Inverter Circuit For Soldering Iron

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).

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Circuit diagram:

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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