Case of the missing power.
In Marty Brown’s book, “Power supply cookbook second edition”, page 36. He gives some formulas for topologies to determine switching current, switching voltage, and rectifier current.
For a half bridge topology he says the current in the power switch is.
Iswitch = 2Pout/ Vin Min and the rectifier current = Iout
Assuming the following.
Full wave center tap out, with single output.
1000W out.
80vdc out.
Ideal diodes no voltage drop.
perfect power supply with 170VDC on split capacitors, from no load to full load.
Iswitch = (2 x 1000W out) / 170vdc = 11.76A
We have 11.76A at 170vdc on the primary = 2000W
Now lets look at the secondary.
Isecondary = 1000Wout / 80VDCout = 12.5A
M Brown says Irectifier = Iout, so Irectifier = 12.5A = secondary current.
With 80vdc out then 80VDC x .707 = voltage on secondary = 56.56VAC
The secondary power is 56.56VAC x 12.5A = 707W
So the transformer has 2000W in and 707W out. Were did the power go?
Remember that in a FWCT output each half of the secondary conducts on each half cycle and supply’s the 56.56VAC and 12.5A for that half cycle. So each winding supply’s the full power output alternately. Or in other words you do not add up both winding’s for this power calculation.
In Marty Brown’s book, “Power supply cookbook second edition”, page 36. He gives some formulas for topologies to determine switching current, switching voltage, and rectifier current.
For a half bridge topology he says the current in the power switch is.
Iswitch = 2Pout/ Vin Min and the rectifier current = Iout
Assuming the following.
Full wave center tap out, with single output.
1000W out.
80vdc out.
Ideal diodes no voltage drop.
perfect power supply with 170VDC on split capacitors, from no load to full load.
Iswitch = (2 x 1000W out) / 170vdc = 11.76A
We have 11.76A at 170vdc on the primary = 2000W
Now lets look at the secondary.
Isecondary = 1000Wout / 80VDCout = 12.5A
M Brown says Irectifier = Iout, so Irectifier = 12.5A = secondary current.
With 80vdc out then 80VDC x .707 = voltage on secondary = 56.56VAC
The secondary power is 56.56VAC x 12.5A = 707W
So the transformer has 2000W in and 707W out. Were did the power go?
Remember that in a FWCT output each half of the secondary conducts on each half cycle and supply’s the 56.56VAC and 12.5A for that half cycle. So each winding supply’s the full power output alternately. Or in other words you do not add up both winding’s for this power calculation.
