Figure 2.1 shows an unregulated AC/DC power supply that was common practice when I started my career.
Fig. 2.1: Simple linear AC/DC power supply
The transformer has two primary windings of 115V which can be connected in parallel or series with the input voltage selector switch. The two 6V secondary windings are wired in series to give a nominal 12VAC output which is then rectified by the bridge rectifier BR and DCsmoothed by the output capacitor, C, to give a typical output voltage of about 14VDC. The bridge rectifier uses four diodes, but there are other options for the secondary rectification using the same transformer and fewer diodes:
Fig. 2.2: Alternative output rectifier options: top: centre-tap, bottom: half-wave
If Vs is the nominal voltage for each secondary winding, then Table 2.1 can be used to work out the average DC output voltage (Vf is the forward voltage drop through a power diode ≈ 0.7V):
For the simple example shown in figure 2.2 with 50Hz mains and 2x 6VACrms secondaries, Table 2.1 becomes:
However, this average DC output value is calculated without the smoothing capacitor and without a load. The larger the capacitor, the closer the measured DC output will be to the peak voltage. On the other hand, the higher the load, the lower the measured DC output voltage will be. To determine the effective output voltage, we need to know the load and output capacitance in order to calculate the output ripple.
Returning to the original, full wave bridge rectified design, we can add the output waveform hown below:
Fig. 2.3: Output capacitor voltage and current waveforms
At the start of each half cycle, the output voltage rises from zero up to the peak voltage C.
Above point B (the residual voltage stored on the output capacitor), it starts to supply the load current and charge the output capacitor. The current (shown in blue) rises sharply. As the secondary voltage drops below its peak, the output capacitor holds the output voltage higher than the AC voltage and the diode bridge become reverse biased and ceases to conduct. The AC current falls to zero. The greyed-out part of the waveform from point C until point D shows where only the capacitor supplies the load current. The input current is thus highly discontinuous with a very high harmonic distortion level.
The line C-D is shown as a straight line on the diagram, although it is in fact an exponential decay curve with the relationship:
However, for practical circuits with large output capacitors, this expression can be approximated by:
With a peak-to-peak ripple voltage of:
And an average DC output voltage of:
So, for a 50Hz mains supply with a 1k ohm load and a 100μF capacitor, we could expect to see a measured DC output voltage of:
With a peak-to-peak ripple of approximately:
Fig. 2.4: Voltage on output capacitor, C, with load R
As the output voltage changes with load and has a high ripple, it is common practice to use a linear regulator to regulate the output and to provide output short circuit protection. For this example, a 12V regulator would be most suitable as the minimum supply voltage would be about 14V, giving 2V headroom for the linear regulator.
Linear power supplies are still used where their advantages outweigh their disadvantages:
Fig. 2.1: Simple linear AC/DC power supply
The transformer has two primary windings of 115V which can be connected in parallel or series with the input voltage selector switch. The two 6V secondary windings are wired in series to give a nominal 12VAC output which is then rectified by the bridge rectifier BR and DCsmoothed by the output capacitor, C, to give a typical output voltage of about 14VDC. The bridge rectifier uses four diodes, but there are other options for the secondary rectification using the same transformer and fewer diodes:
Fig. 2.2: Alternative output rectifier options: top: centre-tap, bottom: half-wave
If Vs is the nominal voltage for each secondary winding, then Table 2.1 can be used to work out the average DC output voltage (Vf is the forward voltage drop through a power diode ≈ 0.7V):
| Rectification Method | No. of Diodes | Output Freq. | Vpeak | VDC, av |
|---|---|---|---|---|
| Bridge Rectifier | 4 | 2Fin | 2√2Vs – 2Vf | 2 Vpeak/ π |
| Centre-Tap | 2 | 2Fin | √2Vs – Vf | 2 Vpeak/ π |
| Half-wave | 1 | Fin | √2Vs – Vf | Vpeak /π |
Table 2.1: Comparison of rectified DC outputs for different rectification methods
For the simple example shown in figure 2.2 with 50Hz mains and 2x 6VACrms secondaries, Table 2.1 becomes:
| Rectification Method | No. of Diodes | Output Freq. | Vpeak | VDC, av |
|---|---|---|---|---|
| Bridge Rectifier | 4 | 100Hz | 15.6V | 10V |
| Centre-Tap | 2 | 100Hz | 7.8V | 5V |
| Half-wave | 1 | 50Hz | 7.8V | 2.5V |
Table 2.2: Results of the calculations in Table 2.1 with 2x 6V secondary windings
However, this average DC output value is calculated without the smoothing capacitor and without a load. The larger the capacitor, the closer the measured DC output will be to the peak voltage. On the other hand, the higher the load, the lower the measured DC output voltage will be. To determine the effective output voltage, we need to know the load and output capacitance in order to calculate the output ripple.
