   ## Power Supply for Integrated circuit (ICs) and Microprocessor

Since a power supply is a vital part of all electronics systems. Most digital ICS, including microprocessor and memory ICS , operate on a ± 5-V supply, while almost all linear ICS (op-amps and special-purpose ICS) require ± 15-V supplies. Therefore, the power supply presented in this section will have ±5 and ±15 V.

### Working of the system

Figure 1-1 shows the block diagram of a typical power supply. The schematic diagram of the power supply that provides output voltage of ±5-V at 1.0A and ± 15 V at 0.500 A is shown in Figure 1-2. In this figure two separate transformers are used because they are readily available; however, it is possible to custom design a single transformer with the same specifications to replace the two. The supply voltages are obtained from a 26.8-V center-tapped (CT) transformer, and the supply voltages are obtained from the 12.6-V CT transformer. The output of these secondaries is then applied to the bridge rectifiers, which convert the sinusoidal inputs into full-wave rectified outputs. The filter capacitors at the output of the bridge rectifiers are charged to the peak value of the rectified output voltage whenever the diodes are forward biased. Since the diodes are not forward biased during the entire positive and negative half-cycle of the input waveform, the voltage across the filter capacitors is a pulsating dc that is a combination of do and a ripple voltage. From the pulsating dc voltage, a regulated dc voltage is extracted by a regulator IC. ### Circuit Description

Consider first how the ± 15-V supply voltages are obtained in the circuit of Figure 1-2. The 7815 is a +15-V regulator, the 7915 is a -15-V regulator, and both can deliver output current in excess of 1.0 A, They will hence perform satisfactorily in the circuit of Figure 1-2 by providing ±15 V at 0.500 A. However, since the drop-out voltage (Vin – V0) is 2 V, the input voltage for the 7815 must be at least +17 V and that for the 7915 must be at least — 17 V. This means that the rectified peak voltage must be greater than +17 V and — 17 V, which in turn implies that the secondary voltage must be larger than 34 V peak or 24 V rms. The voltage across the center-tapped secondary in Figure 1-2 is 26.8 V rms, thus satisfying the minimum voltage requirement of 24 V rms. Also, the peak voltage between either of the secondary terminals and the center-tap (ground terminal) is 18.95 V peak, which is less than the maximum peak voltages of +35 V and -35 V for the 7815 and 7915, respectively.

Note that the voltages across the two halves of the center-tapped secondary are equal in amplitude but opposite in phase. During the positive half-cycle of the input voltage, diode D1 conducts and capacitor C1 charges toward a positive peak value =18.95 V. At the same time, diode D3 is also conducting; hence capacitor C3 charges toward a negative peak value = -18.95 V. This means that the voltage across nonconducting diodes D2 and D4 is 37.90 V peak, which implies that the peak-reverse-voltage (PRV) rating of the bridge rectifiers must be larger than 37.90 V peak or 26.8 V rms. The PRV rating of the bridge rectifier diodes, also known as a working inverse voltage (WIV), is specified on the data sheets. The bridge rectifier, MDA200 (Mot0r0la’s rectifier) in Figure 1-2, has a PRV rating of 50 V, which is higher than needed. This bridge rectifier is, in fact, used here because it is readily available and more commonly used.

During the negative half-cycle of the input waveform, diodes D2 and D4 conduct and charge capacitors C1 and C3 toward the peak voltage of 18.95 V with indicated polarities. Note, however, that the diode pair that conducts during either the positive or negative half-cycle does not do so for the entire half-cycle. The diodes conduct only during the time when the anodes are positive with respect to the cathodes. In other words, when the diodes are forward biased, the capacitors are charged by current pulses. Data sheets give the maximum average rectified current I0max that the diode can safely handle. For the MDA200,Iomax is 2.0 A. In addition, when the power supply is first turned on, the initial charging of the capacitor causes a large transient current called the surge current to pass through the diodes. The surge current IFS flows only briefly and is therefore much larger than the maximum average current I0max. The maximum surge-current IFSM is normally included on the data sheets; it is 60 A for the MDA200.

Finally, the size of the filter capacitor depends on the secondary current rating of the transformer. As a rule of thumb, a 1500-µF capacitor should be used for each ampere of current. The working voltage rating (WVDC) of the capacitor, on the other hand, depends on the peak rectified output voltage and must be at least 20% higher than the peak value of the voltage it is expected to charge to. Capacitors C1 and C3 satisfy these requirements (see Figure 1-2). Capacitors C2 and C4 at the output of 7815 and 7915 regulators, respectively, help to improve the transient response and should be in the range of 1µF.

Next consider the ±5-V supply. The circuit arrangement of the ±5-V supply is identical to that of the ±15-V supply except that here the specifications for the transformer T2 secondary are different. Therefore, the operation and considerations for the ±5-V supply are the same as those presented for the ±15-V supply. The voltage regulators in Figure 1-2 will require heat sinks. Let us examine why. The power dissipated by the 15-V regulators is as follows:

Power dissipated = (dropout voltage)(current)

= (18.95 — I5)(0.5) = 1.98 W

Similarly, the power dissipated by the 5-V regulators is

(8.91 - 5)(1.0) = 3.91 W

Therefore, for the proper operation the regulators must be heat-sinked in order to keep their temperature down. If a regulator is a metal package (TO-3 type), the appropriate heat sink is mounted on the case of the package. However, if the regulator is an epoxy package, a silicon grease may be used on the back of the package, and then the package can be bolted to the chassis of the power supply cabinet with insulating hardware.

Besides the ±15 and ±5-V regulated supply voltages, there is often a need for a 60-Hz square-wave signal, which is used as a time base in scanning the digital displays and as a trigger for sequential and timing circuits. If needed, a 1-Hz (1-s) signal for the real-time clock can be readily obtained from the 60-Hz signal by using a divide-by-60 network. Although not commonly done, a higher-frequency Signal can also be obtained from the 60-Hz signal by using e multiplier. For these reasons, in Figure 1-2 a 60-Hz square-wave signal is produced by using two small-signal diodes and a 555 timer as the Schmitt trigger

### PARTS LISTS

Resistors (all ¼-watt, ± 5% Carbon)

R1 = 10 kΩ

#### Capacitors

C1, C3 = 1500 µF

C2, C4, C6, C8 = 1 µF

C5, C7 = 3000 µF

#### Semiconductors

IC1 = MC7815

IC2 = MC7915

IC3 = MC7805

IC4 = MC7905

IC5 = NE555 timer

D1 – D2 =MDA200 PVR = 50 V, I0max = 2.0 A, IFSM = 60 A

D5 – D8 = MDA970A1 PVR = 50 V, I0max = 4.0 A, IFSM = 100 A

D9, D10 = 1N914 signal diodes

#### Miscellaneous

Transformer T1 = Primary: 117 V, 60 Hz: Hobart P-300

Secondary: 26.8 V CT, 1.0 A

Transformer T2 = Primary: 117V, 60 Hz: Hobart P-305

Secondary: 12.6 V CT, 2.0 V

Fuse 0.750 A slow blow

Switch On-off toggle type

Silicon grease with insulating hardware or four het sink for Voltage regulator

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