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Need a dual supply regulator in a hurry but don’t have any LM317 or 7805 3-terminal regulators handy? This simple circuit can provide regulated supply rails from ±5V to ±12VDC at up to 800mA. I F YOU’RE NOT in the component buying business, then you’ll probably be unaware that there has been a severe world-wide shortage of parts during the last 12 months – particularly 3-terminal regulators. Now since these devices get a guernsey in just about every project designed, we recently decided to see if we could come up with some sort of replacement based on readily available components. Since doing this work, the supply situation has vastly improved but we still feel that the design may be suitable for many applications. The fact that it uses only junkbox parts is a major plus. All the parts for the regulator are built onto a small PC board. This contains everything necessary to convert the AC voltage from a centre-tapped mains transformer to regulated plus and minus DC supply rails, including a bridge rectifier and filter capacitors. It will provide an output voltage of between ±5V and ±12V DC at currents up to 800mA. Circuit diagram Fig.1 shows the circuit diagram for the Dual Regulated Power Supply. It uses four power diodes, an LM358 dual op amp, a zener diode, a couple of transistors and sundry resistors and capacitors. Power is derived from a 12-24V centre-tapped mains transformer. Its output is fed to a bridge rectifier consisting of diodes D1-D4 to produce positive and negative rails which are then filtered using two 470µF electrolytic capacitors. These rails are then fed to the collectors of transistors Q1 and Q2 respectively and are also used to power the dual op amp (IC1). Discrete dual supply voltage regulator By DARREN YATES The assembled PC board can form the basis of a simple variable power supply or can be used to provide fixed regulated supply rails from ±5V to ±12V DC. April 1994  29 R2 10k 6 4x1N4004 A 240VAC N D4 R1 10k D1 5 F1 2A 6-12V Q1 BD139 8 7 100  IC1a LM358 B E D4 .047 1N4148 1k C +VOUT 0V 6-12V D3 470 25VW D2 10 16VW ZD1 4.7V 400mW 470 25VW Fig.1: the regulated positive supply rail is derived by using ZD1 to set a reference voltage on pin 5 of inverting amplifier stage IC1a. This in turn drives current amplifier stage Q1. Inverting amplifier stage IC1b & current amplifier Q2 are used to derive the negative rail. GND PLASTIC SIDE 100k 470 25VW E C 470 25VW 100k -VOUT .047 2 B 1 IC1b 3 100  4 F2 2A B Q2 BD140 E C DUAL REGULATED POWER SUPPLY IC1a and its associated zener diode (ZD1) form the heart of the regulation circuit. This op amp is connected as a bootstrapped-diode reference source and drives current amplifier stage Q1. IC1b, on the other hand, simply functions as a unity gain inverter stage; it drives current amplifier Q2 Zener diode ZD1 functions as the reference element and is part of a positive feedback path around IC1a. This feedback path may not be all that clear at first glance – it starts at the output of IC1a (pin 7) and goes via the 100Ω resistor, the base-emitter junction of Q1, diode D4 and the 1kΩ resistor, before ending at the non-inverting input (pin 5). This loop ensures that the output voltage remains constant. The 10µF capacitor across ZD1 filters out any noise on the line and improves the regulation. Note that it is necessary to include the transistor (Q1) in the feedback loop so that the op amp can compensate for the voltage drop across the base-emitter junction to give the required output voltage. Setting the output voltage for the positive rail is now just a case of selecting the negative feedback network to set the gain of IC1a. This feedback network consists of two resistors (R1 and R2) connected in the usual way; ie, one from the output to the inverting input (pin 6) and the other from the inverting input to ground. 30  Silicon Chip The formula for the output voltage is: Vout = 5.3V x (R2 + R1)/R1 where the 5.3V reference is equal to the voltage across ZD1 plus the voltage across D4 (ie, 4.7 + 0.6 = 5.3V). With the current values, IC1a’s gain is set to two and so the output voltage is set to 10.6V. However, this can be easily PARTS LIST 1 PC board, 04103941, 107 x 53mm 6 PC stakes 2 M205 (2AG) fuse clips 2 M205 2A fuses 2 Micro-U heatsinks 1 centre-tapped mains transformer to suit (see Table 1) Semiconductors 1 LM358N dual op amp IC 1 BD139 NPN transistor 1 BD140 PNP transistor 1 4.7V 400mW zener diode (ZD1) 4 1N4004 rectifier diodes 1 1N914 signal diode Capacitors 4 470µF 25VW electrolytic 1 10µF 16VW electrolytic 2 0.047µF 63VW MKT polyester Resistors (0.25W, 1%) 2 100kΩ 1 1kΩ 2 10kΩ 2 100Ω altered by