bridgecircuit analysisdiodespower electronicsrectifier
For the pictures above, I am confused as how to how they got/derived the formulas for the peak to peak ripple voltage (VrPP) and Vdc. So far, I got the output voltage, which is 14.9 volts. However, I am also confused with that VrPP and Vdc mean as well.
Another question: How exactly do the ripples occur, and what would happen if there wasn't the resistor (RL), or a resistor when there's a capacitor?
Where is the circuit GND reference connected in your simulation circuit? If it is connected to one side of the AC source then this could explain the behavior that you are seeing. The same thing that was noted in this question:
Diode Bridge in Simulator not behaving as expected
Update
Upon seeing the actual simulation circuit I can see that the reference point GND is indeed tied to one side of the AC source. Replace the connection from the bottom side of your AC source over to the connection between D2 and D3 where it should be. You have not constructed a correct bridge rectifier as shown.
I don't get what is the relation between the dc output voltage and Vs
Firstly, the output voltage is not "+,- 15V" - it is +15V dc (relative to ground/earth).
Secondly, the "+" and "-" labels on the transformer secondary winding are meaningless in this context - the secondary voltage is alternating and labeling them as "+" and "-" is confusing.
If your output dc voltage is +15V then, in simple terms, the secondary ac voltage has a peak value of about 15V. If your secondary voltage was less then your dc output voltage would be less.
Diode volt-drop - diodes aren't perfect conductors of electricity in one direction; they do "lose" about 0.7V and this means for a +15V dc output, the ac voltage feeding the bridge is more like 16.4V peak.
To calculate what the transformer secondary ac RMS voltage is, you divide the calculated peak voltage (16.4V) by \$\sqrt{2}\$ and this is 11.6V RMS.
Ripple - if you didn't have a load resistor, the capacitor would become charged to 15V and remain at that voltage. However, the resistor discharges the capacitor over every half cycle of the AC secondary voltage - this is what causes ripple voltage: -
To calculate ripple you need to know what R and C values are and here's another picture with a more detailed analysis: -
Best Answer
The non-technical answer:
Ripples occur because the input is alternating current. 60 times per second the voltage goes from zero to maximum (~1.414 * 115 VRMS, because RMS is sort of an average, rather than the peak, ~163 V), then back to zero and then to -163 V (below zero). The full wave bridge rectifier "flips" the negative portion over to positive, but the input now is going from 0 to peak voltage on each half cycle, 120 time per second.
The job of the capacitor is to store some electricity from the peaks to fill in the gaps where the voltage falls below that peak. With no load at all, i.e. no resistor, the capacitor would charge to the peak voltage and stay there with no ripple. Because the resistor bleeds out some charge, the voltage drops a bit between peaks, as the graph shows, and then the charge is replaced when the input goes up again.
Put simply, a larger capacitor can store more electricity to better fill the gaps. A smaller resistor allows more charge to flow, so the voltage drops more from the peak value.
If your teacher has mentioned it, include in your calculation that there's a voltage drop inside each leg of the rectifier, about 0.6 V for each if "ordinary" silicon (Si) diodes are used (as opposed to Ge, SiC, CuO, Schottky Si, etc.). Since there are two in series in each direction, that's an additional drop of ~1.2 V below the peak output of the transformer.