I have a test 2morrow and I am having problems with the wording of this practice problem... can someone please explain and or walk me through it...
What peak to peak sinusoidal voltage must be connected to a half wave rectifier if the rectified wave form is to have a dc value of 6 V ? Assume that the forward drop across the diode os 0.7 V.
Am I looking for Vpr (peak value of rectified waveform), or Vpp (peak to peak voltage)... Do i have to do anything with Vrms? I am so confused
2006-09-20 10:24:51 · 7 answers · asked by Oo 2 in Science & Mathematics ➔ Engineering
I found the answer to your problem. I am not sure how the book came up with this but it is correct. Vdc=Vs/pi-Vd/2 where pi=3.14 Vd is diode voltage drop. The Vs/pi is easy but I am not sure where the Vd/2 comes from. So the answer is 19.95V. I am 100% sure this is correct if you take into consideration the Vd.
2006-09-20 18:37:32 · answer #1 · answered by DoctaB01 2 · 0⤊ 0⤋
The Vpr will be 1/2 of the p-p input voltage, less the voltage drop across the diode. therefore you would need 2(6.7)V p-p or 13.4 Volts. A half wave rectifier passes either the positive or the negative half of a sine wave, depending on how you hook it up.. Since there is some loss across the diode, you must account for that
2006-09-20 10:33:20 · answer #2 · answered by curious george 5 · 0⤊ 0⤋
You are not the only one...
Can only help with this...
13.4 Volts is the required peak to peak voltage for a motor vehicle.
with a a halfwave rectifier you get half of this so thats 6.7 Volts
Now subtract the forward voltage of the diode. that is the energy consumed by the Rectifier. (You don't think I'm doing this for nothing element) And you end up with 6v
2006-09-20 10:41:30 · answer #3 · answered by COLIN E 1 · 0⤊ 0⤋
Voltage does NOT remain the same. With no capacitor, the DC from a half wave rectifier is about 0.45 times the RMS AC voltage at the input. For a full wave, the DC is twice that or about 0.9 times the RMS AC voltage. In addition, the ripple frequency on the full wave rectifier is twice that of the half wave and easier to filter. When you add a capacitor, the DC output for both half wave and full wave rectifiers is about 1.4 times the AC RMS voltage, but the current capacity and ripple on the half wave is not as good.
2016-03-26 23:33:55 · answer #4 · answered by Anonymous · 0⤊ 0⤋
1. Draw out the input sinusoidal voltage and the output (rectified)aveform.
You want a average(dc) voltage of 6V, Therefore, you form an simple equation of:
6=(1/period)*integrate(Vpksin(wt)). (integrate over half a period)
For normalized form, one can take one period as 2*pi.
Taking into consideration the forward drop of the diode, one can modified it into:
6=(1/period)*integrate(Vpksin(wt))-0.7. (integrate over half a period) since 0.7 is treated as an dc offset.
2006-09-20 14:09:37 · answer #5 · answered by joy_chia 1 · 0⤊ 0⤋
Everyone is leaving off the DC part. you will need your half rectified sinusoidal to have an average of 6VDC not just a peak of 6V So now we are talking RMS.
2006-09-20 12:14:04 · answer #6 · answered by BrianW 3 · 0⤊ 0⤋
Here are several formulas for this type of rectifier.
http://www.visionics.ee/curriculum/Experiments/HW%20Rectifier/Half%20Wave%20Rectifier1.html
2006-09-20 10:30:36 · answer #7 · answered by Glenn N 5 · 0⤊ 0⤋