Constant +/-9V DC Symmetrical Power Supply Circuit (Part 5/13)

In the previous project, a constant 12V DC power supply circuit was designed with a current limitation of 1A. In this project, a symmetrical dual power supply with constant voltage outputs will be designed. A dual symmetric power supply can provide two symmetric output voltages with opposite polarity with respect to a common ground reference. Every electronic circuit needs an adequate power supply at the input for optimal operation. The power supply of any device or circuit must be chosen according to its power requirements. In this project, a regulated power supply circuit is designed which can produce constant voltages of 9V and -9V with a maximum current of 1A.

The power circuit designed in this project uses voltage regulator IC 7809 and 7909 and uses conventional power circuit design steps such as reducing AC voltage, converting AC voltage to DC voltage and smoothing DC voltage to obtain direct input from the grid HERE.

Fig. 1: List of components required for constant +/-9V DC symmetric power supply

Fig. 2: Block diagram of constant symmetric power supply +/- 9V DC

The circuit is built in stages, with each stage aimed at a specific purpose. To step down the 230 V AC, a 12 V – 0 -12 V transformer is used. The terminals of the transformer's secondary coil are connected to a full-bridge rectifier and a wire is taken from the center strip of the transformer to serve as a common ground . The full bridge rectifier is constructed by connecting four 1N4007 diodes together designated as D1, D2, D3 and D4 in the schematics. The cathode of D1 and the anode of D2 are connected to one of the secondary coils and the cathodes of D3 and the anode of D4 are connected to the other terminal of the coil. The cathodes of D2 and D4 are connected, of which one terminal is taken from the output of the rectifier and the anodes of D1 and D3 are connected, of which another terminal is taken from the output of the full-wave rectifier.

A 1A fuse is connected in series to the full wave rectifier output for protection against AC sources. 470 uF capacitors (shown as C1 and C2 in the schematics) are connected across the output terminals of a full-wave rectifier for smoothing purposes. For voltage regulation, ICs LM-7809 and 7909 are connected in parallel with the smoothing capacitors. The output is taken from the voltage output terminals of the regulator ICs. 220 uF capacitors (shown as C3 and C4 in the schematics) are connected to the output terminals of the power circuit to compensate transient currents.

The power circuit operates in stages, each stage serving a specific purpose. The circuit operates in the following steps –

1. AC to AC Conversion

2. AC to DC Conversion – Full Wave Rectification

3. Smoothing

4. Voltage regulation

5. Compensation of transient currents

6. Short circuit protection

AC to AC Conversion

The voltage of the main sources (electricity fed by the intermediate transformer after reducing the line voltage of the generating station) is approximately 220-230 V AC, which needs to be further reduced to the 9 V level. To reduce 220 Vac to 9 Vca, a step-down transformer with center strip is used. The use of the center tap transformer allows the use of positive and negative polarities of the input voltage. The circuit experiences some drop in output voltage due to resistive loss. Therefore, a transformer with a high voltage rating greater than the required 9V needs to be used. The transformer must provide 1A current at the output. The most suitable step-down transformer that meets the mentioned voltage and current requirements is 12V-0-12V/2A. This transformer reduces the main line voltage to +/- 12 Vac, as shown in the image below.

AC to DC Conversion – Full Wave Rectification

The reduced AC voltage needs to be converted to DC voltage through rectification. Rectification is the process of converting AC voltage to DC voltage. There are two ways to convert an AC signal to DC. One is half-wave rectification and the other is full-wave rectification. In this circuit, a full wave bridge rectifier is used to convert 24V AC to 24V DC. Full-wave rectification is more efficient than half-wave rectification as it provides full use of both the negative and positive sides of the AC signal. In the full-wave bridge rectifier configuration, four diodes are connected in such a way that current flows through them in only one direction, resulting in a DC signal at the output. During full-wave rectification, two diodes are forward biased and two other diodes are reverse biased.

Fig. 4: Full Wave Rectifier Circuit Diagram

During the positive half cycle of the supply, diodes D2 and D3 conduct in series, while diodes D1 and D4 are reverse biased and current flows through the output terminal passing through D2, output terminal and D3. During the negative half cycle of the supply, diodes D1 and D4 conduct in series, but diodes D3 and D2 are reverse biased and current flows through D1, output terminal and D4. The direction of current in both directions through the output terminal in both conditions remains the same.

Fig. 5: Circuit Diagram showing the positive cycle of the Full Wave Rectifier

Fig. 6: Circuit diagram showing negative cycle of full wave rectifier

1N4007 diodes are chosen to build the full wave rectifier because they have maximum (average) forward current of 1A and in reverse bias condition can sustain peak reverse voltage up to 1000V. This is why 1N4007 diodes are used in this design for full wave rectification.

Smoothing

Smoothing is the process of filtering the DC signal using a capacitor. The output of the full wave rectifier is not a constant DC voltage. The rectifier output has twice the frequency of the main sources, but contains ripples. Therefore, it needs to be smoothed out by connecting a capacitor in parallel to the output of the full-wave rectifier. The capacitor charges and discharges during a cycle, providing a constant DC voltage as output. Thus, high value 470 uF capacitors (shown as C1 and C2 in the schematics) are connected to the output of the rectifier circuit. As the DC that has to be rectified by the rectifier circuit has many AC spikes and unwanted ripples, to reduce these spikes a capacitor is used. This capacitor acts as a filtering capacitor that shunts all the AC through it to ground. At the output, the remaining average DC voltage is smoother and ripple-free.

