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:
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:
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:
- 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.
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