Returning to the original, full wave bridge rectified design, we can add the output waveform hown below:
Fig. 2.3: Output capacitor voltage and current waveforms
At the start of each half cycle, the output voltage rises from zero up to the peak voltage C.
Above point B (the residual voltage stored on the output capacitor), it starts to supply the load current and charge the output capacitor. The current (shown in blue) rises sharply. As the secondary voltage drops below its peak, the output capacitor holds the output voltage higher than the AC voltage and the diode bridge become reverse biased and ceases to conduct. The AC current falls to zero. The greyed-out part of the waveform from point C until point D shows where only the capacitor supplies the load current. The input current is thus highly discontinuous with a very high harmonic distortion level.
The line C-D is shown as a straight line on the diagram, although it is in fact an exponential decay curve with the relationship:
| Eq. 2.1: |  |
|---|
However, for practical circuits with large output capacitors, this expression can be approximated by:
| Eq. 2.2: |  |
|---|
With a peak-to-peak ripple voltage of:
| Eq. 2.3: |  |
|---|
And an average DC output voltage of:
| Eq. 2.4: |  |
|---|
So, for a 50Hz mains supply with a 1k ohm load and a 100μF capacitor, we could expect to see a measured DC output voltage of:
With a peak-to-peak ripple of approximately:
Fig. 2.4: Voltage on output capacitor, C, with load R
As the output voltage changes with load and has a high ripple, it is common practice to use a linear regulator to regulate the output and to provide output short circuit protection. For this example, a 12V regulator would be most suitable as the minimum supply voltage would be about 14V, giving 2V headroom for the linear regulator.
Linear power supplies are still used where their advantages outweigh their disadvantages:
- As the power supply has only passive components, it is a low noise solution. A well designed linear regulated power supply can have a very smooth output with an output noise level below 5μVRMS. Linear power supplies are still used in high end audio systems and RF amplifiers.
- The same design can be used for very high input voltages by simply selecting different primary side voltage taps (e.g. 208V/380V/480VAC) or very low input voltages (e.g. 12VAC) by using a different transformer. It is still a technical challenge to make a switching power supply that will work well with a 12VAC input
- There are very few components to go wrong, so a well-specified linear power supply can have a working lifetime of more than 20 years.
- They are generally cost effective. However, due to the very high production volume of switching power supplies, the difference between a linear and switching solution is often very small.
The main reasons why linear AC/DC power supplies have been mainly supplanted by switching converters are the following:
- A 50/60Hz transformer is much bulkier and heavier than a transformer designed for switching power supplies. For example, a 10VA mains frequency transformer has a volume of typically 65cm³, whereas a 10W switching transformer can be built into a 2cm³ core – a saving of more than x30 in size and weight.
- 50/60Hz power supplies are inefficient. Power is transferred only at the peaks of the mains cycle – the remaining part of the cycle is not used. On the secondary side, the rectification diodes dissipate a significant amount of power due to the high capacitor charging peak currents. If linear regulators are used to stabilize the output, then the efficiency drops even lower. Overall efficiencies of below 50% are not unusual. In comparison, switching power supplies with efficiencies exceeding 90% are common.
- 50/60Hz power supplies have poor regulation. The output voltage is load dependent and also directly proportional to the input voltage. The hold-up time is short, so the output voltage will be adversely affected by mains brown-outs and any dropped cycles. The transient load response time is also very poor as the power supply must wait until the next AC peak to transfer any extra energy required to cope with a sudden load increase.
- The no load power consumption is too high to meet modern energy efficiency regulations. In addition, the fact that power is transferred only at the cycle peaks means that the power factor is also too low for many applications (a linear power supply has a power factor of typically 0.5 – 0.7)
- The cost of switching power supplies is now so low that a low power linear power supply may actually be more expensive that the more complex switching converter alternative.