changing the value of R1, R2 or ZD1. The negative rail is much simpler to produce because all we need do is invert the output of the positive rail. This is done by feeding the voltage on the emitter of Q1 to the inverting input (pin 2) of IC1b via a 100kΩ resistor. As previously mentioned, IC1b functions as a unity gain inverting amplifier. Its output at pin 1 drives PNP power transistor Q2 via a 100Ω current limiting resistor. As before, the output transistor is included in the feedback loop to ensure that its base-emitter voltage is compensated for. In this way, the negative rail mirrors the voltage on the positive rail. The two .047µF capacitors connected across the base-emitter junctions of Q1 and Q2 reduce the sensitivity of the circuit to noise or glitches and improve the regulation. The final outputs are taken from the emitters of Q1 and Q2 and filtered by two 470µF capacitors. A maximum of 800mA can be supplied by both sections. Construction All of the components for the Discrete Power Supply, including the two 2A fuses, are installed on a PC board coded 04103941 and measuring 107 x 53mm. Before you begin construction, it’s a good idea to check the PC board TABLE 1 1k R2 10k F1 D1-D2 470uF 0V IC1 LM358 D3-D4 470uF AC2 D5 470uF ZD1 R1 10k AC1 .047 100W Q1 10uF 100k +VOUT 0V 100k 470uF -VOUT Q2 Fig.2: install the parts on the PC board as shown here & note that small finned heatsinks should be fitted to Q1 & Q2. Resistors R1 & R2 are selected to set the required output voltage – see text. Fig.3: this is the full-size etching pattern for the PC board. for any shorts or breaks in the tracks. You can do this by carefully checking your etched board against the full-size pattern. Generally, there won’t be any problems here but it’s always a good idea to make sure. Begin the board assembly by installing the two wire links, followed by the resistors, diodes and capacitors. Be sure not to confuse the zener diode with the rectifier and signal diodes. After that, install the IC and power transistors. Be particularly careful with these components – check the orientation of the IC carefully and note that the transistors are installed with their plastic faces towards the adjacent .047µF capacitors. Transformer 5V 12V CT 6V 15V CT 8V 18V CT 12V 24V CT .047 100  F2 DC Output (V) Note also that Q1 is an NPN transistor while Q2 is a PNP type, so be sure to use the correct transistor at each location. Finally, solder in six PC stakes at the external wiring points, install the fuse clips and bolt two small finned heatsinks to the power transistors. There’s no need to isolate the transistors from the heatsinks but don’t let them short against any of the other parts on the board. To test the circuit, you need a centre-tapped mains transformer (or you can use an AC plugpack supply with a centre tap). Table 1 shows the transformer input voltage you need for a given DC output voltage. Wire up the secondary windings of the transformer to the PC board as shown on the overlay diagram and the primary to a mains terminal block. Warning: use extreme caution when installing the mains wiring – 240VAC can kill! The transformer and the PC board should be mounted inside a metal case and this must be securely earthed. Cover all mains connections with heatshrink tubing to avoid the possibility of electric shock. Before applying power, check your wiring carefully for any wrong connections. Once you’re sure that everything is OK, switch on and check the output voltage with your multimeter. If you have used the values shown on the circuit, you should get a reading of about 10.6V on both rails with respect to ground (this will depend on the exact voltage across the zener diode). If need be, you can substitute a trimpot for resistor R2 and trim the output voltage until it is exactly what you require. A variable supply By replacing R2 with a 20kΩ linear potentiometer, you can make a simple dual-tracking variable power supply capable of covering the range from ±5V to about ±12VDC. The circuit could thus form the basis of a very useful benchtop power supply for powering experimental lash-ups. For lower output voltages, you could replace ZD1 with a number of signal diodes. If only one diode is used, the output voltage will be about ±2.4V. Remember, the formula for the output voltage is: Vout = (VZD1 + VD4) x (R2 SC + R1)/R1. RESISTOR COLOUR CODES ❏ ❏ ❏ ❏ ❏ No. 2 2 1 2 Value 100kΩ 10kΩ 1kΩ 100Ω 4-Band Code (1%) brown black yellow brown brown black orange brown brown black red brown brown black brown brown 5-Band Code (1%) brown black black orange brown brown black black red brown brown black black brown brown brown black black black brown April 1994  31