1
\$\begingroup\$

In my book it says that in a diode rectifier the capacitor gets charged close to the peak of the ac voltage.

  1. Is it true the larger the (filter) capacitor the larger the peak current drawn and if so why?

  2. And why is it that when the capacitor gets charged to a value close to the peak of the ac (input) voltage that the current through the rectifier is very large near the peak of the 50-Hz ac (input) voltage ?

\$\endgroup\$
1
\$\begingroup\$

The current, i, through a capacitor, with capacitance C, with a voltage, V, across it is given by:

$$ i = C \frac{dV}{dt} $$

Thus the current is proportional to the value of the capacitance.

\$\endgroup\$
  • \$\begingroup\$ thank you, and why is it then that a larger capacitor also reaches its peak current faster compared to a smaller capacitor? \$\endgroup\$ – Jantje7600 Feb 16 '15 at 17:56
  • \$\begingroup\$ Charge/discharge time constant in seconds is the product of capacitance times the circuit resistance. In this time interval, the capacitor is charged to about 63% of the final voltage, or discharge to 37%. \$\endgroup\$ – GR Tech Feb 17 '15 at 5:38
0
\$\begingroup\$

That is correct. Considering two capacitors with the same ESR, its impedance will be the following:
                                                           Equation
If the capacitance of the first capacitor is greater than the second one, the impedance of the first is lower. The consequence is that for the same voltage, the first capacitor will draw more current from the voltage source.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.