Bridge Rectifier Calculator

A bridge rectifier converts the alternating current output from a transformer into direct current that can be used in a circuit. Four diodes perform this task and regardless of which direction the input voltage swings, two diodes will conduct current so that it always flows in the same direction to the connected load.

This calculator does not just calculate the average voltage like textbooks. It models the entire front end of a DC power supply and takes into account peak values, voltage drops, the capacitor used for energy storage, ripple in the output voltage, current that each diode must withstand, as well as the transformer powering these components.

How the circuit works:

Four diodes are connected in a loop, with the AC source connected to one diagonal and the load to the other diagonal. This connection was described by Leo Graetz in 1897 and is therefore known today as the Graetz bridge.

During the positive half-cycle a set of diodes are forward biased and conduct current. During the negative half-cycle these diodes switch off and another set takes over. As the other pair of diodes is connected in reverse direction, the current flows to the load always in the same direction regardless of the polarity of the half-cycle.

As a result, there is a series of voltage peaks whose frequency is twice the mains frequency. Although the polarity is uniform, the voltage is not stable. A capacitor connected in parallel to the consumer solves this problem. While the diodes conduct current, the capacitor charges up to peak value and then discharges the stored charge to the consumer during the breaks between the voltage peaks.

Basis for calculations:

It all starts with the peak value. Transformers are usually specified by RMS volts but the amplitude of a sine wave is about 1.414 times that.

In bridge circuits there are always two diodes in series with the load so you have to subtract twice the forward voltage from the peak value. In center-tap or half-wave rectifier circuits there is only one diode in the current path, so only one value is lost.

If there is no smoothing capacitor, the DC output voltage will be the average of these amplitudes. For a full wave rectifier this average value is twice the peak value divided by pi and for ideal diodes corresponds to about 0.9 times the RMS input voltage.

In half-wave rectification every other half-cycle is thrown away, halving the average value; this means it is equal to the peak value divided by pi and approximately 0.45 times the RMS input voltage. The difference in this factor is why a full-wave rectifier is preferred.

If a storage capacitor is added the situation changes dramatically. The output voltage no longer corresponds to the average of the amplitudes but remains slightly below the peak value and fluctuates slightly between peaks. The degree of this fluctuation depends on how much charge the load draws during that time.

These two equations are interrelated. The current depends on the voltage and the voltage depends on the ripple. By substituting one equation into another using Ohm's law this relationship can be closed so that the result is obtained with a single calculation without any iterations.

Here n is the number of output pulses within each input period. For a bridge circuit this is 2 while for a unidirectional circuit it is 1. This notation allows all three topologies to be handled with the same model.

Ripple factor, shape factor and other ratios

The ripple factor is the RMS value of the ripple divided by the DC level. For a rectifier without filter it can be expressed as a function of current. At the end all terms cancel out leaving only one constant.

If the values for a bridge circuit are used, one gets a value that can be found in many textbooks: 0.483, or 48.3%. For single-phase rectification it is 1.21, meaning that the ripple is greater than the DC level. By adding a capacitor, the ripple factor falls by two orders of magnitude. No one operates a power supply without a capacitor for this reason.

The efficiency of the rectifier is the proportion of power supplied to the load circuit that exists as direct current. For a bridge circuit, the maximum efficiency is:

In many textbooks this is rounded to 81.2%. In the case of half-wave rectification, only about half of that value can be achieved. This is because with the same peak current strength, twice as many zero crossings have to be balanced out.

Calculation:

We need a transformer for 12 V and frequency of 60 Hz, a typical silicon bridge rectifier, a capacitor to store energy with capacitance of 1000 microfarads and a load with resistance value of 100 ohms.

The peak value is 16.97V which is obtained by multiplying 12 with 1.414. As two silicon diodes cause a voltage drop of 1.4V, the remaining peak value for charging the capacitor is 15.57V.

The filter factor is obtained by dividing 4 x 60 x 0.001 x 100 by the same value and then adding 1, which results in a result of 24/25. The resulting DC voltage is therefore 15.57 x 0.96, or 14.95 V.

