This article will guide you through the entire process of designing and building a basic battery eliminator circuit using 3 methods and with a lot of necessary features.
When a steady and dependable DC power source is needed for electrical applications, battery eliminator circuits come in handy since they eliminate the requirement for conventional batteries.
These circuits translate a transformers AC voltage into a controlled DC voltage that can be used to power electronics.
This is especially helpful for testing, prototyping or any other situation where having a wall powered gadget is more convenient than having to deal with batteries.
Here, we will look at three distinct approaches to building a circuit that eliminates batteries utilizing diodes, capacitors and transformers among other parts.
In order to produce a steady DC output, the AC voltage is rectified and filtered using a different technique in each method.
Circuit Working with Formulas:
Battery Eliminator Circuit using a Single Diode 1N4007:
Parts List:
| Component | Specification | Quantity |
|---|---|---|
| Capacitor | Electrolytic 1000μF 25V | 1 |
| Diode | 1N4007 | 1 |
| Transformer | 0V-12V 5Amp | 1 |
This simple approach rectifies the AC voltage using a single diode 1N4007.
A transformer 0V–12V 5A and a C1 capacitor 1000µF 25V is also included in the circuit.
The main purpose of the transformer is to reduce the mains AC voltage to a lower AC voltage that can be used for rectification.
A pulsing DC signal is produced by the single diodes half wave rectification, which lets only half of the AC waveform pass through.
The pulsing signal is then smoothed by the capacitor to produce a more steady DC voltage.
Formulas: Battery Eliminator Circuit using a Single Diode (Half-Wave Rectification)
Output DC Voltage:
VDC = VACrms / √2
where,
- VACrms the AC voltage of the transformer is represented by the root mean square RMS value .
- The diodes forward voltage drop is represented by VD, which for 1N4007 is about 0.7V.
Ripple Voltage:
Vripple = Iload / f * C
where,
- The load current is represented by Iload.
- The AC supply frequency is either 50 Hz or 60 Hz is represented by f.
- The capacitance 1000µF is represented by C.
How to Build:
To build a Battery Eliminator Circuit using a Single Diode follow the below mentioned steps for connections:
- Connect one wire of transformer secondary to anode of diode D1.
- Connect other wire of transformer secondary to ground 0V.
- Connect cathode of D1 diode to the positive of capacitor C1 and the negative of the capacitor C1 to ground.
Battery Eliminator Circuit using Two Diodes 1N4007:
Parts List:
| Component | Specification | Quantity |
|---|---|---|
| Capacitor | Electrolytic 1000μF 25V | 1 |
| Diode | 1N4007 | 2 |
| Transformer | 0V-12V 5Amp | 1 |
This technique uses two diodes to accomplish a full wave rectification, which is an improvement above the single diode method.
In this setup, the circuit is made to use two diodes to rectify both half of the AC waveform, but the transformer still produces the 0V–12V AC output.
This lowers the ripple in the rectified signal and raises the average DC output voltage.
The output is still being smoothed by the 1000µF 25V capacitor C1 to provide a constant DC voltage.
Formulas: Battery Eliminator Circuit using Two Diodes (Full-Wave Rectification with Center-Tap Transformer)
Output DC Voltage:
VDC = VACrms / √2 – VD
The AC voltage is divided into two halves for a center-tap transformer, each of which provides VACrms / 2 to make the correction.
Every diode drop must be taken into account.
Ripple Voltage:
Vripple = Iload / 2 * f * C
where,
- The load current is represented by Iload.
- The AC supply frequency is either 50 Hz or 60 Hz is represented by f.
- The capacitance 1000µF is represented by C.
How to Build:
To build a Battery Eliminator Circuit using Two Diodes follow the below mentioned steps for connections:
- Connect one outer wire of transformer secondary to anode of diode D1.
- Connect the center wire of transformer secondary to ground 0V.
- Connect other outer wire of the transformer secondary to diode D2.
- Join both the cathodes of the diodes D1 and D2 and connect them with the positive of capacitor C1.
Battery Eliminator Circuit using a Bridge Rectifier (4 Diodes 1N4007):
Parts List:
| Component | Specification | Quantity |
|---|---|---|
| Capacitor | Electrolytic 1000μF 25V | 1 |
| Bridge Rectifier | Diode 1N4007 | 4 |
| Transformer | 0V-12V 5Amp | 1 |
Among the three, the most sophisticated technique makes use of a bridge rectifier made up of four diodes 1N4007.
Compared to the earlier techniques, this configuration provides a more reliable and effective rectification procedure since it converts the complete AC waveform into DC.
The bridge rectifier efficiently rectifies each half of the waveform, while the transformer continues to step down the AC voltage.
The purpose of the capacitor 1000µF 25V is to even out the DC output and provide a steady voltage that may be utilized to power electronics.
Formulas: Battery Eliminator Circuit using a Bridge Rectifier (Full-Wave Rectification)
Output DC Voltage:
VDC = VACrms / 2 − 2 * VD
where,
- VACrms is the AC voltage of the transformer has an RMS value.
- The forward voltage drop of one diode in the bridge is represented by VD about 0.7V for each diode, for a total of 1.4V
Ripple Voltage:
Vripple = Iload / f * C
where,
- The load current is represented by Iload.
- The AC supply frequency is either 50 Hz or 60 Hz is represented by f.
- The capacitance 1000µF is represented by C.
How to Build:
To build a Battery Eliminator Circuit using a Bridge Rectifier follow the below mentioned steps for connections:
- First build a bridge rectifier as the per the above circuit diagram
- Connect the AC inputs of the bridge rectifier with the secondary wires of the transformer
- Connect the positive of the bridge rectifier with the positive of filter capacitor C1.
- Connect the negative of the bridge rectifier with the negative of filter capacitor C1.
Conclusion:
To conclude, Battery eliminator circuits are essential for supplying steady DC power from an AC source and there are differences in the complexity and performance of each approach.
The single diode approach produces a greater ripple voltage but offers a basic solution with basic half-wave rectification.
Compared to the single diode approach, the two diode design improves efficiency and reduces ripple by enhancing the rectification process to full-wave.
The most efficient full-wave rectification is achieved with the bridge rectifier method, which uses four diodes and produces a smoother less ripple DC output.
The particulars of your application, such as the requirement for a smooth DC output and the level of complexity you are ready to accept will determine which approach is best.
Every strategy offers a starting point for comprehending the rectification and filtering methods required to consistently power electronic equipment.