17-01-2013, 01:00 PM
100 Watt Inverter
Watt Inverter.docx (Size: 167.52 KB / Downloads: 260)
Description
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 preamplifed by the two TIP122 transistors.The out puts of the TIP 122 transistors are amplified by four 2N 3055 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.
Inverter Design Tips.
The maximum allowed output power of an inverter depends on two factors.The maximum current rating of the transformer primary and the current rating of the driving transistors.
For example ,to get a 100 Watt output using 12 V car battery the primary current will be ~8A ,(100/12) because P=VxI.So the primary of transformer must be rated above 8A.
Also here ,each final driver transistors must be rated above 4A. Here two will be conducting parallel in each half cycle, so I=8/2 = 4A .
These are only rough calculations and enough for this circuit.
Simple 100W inverter circuit
Here is the circuit diagram of a simple 100 watt inverter using IC CD4047 and MOSFET IRF540. The circuit is simple low cost and can be even assembled on a veroboard.
CD 4047 is a low power CMOS astable/monostable multivibrator IC. Here it is wired as an astable multivibrator producing two pulse trains of 0.01s which are 180 degree out of phase at the pins 10 and 11 of the IC. Pin 10 is connected to the gate of Q1 and pin 11 is connected to the gate of Q2. Resistors R3 and R4 prevents the loading of the IC by the respective MOSFETs. When pin 10 is high Q1 conducts and current flows through the upper half of the transformer primary which accounts for the positive half of the output AC voltage. When pin 11 is high Q2 conducts and current flows through the lower half of the transformer primary in opposite direction and it accounts for the negative half of the output AC voltage.