Basic DC Power Supply Design

Author : Andy Collinson
Feedback: anc@mitedu.freeserve.co.uk

Overview
This article covers the design of basic unregulated DC power supplies. It is not designed to replace the training given by any college or university, but you may find it a useful study supplement.

Load Requirements
Before designing any power supply the load requirements must be known. It is always a good idea to take the worst case scenario when making this decision. For example if your circuit is designed to draw 1 amp at 12 volts, assume that component tolerances are 20% and design to meet these requirements with at least 20-50% reserve current, in this example I would design a power supply which could safely deliver 12 volts at 1.5 amps without overheating.

Transformer Regulation and Efficiency
A transformer is very efficient at converting AC voltages and currents from one value to another. In practice efficiencies of 98% may be achieved, the losses being due to heating effects of the transformer core, winding loss and leakage flux. Transformers have VA ratings which is simply the secondary voltage multiplied by secondary current. Not often published are the regulation figures for a typical transformer. A transformer rated at 20 V , 1 A secondary will only measure 20 volts when it is actually delivering 1 A. The figures below show typical regulation figures for some common VA rated transformers:-

VA rating	6	12	20	50	100
% regulation	25	12	10	10	10

For example a 12 VA rated transformer would have a no load voltage which is 12% higher than the rated value. If the transformer was rated at 12 V, 1 A, then measuring the secondary RMS voltage with a high impedance meter, you would measure approximately 13.44 Volts.

Rectification
This is the process where alternating current is converted to direct current. There are three main types of rectifier circuit, half wave, full wave and bridge.
Half-Wave:
The half wave rectifier circuit is shown in Fig. 1 below:

Fig. 1

The DC output is approximately:

1.41 x V_AC - 0.7

where V_AC is the RMS transformers secondary voltage and 0.7 the voltage drop across the rectifier. A typical waveform is shown in Fig. 2 :

Fig. 2

Full-Wave:
The full wave rectifier circuit is shown in Fuig. 3. This circuit uses a centre tapped transformer, alternate diodes conducting for each half cycle.

Fig. 3

The DC output is approximately:

1.41 x V_AC - 0.7

where V_AC is the RMS transformers secondary voltage and 0.7 the voltage drop across the rectifier. There are twice the amount of "peaks" compared to the half wave rectifier because alternate diodes conduct for each half cycle of the AC input. A typical waveform is shown below (Fig. 4.):

Fig. 4

Bridge Rectifier:
The bridge rectifier is the most popular of rectifier circuits. It uses four diodes arranged in a ring, but complete four terminal bridge rectifiers are also available. The circuit is shown at Fig. 5 below:

Fig. 5

The DC output is approximately:

1.41 x V_AC - (2 x 0.7)

where V_AC is the RMS transformers secondary voltage and 0.7 the voltage drop across the rectifier.As there are two diodes conducting for each half cycle, then there will be two rectifier voltage drops. This is around 0.7 volts but can rise to 1.1 volts per diode for high current rectifier types. A typical waveform is shown in Fig. 6:

Fig. 6

Smoothing Capacitor:
The "raw" DC produced after rectification is Ok to charge a battery or light a lamp but any electronic circuit needs a smooth DC supply. In the case of audio circuits, particularly amplifiers, any unfiltered DC will be heard as a "hum" in the equipments loudspeakers. The hum is proportional to the AC power supplies frequency. Smoothing is accomplished by placing a large value capacitor in parallel with the load, as seen in Fig. 7 below.

Fig. 7

The resulting waveforms are drawn in Fig. 8 below. The "brown" waveform represents the filtered DC feeding the load resistor.

Fig. 8

Ripple Voltage
The rectifier diodes will charge up the filter capacitor, C1 to the peak DC value, and between non conducting cycles of the diodes, will discharge into the load resistor. This creates the sawtooth waveform known more commonly as ripple voltage. The value of the ripple voltage is dependant on load current, power supply frequency and capacitor value. Approximate ripple voltage is calculated using:

where V is ripple voltage (mV), I is DC load current (mA), f is frequency of AC supply and C is smoothing capacitor value (F). For countries using a 50Hz supply like the United Kingdom, then the following simplified equation will also give the same results:

where V is ripple voltage (V), I is load current (mA) and C is smoothing capacitor (uF).

Example: The bridge rectifier circuit above had a load current of about 191mA. Feeding this value into the first equation results in 191/(2 x 50 x 2200 e-6) =868.1 mV and the bottom equation (10 x 191)/2200 =0.868V or 868mV

Average DC Value
The mean or average dc value is the value measured by a meter or multimeter. With an oscilloscope you will see the peak to peak waveforms and the oscilloscope horizontal cursor can be used to measure the dc value. The mean dc value for a full wave or half wave unregulated supply is difficult to predict because as the load current changes, so does the peak to peak ripple voltage. However, with reference to the half wave rectified diagram below :-

The mean or average dc value (shown in yellow) lies midway between the peak to peak ripple (shown as a blue dash line). For half wave rectified supplies the average dc value (with light shown) is calculated using :

For full wave or bridge rectified circuits the average dc voltage is calculated by:

V_p is the peak voltage value (also the maximum ripple voltage), C is capacitance (F), I is load current (A) and f is the supply frequency (in Hz)

Diode Conduction and Peak Diode Current
This is not often seen but is illustrated with this Tina diagram below. The circuit is shown in Fig. 7. It shows an output load voltage (in brown) with a little over 0.7 volts peak to peak ripple. It represents a full wave rectifier circuit with a 2200u capacitor and two rectifier diodes, D1 and D2 have their current waveform superimposed onto the graph shown in Figure 9 :-

Fig. 9

As can be seen, both diodes conduct rapidly on the leading edge of the ripple waveform charging the smoothing capacitor. D1 will charge the capacitor on the positive half of the input cycle, the capacitor then discharges through the load before the next half cycle where D2 will charge the smoothing capacitor and so on. The time interval T1 is the time each rectifier diode conducts. The peak current through each rectifier is much higher than the load current and can be calculated with the following equation:-

where T₁ = diode conduction time
           T = 1/f =1/50 for UK 50Hz line voltage
        I_dc = average load current
     I_peak = peak current through the conducting rectifier

Load Regulation
Because the output voltage will drop, when any load current is drawn, load regulation is the term used to describe this fall. The load regulation of a power supply is defined as the percentage change in output voltage when the load current is increased from zero to full rated output.

Summary
Designing an unregulated supply is not difficult providing you know the load requirements. Start with the load, allow for maximum current and make sure that the transformer has the correct VA rating to supply this power at the rated load. If the supply is used for an audio circuit than adequate filtering must be used, to avoid mains hum in the audio circuit.

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