Bridge rectifier AC - DC Converter with capacitive input filter

Discussion in 'Homework Help' started by User_axpro, Apr 15, 2013.

1. User_axpro Thread Starter New Member

Apr 15, 2013
8
0
After reding a few other similar threads im stillabit stuck on designing a bridge rectifier circuit ( I blame bad teaching with lack of understanding), Here it go's:
I have been asked to design a circuit to the following parameters to help with my understanding:

Load voltage of 50v, load resistance is 20Ω, ripple factor of 0.01 and input voltage of 240v 50Hz.
• I need to determine the minimum current rating for the transformer secondary
• PIV rating
• Value of series resistor
• Minimum capacitor value and operating voltage.
A) i grasp the concept of the minimum current as that will be 50/20 = 2.5 amps.
B) Right so PIV rating, to figure this out we require VS,would VS just be Vd.c = 50v so V.Ac = 50/0.9 = 55.55v which is rms right ?
We have been taught in lesson strangley vs = 1.5 x the 50v quoted in the question but i cannot decide is the would be a peak value (x0.414 + safety margin and diode drop? )

assumming it is 50/0.9 vs = 55v rms.
so PIV rating = 55 x √2 = 77.78v.
Safety margin for PIV rating = 77.78v x 2 = 155.56v.
C) Rs = (VS peak - 1.4 -vl ) / I
Rs = (77.78v - 1.4 - 60) / 2.5
Rs = 6.552Ω

D) Remembering 0.289 is a constant and rectified requency = 50Hz*2
C = 0.289 / (F*(RS+RL)*R)
C = 0.289 / (100*(6.552+20)*0.01)
c = 10884 μF
Voltage rating = working voltage + safety margin so would be around 100v

The only real difficulty is working out the correct value for Vs, however when i simulate my results on a simulation package the value for Vd.c output is wrong. is this a limitation of simulation?
*Edit: Simulation package used was livewire and multisim yet same results.

Last edited: Apr 15, 2013
2. User_axpro Thread Starter New Member

Apr 15, 2013
8
0
After simulating yet again i have came to the conclusion that this is the correct process, feel free to post your replys,
If wondering how i have came to the conclution that V.DC = 0.9 A.C:
Assuming Vs is in rms
Vpk=sqr(2) Vrms
Vdc = 2Vpk/pi = [2sqr(2)Vrms]/pi = 0.9Vs