The load current is 14.95/100 = 0.149 A. During the time interval of 1/120 seconds between peaks this current draws 0.149/120 coulombs from the capacitor. Dividing by 1000 microfarads gives a ripple voltage of 1.25 V.

Therefore a transformer with a rated voltage of 12V supplies a current rail with a nominal voltage of 15V on which a sawtooth wave with an amplitude of 1.25V is superimposed. Anyone who has already experienced that a 12V transformer shows 17V when measured at no load will confirm this calculation in practice.

Selection of smoothing capacitor:

If you simplify the formula for noise, then you can directly calculate the capacitance of the capacitor. You choose an acceptable level of noise and calculate as follows:

The following table shows the results of a calculation for a full wave rectifier with a load current of 150 mA at 60 Hz. Note that the effect is rapidly diminishing. Increasing the capacitance from 470 to 1000 microfarads reduces the ripple by almost one volt, while increasing it from 2200 to 4700 only gives a reduction in ripple of about one third of the first change.

If a linear regulator is connected after the rectifier, it's important to consider not the average value but the minimum of noise. The voltage at the bottom must exceed a value that corresponds to the output voltage of the regulator plus the dropout voltage. Otherwise, in each period the regulator will lose control and let the noise through unfiltered.

Selection of diodes.

There are three important ratings. The maximum reverse voltage is the voltage that must be blocked by a diode when it is blocking, acting against its direction of conduction. In a bridge circuit this corresponds to one maximum value while in a center-tap circuit there are two maximum values. This is because the entire winding voltage will be applied across both ends of the diode when it is not conducting. A half-wave rectifier circuit with energy storage capacitor also requires two maximum values. This is because the capacitor fixes one pole at a value close to the positive maximum and the input voltage changes toward the negative maximum.

The maximum current in the forward direction is another nominal value. In a bridge circuit, each diode alternately conducts during half of the period and thus carries half of the DC load current. With a single half-wave rectifier circuit it carries the entire load current.

The third rating is the surge voltage withstand value, which can be the most insidious. When switching on, the energy storage capacitor appears empty and short-circuited, so that the initial current can rise to several tens of amperes within a few milliseconds. In the data sheets this is referred to as IFSM. Check this rating depending on the capacitance of the capacitor. If there is little room for manoeuvre, a small resistor or an NTC thermistor can be connected in series.

Selection of transformer power

Rectifiers with a capacitor input are not a simple load for transformers. The current flows only briefly and as large pulses near the peak values of each half cycle so that the winding's RMS current is significantly higher than the DC output of the power supply.

For a bridge rectifier with capacitor input in conventional transformer construction the RMS voltage of the secondary winding is chosen to be about 0.71 times the DC output voltage, the RMS current of the secondary winding is about 1.61 times the DC current and the apparent power (VA) is about 1.14 times the DC power. For inductor filter these factors change to a ratio of 1.11 for DC voltage and 1.06 for DC current. On the other hand, the current flow becomes more uniform which reduces transformer loading considerably.

The scope of this model is:

It is assumed that the discharge process of the capacitor between peaks is linear. In reality this process is exponential and the linear approximation is only accurate if the noise is less than about one tenth of the initial value. If this limit is exceeded, the calculation tool will display a warning and the actual noise will be slightly larger than the displayed value.

This model does not take into account the ESR (equivalent series resistance) of the capacitor, the recovery time of the diode, the winding resistance of the transformer or voltage regulation under load. As the voltage of a real transformer drops by several percent from no-load to full-load conditions, the DC values given here should be considered as an upper limit and safety margin should be planned in.

This is a single phase model. For three-phase rectifiers the number of ripple peaks per period is six and the original waveform is much smoother, so the constants given here cannot be applied directly.

Rectifier circuits connected to the mains can cause lethal voltages, and charged energy storage capacitors are dangerous even after long periods without power. These values should be considered as estimates for design purposes only and not used in place of appropriate testing or local installation codes.