This project is the next step of the following project: https://easyeda.com/sasadangelo/transformer In it, we have a Power Source 230V AC 50 Hz (typical power source in Italy) that is converted in 12V AC by a transformer with a ratio of 0.053 (12/230). The sinusoidal signal goes in input to a Graetz bridge that let's pass the positive wave and invert the negative wave. The output signal is 100Hz and at each instant, the output voltage has a 1.4 drop because it passes trough two diodes. Vrms in output to the transformer is 12V AC then Vpt=Vrms*SQRT(2)=16.97 (Volt peak transformer) considering that this receives a 1.4 drop in the output the Graetz Bridge we will have Vpout=Vpt-1.4V=15,57V (Volt peak output) that means Voutrms=Vpout/SQRT(2)=11V (Volt arms output). That is the theory, the reality is a bit different because there is some tolerance and diode works a bit different from the ideal. First of all, looking at the project we will see that Vrms=12.115V instead of 12V. Looking at the Graetz Bridge datasheet we will see: https://www.diodes.com/assets/Datasheets/ds21219.pdf Vfm=1.05 (max) instead of 1.4 and its value depends on the current, this means that the value could be different if we use a 100k, 10k, 1k, 100, 10 ohm resistors. Vpt=12.115*SQRT(2)=17.13V instead of 15.57. However, if you look at the project and the waveform volProbe1 you'll notice that the wave is almost 0.5V less on the peak and the minimum value is -0.5V. The reason is that the other terminal is -Vdrop diode less than GND. Vpout=Vpt-1.05=16.1V and Voutrms=Vpout/SQRT(2)=11.37V very close to the value 11.49V provided by the multimeter. Looking at volProbe2 we will see a Vpout=16.2 instead of 16.1 calculated. However, we have to consider the tool tolerance, see discussion here: https://easyeda.com/forum/topic/Voltage-drop-on-graetz-81f36e9708af49e3b49753ca91f85c85 The question now is how the Vpout and Voutrms change for 10, 100, 1k, 10k, 100k resistors. 10-ohm resistor -> Vpout=15.45V Voutrms=10.72 Iout=1A -> Vdrop=1.42V 100 ohm resistor -> Vpout=15.73V Voutrms=10.98 Iout=100mA -> Vdrop=1.14V 1k ohm resistor -> Vpout=16.08V Voutrms=11.24 Iout=11mA -> Vdrop=0.79V 10k ohm resistor -> Vpout=16.2V Voutrms=11.49 Iout=1.1mA -> Vdrop=0.67V 100k ohm resistor -> Vpout=16.59V Voutrms=11.74 Iout=117uA ->Vdrop=0.3V As we can as the current increase (the resistor decrease) the Voltage drop on the diodes increases and we know this from the theory. Looking at the numbers the drop goes beyond the 1.05 declared in the datasheets and arrive to the 1.4 of the ideal diodes.