mading2018
Member
Hello,
I have some concern regarding an equation for calculating power losses for half  and full diodebridge.
Can someone please check if this equation is correct or not? See the attached image.
"
Fig. 1. Typical diode fullwave bridge.
Fullwave diode bridges are found in many electronic systems (Fig. 1). At typical AC power line voltages, the drop across the diodes has little impact on the rectified output voltage and diode power dissipation. However, diodes in low voltage, high power rectification applications dissipate significant power and the inherent diode drops take a significant bite out of the operating voltage. For example, power dissipation of the diode bridge in Fig. 1 is:
Equation: P = 2 × 0.6 V × IL (1)
Where:
IL = Line current in amperes
P = Power dissipation in watts
0.6 V = Typical voltage drop for one conducting rectifier diode"
What I understand from this equation, this gives total power losses for a whole bridge, by multiplying with 2.
If I had multiplying with 1, I would obtain half of the power losses (which would be for a halfdiode bridge).
I really need a confirmation of this soon...
Ref:
http://www.powerelectronics.com/pmi...ercutspowerdissipationimprovesefficiency
I have some concern regarding an equation for calculating power losses for half  and full diodebridge.
Can someone please check if this equation is correct or not? See the attached image.
"
Fig. 1. Typical diode fullwave bridge.
Fullwave diode bridges are found in many electronic systems (Fig. 1). At typical AC power line voltages, the drop across the diodes has little impact on the rectified output voltage and diode power dissipation. However, diodes in low voltage, high power rectification applications dissipate significant power and the inherent diode drops take a significant bite out of the operating voltage. For example, power dissipation of the diode bridge in Fig. 1 is:
Equation: P = 2 × 0.6 V × IL (1)
Where:
IL = Line current in amperes
P = Power dissipation in watts
0.6 V = Typical voltage drop for one conducting rectifier diode"
What I understand from this equation, this gives total power losses for a whole bridge, by multiplying with 2.
If I had multiplying with 1, I would obtain half of the power losses (which would be for a halfdiode bridge).
I really need a confirmation of this soon...
Ref:
http://www.powerelectronics.com/pmi...ercutspowerdissipationimprovesefficiency
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