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.
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.
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.
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.
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