Unregulated Power Supply Design

Article : Andy Collinson
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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.

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 -- this is strictly true only if the attached load is purely resistive (i.e has a power factor of 1.0). A reactive load containing capacitors or inductors (which one would expect for such a power supply) has a low power factor (i.e. less that 1.0) and thus de-rates the transformer's power capacity to the stated VA multiplied by the power factor because it draws more current than a purely resistive load. So, when choosing a transformer for a reactive load, one needs to divide the load in watts by the load's power factor to arrive at the VA needed which has sufficient "headroom" to accommodate the low power factor.

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, when measuring the secondary RMS voltage with a high impedance meter, you would measure approximately 13.44 Volts.

AC Waveforms
Before looking at rectification, some general information about AC waveforms. Figure 1 below shows two cycles of a sinusoidal waveform. The vertical axis shows amplitude and the peak to peak value (Vpp), shown by the pink arrow is 20 V pp. The peak value (VPK) is half the peak to peak value and is shown by the red arrow. The horizontal axis shows time.

One complete sinusoidal cycle consists of a positive "peak" and a negative peak or "trough". One cycle is also 360° or 2π radians, a half cycle (VPK or -VPK) is 180° or π in radians. Usually waveforms are displayed with horizontal axis in units of time.

The waveform below shows two cycles each with a duration of 1ms. As frequency is the reciprocal of time, then this waveform has a frequency of 1kHz. Figure 1

RMS Value
The RMS or ROOT MEAN SQUARED value is the equivalent DC ( voltage or current ) which would provide the same energy to a circuit as DC voltage or current. In other words, if an AC sine wave has a value of 10 Volts RMS it will provide the same energy to a circuit as a DC supply of 10 volts.

 VRMS = VPK or VPk * 0.707 √ 2
Average Value
The AVERAGE value is normally taken to mean the average value of only half cycle of a sine wave. The average value of a complete sine wave is of course zero, as both halves are symmetrical about zero. Using only half a cycle, the average value (voltage or current) is always 0.637 of the peak value.
 VAVG = 2* VPK or VPk * 0.637 π
Peak Value
The Peak value of a sine wave is the maximum positive peak. Defined in terms of RMS voltage its value is: VPk = √ 2 * VRMS or VPk = 1.414 * VRMS Peak to Peak Value
The peak to peak value, Vpp, (sometimes wrote as VPk-Pk) is simply twice the peak value. The peak to peak value is the waveform that is displayed on an oscilloscope. RMS values are displayed by an AC multimeter.

Periodic Time
The time taken for one complete sinusoidal cycle, (both positive and negative peaks) is known as the periodic time, (T). The frequency, (F) of the wave is the reciprocal of 1 cycle. Conversely, the reciprocal of frequency gives the periodic time.
 T = 1 or F = 1 F T
The Mains (Line) Supply
In the UK, the domestic electricity supply (mains) is a sine wave with nominal value 240V RMS and a frequency of 50Hz or 50 cycles per second. In The USA the line voltage is 120V RMS at a frequency of 60Hz.
Examples
The Peak value of the UK mains is therefore 240 * 1.414 = 339.36V
The periodic time for 1 cycle of a 50Hz sine wave is 1/50 = 20ms

Rectification - Unfiltered Power Supplies
This is the process where alternating current is converted to direct current. Unfiltered, means that there is no smoothing capacitor present and the dc output will contain "ripples" at the line (mains) frequency. There are two types of rectification, half wave and full wave, also known as a bridge rectifier.

Half-Wave
The half wave rectifier circuit is shown in Fig. 2 below: Figure 2

The DC output across a resistive load, is approximately the value of a half cycle, less one diode drop. Rectifier diodes have a forward voltage that varies from about 0.7V to 1.1 Volts in high current rectifiers. Conduction occurs for only one cycle, so is not very efficient, also without a smoothing capacitor, the output is quite "lumpy". Often these are used in cheap car battery chargers where the quality of the supply is not too important.

Full-Wave using Centre Tapped Transformer
A full wave rectifier circuit using a centre tapped transformer is shown below in Figure 3. This circuit uses just two diodes each one conducting on alternate half cycles. The positive side is marked with a "+" and the output waveform shown in figure 5. Notice that the output ripple is now doubled. Figure 3

Full-Wave or Bridge
The bridge rectifier is the most popular rectifier circuit. It uses four diodes arranged in a ring, but complete four terminal bridge rectifiers are also available. The circuit is shown in figure 4 below: Figure 4 There are twice the amount of "peaks" compared to the half wave rectifier because alternate diode pairs conduct for each half cycle of the AC input. A typical waveform is shown below (Figure 5): Figure 5 The blue trace is the peak to peak voltage of the transformers secondary winding, and the red trace is the unfiltered DC voltage. The DC output is approximately: 1.41 x VRMS - (2 x 0.7)

Rectification - Filtered Power Supplies
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 equipment's loudspeakers. The hum is proportional to the AC power supply's frequency. A filtered or smoothed supply is achieved by placing a large value electrolytic capacitor at the rectifiers output, as shown in Fig. 6 below.

Figure 6 The resulting waveforms are drawn in Fig. 7 below. The "brown" waveform represents the filtered DC feeding the load resistor.

Figure 7 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 dependent on load current, power supply frequency and capacitor value. Approximate ripple voltage is calculated using:

 Δ V = I 2fC

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:

 Δ V = 10I C

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. An oscilloscope shows 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 Figure 8 below:-

Figure 8 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 is calculated using :

 Vdc ≈ Vp − I 2Cf

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

 Vdc ≈ Vp − I 4Cf
Vp 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 2200uF capacitor and two rectifier diodes, D1 and D2 have their current waveform superimposed onto the graph shown in Figure 9 :-

rectifier diodes, D1 and D2 have their current waveform superimposed onto the graph shown in Figure 9 :-

Figure 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:-

 Ipeak = T Idc T1
where:
 T1 = Didode conduction time T = 1/f f being line frequency Idc = Average load current Ipeak = Peak current through the rectifier