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Single diode (or half-wave) rectifier will produce on a resistive load an average voltage Vdc of 0.45xVac and average dc current Idc of 0.64xIac ..
Full wave bridge on resistive load gives you (average) Vdc=> 09Vac and Idc=>0.9Iac ..
In both cases, when you add a smoothing cap the voltage (with no load) rises to Vdc=> 1.41xVac .. and currents (1diode) Idc=> 0.28xIac and (4 diodes bridge) Idc=> 0.62xIac ...
This schematic shows a half-wave power supply with a load sufficient to draw down the supply level to 110 VDC.
Some amount of impedance is usually in the power supply.
I installed a 3 ohm resistor to represent this.
Anyway it limits current to a reasonable amount, because much more current would be liable to trip a circuit breaker.
Technically speaking, a single-phase (one-diode) rectifier will charge the DC-output capacitor to the peak voltage of 110 VAC, 160 V DC approx. If you discharge this capacitor by a load current, the voltage will decrease. This effect depends upon the capacitor value and load current.
You can use a two-phase(bridge) rectifier to reduce the voltage decrease by load current.
As you presumably prefer to use no AC insulation transformer, the complete device is VERY DANGEROUS.
I will never recommend to anyone to even test it. Please ALWAYS use insulation transformers, to separate AC mains from the rectifier ad the load.
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