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Half Wave Diode Rectifiers
Half wave, full wave, and bridge rectifiers peak values and time interval equation(s)?
I need to calculate the Peak voltage and time interval between pulses for a Half wave, full wave, and bridge rectifier. all three start with an input value for 120V sin wave at a frequency of 60 HZ and all three have 1 watt diode(s). and i need to take voltage drop into account.
Now i'm confused what equations to use for all three? do they all have the same equations? or are there different equations for each one?
i know to find peak voltage of a sin wave it's the v/sqrt(2)
but i think there is more to that because it doesn't seem right for all three to have the same peak voltage to me.
And to find the time interval between pulses i have no idea what equation(s) to use at all.
any help at all would be appreciated, even just a starting equation would be a big help.
120 V rms (AC) has a peak value of sqrt(2)*120 V. A diode's worth of voltage drop is whatever your teacher said to use; if there was no word it's typically 0.5 to 0.7 V for Si diodes. So the peak voltage output is sqrt(2)*120 V minus the number diodes' worth of voltage specified below.
A half-wave rectifier conducts only half the cycle and the voltage drop is 1 diode's worth.
A full-wave rectifier conducts the whole cycle and the voltage drop is also 1 diode's worth.
A full-wave bridge rectifier conducts the whole cycle and the voltage drop is 2 diode's worth.
For half-cycle conduction, pulses occur at 60 Hz, so the pulse period is 16.67 ms.
For full-cycle conduction, pulses occur at 120 Hz, so the pulse period is 8.33 ms.
If by "time interval between pulses" you mean not the pulse period but the zero-voltage portion between pulses, that's pretty complicated. Without any diode drops, it's just 8.33 ms in the half-cycle rectifier and zero in the others. With diode drops, you have to add the time interval when the absolute value of a 120 V sine wave is less than the total diode drop. It's doable. Just find the arcsine of the total diode drop/120, then multiply that by 2*16.67 ms/(2pi).
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Half Wave Rectifier
Probability.. Can someone help me? Please?
Suppose the input-output relationship for a half-wave diode rectifier can be modeled as
Y=aX^2, X>=0
=0, X<0
where a>0 is given. If the input has a zero-mean, unit-variance Gaussian density function, find Py(y), the density of the output voltage.
X ~ N(0, 1) and X^2 ~ Chi-Square with 1 Degree of Freedom
Therefore,
P(Y <= y) = P( X^2 <= y/a ) = 1/Γ(1/2)∫[t^(-1/2)e^(-t), {t, 0, y/2a} ] = Erf{ √[ y/(2a) ] }
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