## anonymous 4 years ago explain the difference if any in ripple voltage between the full-wave and half-wave rectifier under load ?

1. anonymous

First we should derive the equation for ripple peak-to-peak voltage, so that we can see how full-wave and half-wave affect it. Lets assume that the ripple voltage has a triangle wave form and it takes half a period to fall from max value to min value (from higher peak to lower peak). So this is the derivation for a full-wave rectifier. Voltage across the capacitor is the same as the ripple voltage so we can write $\ U_C=\frac{Q}{C}$ Where Q - charge, C-capacitance Q is just: $\ Q=It$Where I-current, t- time Since our fall from upper peak to lower peak is half a period we can write $\ Q=I\frac{T}{2}$Where T- time period of our wave. Substituting this last expression into our 1st one we get:$\ U_c = \frac{IT}{2C}$and we know that $\frac{1}{T} =f$f- frequency So our final expression for ripple peak-to peak voltage is: $\ U_C = U_{Rpp}=\frac{I}{2fC}$ This expression is true for a full-wave rectifier. If you did the same calculation for a half-wave rectifier you would see that the time it takes to fall from upper peak to lower peak equals the full period T of our input AC signal and the expression for ripple PP voltage would be: $\ U_{Rpp}=\frac{I}{fC}$. In conclusion: Half-wave rectifier has a higher peak-to-peak voltage the full-wave rectifier. So which method you think is better for rectifying AC to DC?

2. anonymous

one more thing is that how u will define vp-p voltages for capacitor ? how capacitor gets charge fast before its RC time >??

3. anonymous

DUDE ripple cn calculate by vr(p-p) divided by vdc (dc input)

4. anonymous

We assume that the fall from higher peak to lower peak is linear because RC constant is large in comparison to the period of the ac waveform. I don't quite understand your 2nd question, but if I understood it correctly you are asking how the capacitor charges up so fast every cycle. When t=0 capacitor is empty and when you close the circuit it is starting to charge up. When it gets charged up it doesn't go back to no charge at all, it only looses some of it's charge (used for smoothing out the signal) and it gets charged up back quickly.

5. anonymous

gogind y capacitor get full charge in same time while v r applying different input source ?????

6. anonymous