A half wave rectifier circuit is with 1K ohm load operates from 120V 60Hz household supply through a 10:1 stepdown transformer. The dive drops .7 Vwhen forward bisaed. A capacitor C is chosen to porvide a peak to peak ripple voltage of 10%of the peak output voltage. A) Calcualte the peak input voltage and peak voltage after the transformer. B) Calculage the peak output voltage C) Calculage the capacitance C.
2006-08-04 12:24:58 · 8 answers · asked by Mr.answers 2 in Science & Mathematics ➔ Engineering
Here is the corrected question, sorry about the spellings!
Mr.answers
6 minutes ago
An Engineering question!!!?
A half wave rectifier circuit is with 1K ohm load operates from 120V 60Hz household supply through a 10:1 stepdown transformer. The diode drops .7 V when forward biased. A capacitor C is chosen to provide a peak to peak ripple voltage of 10% of the peak output voltage.
A) Calcualte the peak input voltage and peak voltage after the transformer.
B) Calculage the peak output voltage
C) Calculage the capacitance C.
2006-08-04 12:33:53 · update #1
A)
120V refers to the RMS voltage
Vpeak=Vrms*sqrt(2) = 120*sqrt(2) = 170V approximately
since the transformer is stepping down by a factor of 10, that means the peak voltage on the other side of the transformer is
170/10 = 17V
B)
Since the diode drop is 0.7V, the peak output is approximately
17V-0.7V=16.3V
C)
The ripple voltage is given by
Vripple/Voutpeak=1-e^(t/( Rload * C ))]
where t is the discharge time of the capacitor. In a 60Hz half-wave rectifier, t is 16.67ms.
solve for C using
Vripple/Voutpeak=10% = 0.1
t = 16.67ms
Rload=1Kohm
e=natural logarithm base = 2.71828 . . .
which should give you (if I did the calculation correctly, you may want to check me)
C=1.58E-4 F = 158uF
2006-08-04 13:53:02 · answer #1 · answered by hypa_dude 2 · 2⤊ 0⤋
Sounds like a simple Ohm's law question. I'm off work for the weekend. Sorry. Done with that noise for a couple days.
2006-08-04 13:12:14 · answer #2 · answered by Anonymous · 0⤊ 1⤋
a) the first one A
2006-08-04 12:28:03 · answer #3 · answered by jack jack 7 · 0⤊ 1⤋
The information provided is insufficient.
We need to know the load current.
2006-08-04 13:51:01 · answer #4 · answered by dmb06851 7 · 0⤊ 1⤋
Are you nipping a little, or did you just doze off during your English classes?
2006-08-04 12:42:59 · answer #5 · answered by Goat 2 · 0⤊ 0⤋
1.- 1697.04 Volts, 169.70 volts.
2.- 162.70 Volts
3.- 1200 uF
2006-08-04 13:40:52 · answer #6 · answered by dianameza 4 · 0⤊ 1⤋
Hey - watch your language. There are kids reading this forum!
2006-08-04 12:28:19 · answer #7 · answered by Jay 6 · 0⤊ 1⤋
I am confused by your question, because there are a lot of mis-spellings.
2006-08-04 12:28:26 · answer #8 · answered by ricoinc 2 · 0⤊ 1⤋