Fig. 7: Smoothing Capacitor Circuit Diagram

To provide regulated +/- 9V at the output, ICs LM-7809 and 7909 are used. These ICs are capable of supplying current up to 1A. IC 7809 is a positive voltage regulator that provides stable +9V output with a positive 12V input supply. To obtain a negative voltage at the output, a 7909 negative voltage regulator is used. It provides -9V at the output on a -12V input. IC 7809 provides an output voltage in the range of 8.6V to 9.4V with the input voltage range of 11.5V to 24V whereas IC 7909 provides an output voltage in the range of -8 .6V to -9.4V with the input voltage range of -11.5V to -23V. Common ground is provided by the center terminal of the transformer. Both regulator ICs are capable of regulating the load on their own. They provide regulated and stabilized voltage at the output, regardless of fluctuation in input voltage and load current.

The LM7809 and 7909 ICs have the following tolerable power dissipation internally –

Pout = (Maximum IC operating temperature)/ (Thermal Resistance, Junction-Air + Thermal Resistance, Junction-Case)

Pout = (125) / (65+5) (values ​​according to technical data sheet)

Pout = 1.78W

Therefore, both voltage regulator ICs can internally sustain a power dissipation of up to 1.78W. Above 1.78W, the ICs will not tolerate the amount of heat generated and will start to burn. This can also cause a serious fire hazard. Therefore, heat sink is necessary to dissipate excessive heat from ICs.

Compensating Transient Currents

At the output terminals of the power circuit, 220 uF capacitors (shown as C3 and C4 in the schematics) are connected in parallel. These capacitors help in quick response to load transients. Whenever the output load current changes, there is an initial shortage of current, which can be met by this output capacitor.

The output current variation can be calculated by

Output current, Iout = C (dV/dt) where

dV = Maximum allowable voltage deviation

dt = transient response time

Considering dv = 100mV

dt = 100us

In this circuit, a 220 uF capacitor is used, so

C = 220uF

Iout = 220u (0.1/100u)

Iout = 220mA

In this way it can be concluded that the output capacitor will respond to a current change of 220mA for a transient response time of 100 us.

Fig. 8: Circuit diagram of transient current compensator

Short circuit protection

A diode D5 is connected between the voltage input and output terminals of IC 7809 to prevent the external capacitor from discharging through the IC during an input short circuit. When the input is shorted, the cathode of the diode is at ground potential. The anode terminal of the diode is at high voltage as C3 is fully charged. Therefore, in this case, the diode is forward biased and all the capacitor discharge current passes through a diode to ground. This saves IC 7809 from reverse current. Similarly, a diode D6 is connected between the voltage input and voltage output terminals of IC 7909, which protects the IC from discharging capacitor C4 through the regulator when the input is shorted.

Fig. 9: Short circuit protection circuit diagram

The following precautions must be taken while assembling the circuit –

• The rated current of the step-down transformer, bridge diodes and voltage regulator ICs must be greater than or equal to the current required at the output. Otherwise, it will not be able to provide the required current at the output.

• The rated voltage of the step-down transformer must be greater than the maximum required output voltage. This is due to the fact that the 7809 and 7909 ICs experience voltage drop of around 2 to 3 V. Therefore, the input voltage must be 2 V to 3 V greater than the maximum output voltage and must be at the limit of input voltage of the regulator ICs. .

• Capacitors used in the circuit must have a higher voltage rating than the input voltage. Otherwise, the capacitors will start leaking current due to excess voltage on their plates and will explode.

• A capacitor must be used at the output of the rectifier so that it can deal with unwanted noise from the mains. Likewise, the use of a capacitor at the output of the regulator is recommended to deal with rapid transient changes and noise at the output. The value of the output capacitor depends on the voltage deviation, current variations and the transient response time of the capacitor.

• A protection diode should always be used when using a capacitor after a voltage regulator IC, to prevent the IC from countercurrent during capacitor discharge.

• For high output load activation, a heat sink must be mounted in the regulator holes. This will prevent the IC from exploding due to heat dissipation.

• As the regulator IC can only draw current up to 1A, a 1A fuse needs to be connected. This fuse will limit the current in the regulator to 1A. For currents above 1A, the fuse will blow and this will cut off the input power to the circuit. This will protect the circuit and regulator ICs from currents greater than 1A.

Once the circuit is assembled, it can be tested with a multimeter. Measure the output voltage at the terminals of the 7809 and 7909 ICs. Then measure the voltage outputs when the loads are connected.

In regulator IC 7809, the input voltage is 12V and the output voltage is 9.04V. With a 20 Ω resistance load, the output voltage reads at 8.03 V showing a voltage drop of 1.01 V. The output current is measured at 400 mA, so the power dissipation in the resistance load of 20 Ω is as follows –

Pout = (Vin – Vout)*Iout

Pout = (12–8.03)*0.4

Pout = 1.58W

In regulator IC 7909, the input voltage is -12V and the output voltage is -9.18V. With a 20 Ω resistance load, the output voltage reads -9.11 V showing a voltage drop of 0.07 V. The output current is measured at 455 mA, so the power dissipation in the resistance load of 20 Ω is as follows –

Pout = (Vin – Vout)*Iout

Pout = (-12 – (-9.11)*0.455 (energy dissipation cannot be negative)

Pout = 1.3 W

From the above tests, it can be seen that the power dissipation is always less than 1.78W (internal tolerable limit of ICs 7809 and 7909). Still, it is recommended to use a heat sink to help cool the IC and increase its useful life.

The power supply circuit designed in this project can be used to power chipsets that require a negative power supply, such as operational amplifiers, bipolar amplifiers, and multivibrator circuits. The circuit can also be used as a 9V 1A power adapter.