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Calculating Full Wave Rectifiers
Most power supplies are made of three sections: power transformer, rectifier and filter.
The transformer provides separation from the power line, voltage step-up for the rectifier tube or bridge rectifier and filament winding(s).
The rectifier section is used to convert the a-c voltage into d-c voltage. As explained in the tutorial, the full wave bridge rectifier is the prefered rectifying method.
All measured ac voltages are rms voltages, as measured with an voltmeter.
where Esec = transformer secondary rms voltage.
Peak Secondary Voltage Epeak = Esec * 1.414
Peak Inverse Voltage PIV PIV = 2 * Epeak
In a center-tapped full-wave rectifier the PIV is twice the peak value of the output voltage.
Output rms Voltage Eout = Esec/2
Peak Rectified Voltage Vpeak = Epeak / 2
Include diode voltage drop: (bias 2 * 0.7V)
Vpeak = (Epeak / 2) - Vbias
Esec = transformer secondary rms voltage.
Peak Secondary Voltage Epeak = Esec * 1.414
Peak Inverse Voltage PIV PIV = Epeak
Peak Rectified Voltage Vpeak = Epeak - Vbias
Output Voltage of an Capacitor-Input Filter
Vdc = (1-(0.00417/Rl*C)) * Vpeak
RMS Ripple Voltage Vr = (0.0024/Rl*C) * Vpeak
Output Voltage Ripple Factor r = Vr/Vdc
Percent Ripple = r * 100
The following data is known:
Calculate the PIV, output dc voltage, ripple factor and percent ripple.
Let's say, we have a 12kOhm load resistance and want maximum 1% percent ripple.
Filter Capacitor C = 0.0024 / Rl * r
The choke input filter is shown in Fig.3
RMS Ripple Output Voltage of the LC Filter Vr(out) = (Xc/(Xl-Xc)) * Vr(in)
To the dc voltage, the choke presents a winding resistance Rw in series with the load resistance Rload. This resistance produces an undesirable voltage drop and therefore Rw must be small compared to load resistance.
LC Filter DC Output Voltage Vdc(out) = (Rload/(Rw + Rload) * Vdc(in)
Calculate output dc voltage, rms ripple voltage, ripple factor.
The PIV rating of the bridge rectifier diodes is half that required for the center-tapped configuration.
Capacitor Input Filter
Where: Rl = load resistance in Ohms C = filter capacitor in Farad
The lower the ripple factor, the better the filter action.
Calculating a Center-Tapped Capacitor Input Rectifier
Transformer secondary voltage is 400V
Filter capacitor is 47uF
Diode voltage drop is 1.4V
Output Load is 5 kOhm
Epeak = 400 * 1.414 = 566V
PIV = 566 * 2 = 1132V
Peak output voltage Vpeak = 566 / 2 = 283V
Peak Rectifier Output Voltage = 283 - 1.4 = 281.60V
Filter Output Voltage Vdc = (1-(0.00417/5k *47uF)) * 281.60 = 276V
RMS Ripple Voltage = (0.0024/5k*47u) * 281.60 = 2.87V
Ripple factor r = 2.87 / 276 = 0.0104
Percent Ripple = 0.0104 * 100 = 1.04%
Calculating Required Filter Capacitance
C = 0.0024 / 12000 * 0.01 = 20uF
This is the minimum value for C. Increasing this value will decrease percent ripple.
A very large filter capacitor however, produces an initial surge current that could destroy the rectifier diodes if the forward current rating is to low. In this case, a current limiting resistor should be used.
Choke Input Filter
The inductive reactance and the capacitive reactance form an ac voltage divider which determines the magnitude of the ripple voltage.
Calculating a Full-Wave Choke-Input Filter
The following data is known:
DC Input peak voltage is 185V
Choke Impedance is 2H
Choke Ohm Resistance is 120 Ohm
Filter capacitor is 47uF
Output Load is 5 kOhm
Vdc(in) = Vavg = 2 * Vpeak / Pi = 118V
RMS ripple voltage Vr = 0.308 * Vpeak = 57V
Dc Output Voltage = (Rload / (Rw + Rload) = 5000 / 120 + 5000 = 115V
For ripple calculation, we need Xl and Xc.
Xl = 2 Pi f L = 6.28 * (120Hz*2H) =1507 Ohm
Xc = 2 Pi f C =6.28 *(120Hz*47uF) =28 Ohm
RMS ripple output voltage Vr(out) = (Xc/(Xl-Xc)) * Vr(in) = (28 / (1507 - 28)) *57 = 1.07V rms
Ripple factor r = 1.07 / 118 = 0.009 Percent ripple is = r*100 = 0.9%
Please report any errors or omissions.
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