Hi Pommie,
What is the source of the 12V? Is it a transformer?
If you are using a transformer as the voltage/current source, the output voltage will also drop as a result of the current loading on the transformer.
What value and type of reservoir capacitor are you using?
As you no doubt know, in a rectifier/reservoir capacitor circuit, the current is supplied to the reservoir capacitor in huge gulps, at the peaks of the rectified voltage camel-humps. The size of the current depends, to a first approximation, on the source impedance, band-gap voltage of the rectifier diodes, dynamic impedance of the rectifier diodes, and the impedance of the reservoir capacitor.
By the way, the band-gap voltage of silicon semiconductor drops by approximately 2mV/Deg C. So, with a diode operating at a typical junction temperature of 125 Deg C, the band-gap voltage would be reduced by 200mV from the band-gap voltage at 25 Def C.
The peak current, rather than the reservoir capacitor drain current, should be used to determine the rectifier diode forward drop, or MOSFETs Vds, rather than the reservoir load current. A good working estimate of the peak current is ten times the reservoir load current, 100A in this case.
If you use MOSFET active rectifiers, this high current needs to be taken into account. Using MOSFETs, would give you a forward voltage drop (Vds) of,
Vds= Ipeak * Rdss * 2
Where,
Vds : Voltage between drain and source of the MOSFET in Volts.
Rds : Resistance between drain and source of the MOSFET in Ohms.
2 : MOSFET channel temperature/Rds factor
Take a 0.010 Ohm Rds MOSFET,
Vds =100A * 0.01 * 2 = 2V.
Then multiply that by 2, because you would have a minimum of two MOSFETs in series to form a bridge rectifier (I think this is correct, but have not checked), and you get 4V, so MOSFET active rectifiers may not be the answer.
If you do decide to use schottky rectifiers, I would advise going for as higher current rectifiers as you can reasonably afford/fit, say 100A upwards. The next best thing is to parallel diodes to get a lower forward drop.
The ripple voltage across the reservoir capacitor will also subtract from the average output voltage. To a first approximation, the peak-to-peak ripple voltage in Australia (240V, 50Hz) can be calculated by,
Vripple = (I * 0.01)/ Cres
Where,
Vripple: Peak to peak ripple voltage across the reservoir capacitor in Volts
I: Load current flowing out of the reservoir capacitor in Amps
Cres: Reservoir capacitor capacitance in Farads
spec
DATASHEETS
https://www.onsemi.com/pub_link/Collateral/MBR1635-D.PDF