I am making a rectifier for a single N-Scale train building. It takes DC, but the train controller gives out 19VAC. I have 4 1N4001s and some capacitors, but probably not the right size/uf. I only need to make the 19VAC into 19VDC for this one model. I do not know its current draw however.
using Q = C *V, take the derivative with respect to Time, and get
dQ/dT = C * dV/dT + dC/dT * V
Now define dV/dT to be zero (the capacitance is constant), and we have the useful ripple prediction formula
dQ/dT = C *dV/dT
Now define dQ/dT to be the current (I) and we have
I= C * dV/dT
So what? Let I = 1 ampere, dV = 1 volt, dT = 1/(2*50Hertz) = 0.01 second
What must the smoothing capacitor be?
1 amp = C_farads * 1 / 0.01
and we have
1= C* 100,
C= 1/100 or 10,000uF
Example with ONE UF and 1 AMPERE load: