electricpete
Electrical
- May 4, 2001
- 16,774
Here's my analysis of items a) through d):electronics tutorials said:Rectification Example No1
A single phase half-wave rectifier is connected to a 50V RMS 50Hz AC supply. If the rectifier is used to supply a resistive load of 150 Ohms. Calculate the equivalent DC voltage developed across the load, the load current and power dissipated by the load. Assume ideal diode characteristics.
First we need to convert the 50 volts RMS to its peak or maximum voltage equivalent.
a) Maximum Voltage Amplitude, VM
VM = 1.414*VRMS = 1.414*50 = 70.7 volts
b) Equivalent DC Voltage, VDC
VDC = 0.318*VM = 0.318*70.7 = 22.5 volts
c) Load Current, IL
IL = VDC ÷ RL = 22.5/150 = 0.15A or 150mA
d) Power Dissipated by the Load, PL
PL = V*I or I2*RL = 22.5*0.15 = 3.375W ≅ 3.4W
… [neglected diode voltage drop]
[ul]
[li]a) Vmax = sqrt2 times the rms of the sinusoid [ I AGREE][/li]
[li]b) "equivalent dc voltage".
Vdc = int(VM*sin(w*t) from wt=0..Pi) / (2*pi) = [-cos(Pi) – -cos(0)] / (2*pi) = 2 / (2*pi) = 0.318*VM – [ I AGREE, although I would stay away from the terminology "equivalent" without clarifying for which purposes it is equivalent][/li]
[li]c) load current – they call it IL but what they’re calculating is the average (dc) value IDC = <i(t)>. IDC = VDC / RL [I agree, with the caveat that their IL is really IDC][/li]
[li]d) Power dissipated by the load PL = VDC*IDC or IDC^2*RL = 22.5*0.15 = 3.375W ≅ 3.4W [I disagree]. That is the power dissipated by the DC component of the voltage and current. There are other components in the fourier decomposition of the voltage and current that also cause power dissipation in the resistor.[/li]
[/ul]
Compare a half-wave rectified voltage feeding a resistive load to a sinusoidal voltage feeding the same resistive load, where the peak voltage of both are the same (i.e. the sin would represent the input to the hwr). The hwr circuit provides half the power of the sin circuit (with diode voltage neglected, same assumption as stated on the webpage). That conclusion is evident because the instantaneous power of the hwr circuit is the identical to that for the sinusoid circuit during the half-cycle that the hwr is conducting, while it is zero for the other half. In a given period of time, the energy delivered by the hwr circuit is half that of the sin circuit. Since average power is energy per time, the average power in the hwr circuit is half that of the sin circuit.
For the particular example listed:
Power of the sin circuit would be Psin = Vrms^2 / R = 50^2 / 150 = 16.667 watts.
Power of the hwr circuit would be Phwr = 0.5*Psin = 8.333 watts.
My answer for part d is 8.3 watts which does not agree with their answer 3.4 watts.
I say the link is wrong (It has become a source of controversy in another thread). What do you say?
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(2B)+(2B)' ?