The bridge rectifier, or a full-wave bridge rectifier, is made up of four separate p-n junction diodes, an alternating supply, and a load resistor. Bridge rectifiers or diode bridge rectifiers have four diodes that create a closed loop, referred to as a bridge. The fundamental benefit of the bridge rectifier circuit is that it does not require a central tape transformer, which decreases its size.
The input of one side of the bridge is connected to the single winding. The load resistor is on the other side of the bridge, as shown in the bridge rectifier diagram below:
The following different terminologies and units are used for the calculations around alternating voltages:
These values can be easily converted to each other. The way formulas for these conversions differ upon the the waveform that applies (sinus, triangle, square). The relevant basic conversions are:
Vp = Vpp / 2 Vdc = Vavg = ( 2 * Vp ) / π Vrms = Vp / √(2) Vrms = Vpp / (2 * √(2)) Vpp = 2*√(2) * Vrms Vdc = Vavg = Vp / 2 Vrms = Vp / √(3) Vp = Vpp / 2 Vdc = Vavg = Vp Vrms = Vp
After the rectifier the output voltage is a positive pulsed DC as shown in the picture below.
The output voltage of the rectifier is lower +/- 1.4V lower then the input voltage. This is due to the forward-biased (conducting) voltage over two diodes. Due to voltage over the diodes the maximum peak value is:
Vp_out = Vp_in - 2 * 0,7V The term Vin defines the voltage coming from the secondary windings of the transformer (or input voltage). The average Vdc value can be calculated with the formula below:
Vdc_out = Vavg_out = 2 * Vp / π Using Ohm’s law to derive the current, we should note that two types of resistance will limit the current, the load resistance (RL), and the forward resistance of the diode (Rf). Note: We can find the forward resistance using the diode’s I-V characteristic. The maximum current can be calculated with:
Imax = Vmax / ( 2 * Rf + Rl ) Where Rl is the load resistance, Rf is the forward resistance of the diodes. Vmax is the maximum AC voltage.
The value of the ripple voltage will depend on a number of factors:
We can improve the average DC output of the rectifier while at the same time reducing the AC variation of the rectified output by using smoothing capacitors to filter the output waveform. Smoothing or reservoir capacitors connected in parallel with the load across the output of the full wave bridge rectifier circuit increases the average DC output level even higher as the capacitor acts like a storage device.
For a full wave rectifier the ripple voltage can be calculated with the following formulas:
Vr = Vp / ( fr * Rload * C ) Vr = Iload / ( fr * C ) Where:
For a full wave rectifier with filter the ripple factor can be calculated with the following formula:
Ripple factor = 1 / ( 4√3fCR ) Calculate the ripple factor of the output waveform as the ratio of the ripple voltage or current (also known as the RMS voltage) to the DC voltage.
Ripple factor = √ ( ( Vrms / Vdc )² - 1 ) Ripple factor = √ ( ( Irms / Idc )² - 1 )
Please note: the calculations below only work if the RC time is » (x10) then the period T. E.g. in case of 100 hz (T=10ms) the RC should be in the range of 0,1 (100 ms) = 100 Ohm * 1000uF.
For a full wave rectifier with filter the output voltage can be calculated with the following formula:
Vdc = Vp - ( Idc / 4fC ) Substituting Idc with Vdc / R this becomes:
Vdc = Vp * 4fRC ) / ( 1 + 4fRC ) Based on the above formulas, the table below calculates the required tranformer voltage to achieve a specific DC voltage.
| Preferred Vdc | Stabilizer drop (V) | Vp | Rectifier Drop (V) | Trafo Vpp | Trafo RMS |
|---|---|---|---|---|---|
| -3 | -1,4 | ||||
| 5 | 8 | 8,0 | 9,4 | 18,9 | 6,7 |
| 9 | 12 | 12,1 | 13,5 | 26,9 | 9,5 |
| 12 | 15 | 15,1 | 16,5 | 33,0 | 11,7 |
| 15 | 18 | 18,1 | 19,5 | 39,0 | 13,8 |
| 16 | 19 | 19,1 | 20,5 | 41,0 | 14,5 |
| 18 | 21 | 21,1 | 22,5 | 45,0 | 15,9 |
| 20 | 23 | 23,1 | 24,5 | 49,0 | 17,3 |
| 24 | 27 | 27,1 | 28,5 | 57,1 | 20,2 |