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How to design a +5V and -5V Dual Power Supply Circuit?

For balanced operation, the majority of analog electronic circuits require two power supply rails; this is especially important when building operational amplifier circuits. In addition to the positive supply voltage, digital systems such as A/D converters, op-amps, and comparators require negative supply voltage. In all of these circumstances, the required current will be modest, but providing a -5V supply is typically expensive and inefficient when a significant number of discrete and integrated-circuit components are utilized. In this article, we will learn how to construct a simple, USB-powered twin 5V power supply circuit with a low current.

Although there are numerous ways to split a single voltage, the virtual ground potential of the resulting voltages is not consistent. If we utilize two batteries to obtain dual polarity voltage, one battery would eventually deplete more quickly than the other, making it harder to maintain balanced dual polarity voltage. When a resistor is used as a potential divider, some power is lost as heat and the voltage is not stable. To solve these issues, we will use the ICL7660 CMOS Voltage converter IC .

ICL7660

ICL7660 and ICL7660A are monolithic CMOS charge pump voltage converters that convert +1.5V to +10.0V input voltage ranges to -1.5V to -10.0V output voltage ranges.

Except for the two external capacitors, the ICL7660 and ICL7660A contain all the necessary circuitry to construct a negative voltage converter. The operation of the gadget is best explained by the ideal voltage converter theory that follows.

During the initial half-cycle, both S1 and S3 switches are closed (Note: Switches S2 and S4 are open during this half cycle). The voltage V+ is applied to the capacitor C1. Closed during the second half cycle of operation are switches S2 and S4 (Note: Switches S1 and S3 are open during this half cycle). The voltage on the capacitor C1 shifts by V+ volts in the negative direction. The Charge is then transferred from C1 to C2, providing the switches on C2 are perfect and there is no load on C2. Consequently, the inverted V+ voltage is accessible across C2. The ICL7660 and ICL7660A operate similarly to this voltage converter’s ideal operation.

ICL7660 Application Tips:

• To prevent device latch-up, the capacitor C2 must be positioned close to IC2 Do not exceed 10V for the ICL7660 and 12V for the ICL7660A.
• Do not connect the LV connector to GROUND when the supply voltage exceeds 3.5V.
• In the case of polarised capacitors, the ‘+’ terminal of C1 must be connected to pin 2 of the ICL7660 and ICL7660A, while the ‘+’ terminal of C2 must be connected to GROUND.
• Use low-value ESR capacitors in place of C1 and C2 for optimal performance.
• If the distance between the USB port and the circuit is considerable, a buffer capacitor can be placed across the input supply.
• This circuit’s output current is limited to 40mA. For current needs up to 100mA, the IC MAX660 can be substituted for U1.

5v Power Supply Circuit and Working:

Below is the entire 5v power supply circuit diagram utilizing ICL760. The +5V input voltage can be received from any USB port on a laptop or computer, or from a charger or adapter.

The Circuit is composed of ICL7660 (U1) and two capacitors (C1 and C2). USB’s 5V output is connected to pin 8 of U1 The voltage inverter component is comprised of the IC U1 and capacitors (C1 and C2), which convert +5V to -5V. Pin 5 of U1 provides access to the converted -5V supply. Thus, the dual 5V power supply is accessible via the J2 connector.

We have simulated the circuit in Proteus before building it on hardware:

Testing Dual (±) 5V USB Power Supply Circuit:

Assemble the circuit on the PCB/breadboard according to the given circuit schematic. Place the capacitor C2 as close as possible to the integrated circuit U1. If the circuit is connected to a PCB, the IC should be affixed to a suitable IC base. Once the 5v power supply circuit is constructed, it should be like this.

To test the circuit, connect the USB to a laptop or power bank or any USB to power the circuit. Check the output voltage on J2 using a multimeter with reference to the ground. In the testing video given below, the multimeter is connected to the positive rail when it shows 4.9V. Then the multimeter is connected to the output of the IC (i.e., pin 5 of ICL7660), then it shows -4.7V.

Conclusion

Hope this blog helps you to understand how to build a dual power supply circuit using the ICL7660 IC. We, MATHA ELECTRONICS will come back with more informative blogs.