Diodes, rectifiers and electronically applied math
I have questions answer any or all of them of them if you can
(1) When calculating the Voltage average of DC current that has been generated as a result of a AC current passing trough a rectifier how do you account for voltages that have have been limited as an example, 10 to 20 volts where the 0-10 is not included in the waveform, as opposed to 0-20 volts where the entire range 0-20 is in the waveform.
example: the equation used to calculated voltage is Vavg=2(Vpeak)/π(Pi) or
2 times the voltage peak divided by Pi
with a DC voltage that starts at 0V and peaks at 100V in a waveform the math to calculate the voltage average is :
63.7V= 2(100Vp)/Pi which is correct
But in a voltage that ranges from 10V-20V using the same equation looks like:
which is incorrect the correct answer is 16.4
how do you account for the elimination of 0-10 volts in the second problem and calculate the average for a waveform that has passed through a rectifier that has a waveform of 10-20 Volts my text offers no explaination but it does give an answer which is 16.4.
another question which I do not know if mathmatical mechanics would be the same ordifferent from solving the second problem is -15 volts to 25 volt with an answer of Vavg=10.5V, how do you arrive at 10.5 Volts?
(2) A center tapped tranformer used in a full-wave rectifier having a average voltage of 120V(out), what is the Vp across each half of the of secondary circuit.
Since the current has not yet been filtered it still has a wave form and the peak has be calculated using the reverse of the equation 2(Vpeak)/ π in order to determine Vpeak as in
which in an ideal circuit would account for Vpeak across each side of the circuit due to the alternating voltage on each side of the circuit through the two diodes in the circuit as they alternately become foward biased and negatively biased. but the diode drop is being taken into account.
anyway I thought that the only one diode drop would diode be taken into account well have a look:
the voltage drop is a result of the potential barrier of the diode
the voltage alternately is applied to two sides of the circuit with
V(secavg)-Diode Drop(0.7V)=V(outavg) &
V(secpeak)-Diode Drop(0.7V)= V(outpeak)
so why in my text would they apply the equation
when only one diode drop is accountable for a voltage drop from
V(sec) to V(out) and why would the subtraction occur within the factoring instead of after. It makes absolutely no sense to me at all why on the way out only one diode drop is accounted for but on the way back in a diode drop from each diode is accounted for.
(3)When diodes in a bridge rectifier are all swithced from their original position and changed to the opposite polarity how does it affect the waveform.
(4)The rms voltage is 20V in a bridge rectifier, what is the PIV?
( I have an answer for this one but nothing to check it against)
Last edited by Seth.Galassi; 01-26-2011 at 10:23 PM.
|applied, diodes, electronically, math, rectifiers|
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