange is that the discharge curve for a constant current should be faster than for a resistive load because the current tapers off as the voltage decreases. This would seem to imply that you would need a smaller capacitor for a resistive load, to intersect the 7V mark at the same time. But in reply #4 it's given as 280mF. That is greater than the value I came up with. Why?
Because I calculated with a 20% drop and you for 3V? Because I understood his 10V as a capacitor rating - with a little bit two small headroom - his 7-9V for a range of a fixed value he wants to operate at and his question as a failure to understand that he will not get a constant voltage from a cap.
Anyway:
I calculated
0.8U0 = U0exp(-t/(RC))
=> ln(0.8) = -t/(RC)
<=> C = -t/(Rln(0.8))
with R=U_0/I_0 = 8V/50mA = 160Ohm:
C = -10s/(160Ohmln(0.8)) = 280mF
with U_10s = 6.4V
If you calculate that with a constant I of 50mA, you get
C = 50mA/(1.6V/10s) = 313mF
So as you expected, with a constant current you need a bigger cap.