How to build Inverter Circuit
Here is a simple but inexpensive
inverter for using a small soldering iron (25W, 35W, etc) In the absence of
mains supply. It uses eight transistors and a few resistors and capacitors.
Transistors Q1 and Q2 (each BC547) form an astable multivibrator that produces
50Hz signal. The complementary outputs from the collectors of transistors Q1
and Q2 are fed to pnp Darlington driver stages formed by transistor pairs Q3-Q5
and Q4-Q6 (utilising BC558 and BD140).
The outputs from the drivers are fed
to transistors Q7 and Q8 (each 2N3055) connected for push-pull operation. Use
suitable heat-sinks for transistors Q5 through Q8. A 230V AC primary to
12V-0-12V, 4.5A secondary transformer (T1) is used. The centre-tapped terminal
of the secondary of the transformer is connected to the battery (12V, 7Ah),
while the other two terminals of the secondary are connected to the collectors
of power transistors T7 and T8, respectively.
When you power the circuit using
switch S1, transformer X1 produces 230V AC at its primary terminal. This
voltage can be used to heat your soldering iron. Assemble the circuit on a
generalpurpose PCB and house in a suitable cabinet. Connect the battery and
transformer with suitable current-carrying wires. On the front panel of the
box, fit power switch S1 and a 3-pin socket for connecting the soldering iron.
Note that the ratings of the battery, transistors T7 and T8, and transformer
may vary as these all depend on the load (soldering iron).
Circuit
diagram:
Parts:
- P1-P2 = 47K
- R1-R2 = 1K
- R3-R4 = 270R
- R5-R6 = 100R/1W
- R7-R8 = 22R/5W
- C1-C2 = 0.47uF
- Q1-Q2 = BC547
- Q3-Q4 = BC558
- Q5-Q6 = BD140
- Q7-Q8 = 2N3055
- SW1 = On-Off Switch
- T1 = 230V AC Primary 12-0-12V
- 4.5A Secondary Transformer
- B1 = 12V 7Ah
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