soldering iron

12 Volt Inverter for Soldering Iron

Here is a simple 12 V 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 T1 and T2 (each BC547) form an astable multivibrator that produces 50Hz signal. The complementary outputs from the collectors of transistors T1 and T2 are fed to pnp Darlington driver stages formed by transistor pairs T3-T5 and T4-T6 (utilising BC558 and BD140). The outputs from the drivers are fed to transistors T7 and T8 (each 2N3055) connected for push-pull operation.

Use suitable heat-sinks for transistors T5 through T8.
A 230V AC primary to 12V-0-12V, 4.5A secondary transformer (X1) 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 general purpose 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.

Soldering Iron Voltage Inverter Circuit Diagram

soldering iron voltage inverter

Note that the ratings of the battery, transistors T7 and T8, and transformer may vary as these all depend on the load. Source


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  • Ankit

    I did the circuit as shown above and it is displaying fluctuating 230 V in Multimeter.That 230 volts doesn’t remain constant and after some time it goes to zero.Afterwards i connected the soldering gun to the output of transformer,but its not getting heated at all.Please help

  • amit

    my 22ohm resistor just blasted…. what do I do????

    • Jim Keith

      Appears that either your resistors have insufficient power rating and /or the flip-flop is not oscillating.

  • Amit

    okay thanks sir….
    also can u please tell me, what should be the transformer ratings???
    I asked fr a 12-0-12 to 230v step up centre tapped transformer but he then asks for its current rating….
    what current rating should I tell him???

  • Jim Keith

    Your electronics vendor is correct. Under worst case condition (battery charging @ 14V), the collector current through the 22Ω resistor is 0.55A and the Vce sat of the BD140 is about 1V.
    Pc = EI/2 (for 50% duty cycle) = 0.275W
    ¢j-a (Thermal resistance junction to ambient) without a HS is 100°/W per the data sheet.
    Therefore junction temp rise is 27.5°.
    Assuming a hot ambient temp (40°C), junction temp = 40 + 27.5°C.
    Hot for your finger, but cool for the transistor that is rated for a max junction temp of 150°C. Under such conditions, without a heatsink, the bulk of the thermal resistance is between the transistor case and the air, so the case surface runs close to the junction temperature.

  • Amit

    sir do we really need a heat sink for bd 140/139???
    the vendor I went to said we don’t need a heat sink for them….

  • Amit

    Thank you sir… can u please give me the calculations for the above inverter??? Please sir, i am doing a project on this and i am running out time..

    • Jim Keith

      This is a square wave inverter that is used for a very non-critical resistance heating application. The only possible calculations is the frequency of the astable flip-flop and the transformer turns ratio.

      The half-cycle period is roughly the RC product of C1 * Vr2 and C2 * Vr1. It appears to me to be running substantially higher than 50 or 60HZ which is OK because steel laminated transformers perform well up to about 1kHZ. The half-cycle timings must be matched to prevent the transformer from saturating due to the DC (unbalanced) component.

      The voltage calculation is the turns ratio of the transformer minus the resistance drop of the copper windings. To obtain the exact turns ratio, take the transformer out of the circuit and connect the primary across the 230V mains, then measure the primary voltage and the unloaded secondary voltage to the center-tap –this ratio is the turns ratio.

      Working the turns ratio backwards from the 12V square wave, you can calculate the open circuit transformer output of the inverter.

      Measure the winding resistances and factor in the IR drop on each winding using the measured soldering iron load current. Multiply this by the turns ratio to obtain the primary current and factor in the IR drop of the primary. The whole mathematical expression gets cumbersome, but should get you close to actual performance.

      The soldering load is non-critical –especially if it is temperature regulated, so a ±20% voltage error will still work OK.

      A circuit enhancement that you may wish to incorporate is the addition of back diodes across each output transistor (collector to ground). This provides a better path for transformer reactive currents to flow at the end of each half-cycle –otherwise, the reactive current will reverse bias and zener the base-to-emitter junctions of the power transistors and cause additional power loss and transistor stress. The reactive currents are appreciable because in a simple low frequency power transformer like this the secondary halves are not tightly linked together –this causes a series leakage reactance that causes the primary current to continue to flow for a short period after the transistor turns off.

      A small capacitor (perhaps 0.1uf) may be connected across the entire primary (24VCT) to help reduce possible voltage transients that could damage the power transistors.

      Note that I have not built this circuit, so you are on your own –hope that this is helpful.

  • Amit

    Im sorry im not good at electronics… Can anyone please tell me Why didnt we use a microcontroller here???
    Or is not using a microcontroller advisable here??

    • P. Marian

      A microcontroller can be used but it is not necessary in this case (it’s just a 12V DC to 230V AC converter). A µC is useful when a LCD display is required or to adjust the temperature by pressing switches.

  • Isaac

    Is it possible to use an NPN darlington pair instead of a PNP darlington pair?

  • welcome

    what are the power ratings of the components in this circuit? such as resistors, capaciors and the transistors. any one answer me please. i want to build this circuit.

  • jarot.saputra

    is this circuit compatible with television psu? i it can use to supply a television?

    • Defne

      You Will have to calibrate the pedal. If you bweorord the code from this site, the calibration was for MY pedals and varies a lot from set to set. The trick is to write a small code snippet that will display the lowest value, when the pedal is in the heel position, and another that will display the highest value when the pedal is in the toe position. Something like this://lowest valueIf (analogRead(pedalPin) highValue) HighValue = analogRead(pedalPin);Serial.print(lowValue);Serial.print( );Serial.println(highValue);

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