Voltage Inverter Using Switch-Mode Regulator

This circuit uses a step-up switch-mode regulator, which is usually used to produce a positive supply, to generate a regulated negative output voltage. The device used here is the MIC4680 from Micrel (www.micrel.com), but the idea would of course work with similar regulators from other manufacturers. Because of coil L1, which performs the voltage conversion by the intermediate storage of energy in the form of a magnetic field, the output is effectively isolated from the input. We can therefore connect the right-hand side of L1 to ground rather than to the positive output without causing a large current to flow.

Then we connect the ground pin of the regulator IC and all the components connected to it as the negative voltage output, isolated from ground. The components on the output side of the regulator are connected as usual: flywheel diode D1, coil L1 and the voltage divider formed by R1 and R2. These last two components set the output voltage, according to a formula given in the data sheet. Example component values for the MIC4680 used here are given in the table. The input voltage should lie within the permitted range for the regulator used, and must in any case be at least as great in magnitude as the desired output voltage (here +5 V or +12 V), so that the step-down regulation technique can work.

It is important to take care when building this circuit to mount the regulator using an insulator, since generally the GND pin of the device is connected to the heatsink tab. Also, the ON/OFF control input cannot be driven using a normal logic signal, since the regulator’s ground reference is the output voltage rather than ground itself. If the ON/OFF function is required, a level shifter or optocoupler must be used.

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100W Inverter Circuit Schematic

Here is a 100 Watt inverter circuit using minimum number of components. I think it is quite difficult to make a decent one like this with further less components.Here we use CD 4047 IC from Texas Instruments for generating the 100 Hz pulses and four 2N3055 transistors for driving the load. The IC1 Cd4047 wired as an astable multivibrator produces two 180 degree out of phase 100 Hz pulse trains.

These pulse trains are preamplified by the two TIP122 transistors.The out puts of the TIP 122 transistors are amplified by four 2N3055 transistors (two transistors for each half cycle) to drive the inverter transformer.The 220V AC will be available at the secondary of the transformer. Nothing complex just the elementary inverter principle and the circuit works great for small loads like a few bulbs or fans.If you need just a low cost inverter in the region of 100 W, then this is the best.

Circuit diagram:

100 Watt Inverter Circuit Diagram

Parts:
P1 = 250K
R1 = 4.7K
R2 = 4.7K
R3 = 0.1R-5W
R4 = 0.1R-5W
R5 = 0.1R-5W
R6 = 0.1R-5W
C1 = 0.022uF
C2 = 220uF-25V
D1 = BY127
D2 = 9.1V Zener
Q1 = TIP122
Q2 = TIP122
Q3 = 2N3055
Q4 = 2N3055
Q5 = 2N3055
Q6 = 2N3055
F1 = 10A Fuse
IC1 = CD4047
T1 = 12-0-12V
Transformr Connected in Reverse

Notes:

• A 12 V car battery can be used as the 12V source.
• Use the POT R1 to set the output frequency to50Hz.
• For the transformer get a 12-0-12 V , 10A step down transformer.But here the 12-
• 0-12 V winding will be the primary and 220V winding will be the secondary.
• If you could not get a 10A rated transformer , don’t worry a 5A one will be just
• enough. But the allowed out put power will be reduced to 60W.
• Use a 10 A fuse in series with the battery as shown in circuit.
• Mount the IC on a IC holder.
• Remember,this circuit is nothing when compared to advanced PWM
• inverters.This is a low cost circuit meant for low scale applications.

Design tips:

1. The maximum allowed output power of an inverter depends on two factors.The
2. maximum current rating of the transformer primary and the current rating of the driving
3. transistors.
4. For example ,to get a 100 Watt output using 12 V car battery the primary current will be
5. ~8A ,(100/12) because P=VxI.So the primary of transformer must be rated above 8A.
6. Also here ,each final driver transistors must be rated above 4A. Here two will be
7. conducting parallel in each half cycle, so I=8/2 = 4A .
8. These are only rough calculations and enough for this circuit.

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500W Low Cost 12V to 220V Inverter

Attention: This Circuit is using high voltage that is lethal. Please take appropriate precautions

Using this circuit you can convert the 12V dc in to the 220V Ac. In this circuit 4047 is use to generate the square wave of 50hz and amplify the current and then amplify the voltage by using the step transformer.

How to calculate transformer rating
The basic formula is P=VI and between input output of the transformer we have Power input = Power output
For example if we want a 220W output at 220V then we need 1A at the output. Then at the input we must have at least 18.3V at 12V because: 12V*18.3 = 220v*1
So you have to wind the step up transformer 12v to 220v but input winding must be capable to bear 20A.

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