Self-starting inverter

In this article, I will explain a circuit that can be used to automatically start an inverter when a load is connected to it and turn it off once the load is disconnected. This circuit works by biasing the output of the inverter with a low-current signal from the battery and waiting for a change in that signal. The circuit then uses a current transformer to monitor the current through the load. If there is no current in the load for a few seconds, the inverter is switched off.

The inverter

The inverter I am using is a modified square wave inverter. The voltage between the live and neutral terminals at the output looks like this:
However, this diagram misses out on an important fact: the neutral of the output is not connected to the negative battery terminal. In fact, the modified square wave is generated by alternately switching the live and neutral terminals of the output to an internal high-voltage rail. As a result, both live and neutral are always positive relative to the negative battery terminal:

Schematic and operation

The schematic of the circuit is shown below

When the inverter is off, none of the transistors in the inverter's output circuit are on. Thus, the output of the inverter is in a high-impedance state. The live output of the inverter is biased to battery voltage through R1, R2, and D1. Since no load is connected to the inverter and the inverter does not sink any current, Q1 is off, and as a result, Q6 is also off. This allows current to flow through R4 and R7, turning Q4 on.

When a load is connected to the inverter, the load allows a small amount of current to pass from the L_IN terminal to the N_OUT terminal. This current flows through Q1, R2, D1, the current transformer T1, R6, and Q4. This turns on Q1, allowing C3 to charge up through R3. Current will also flow through R10, turning on Q6. Since C2 starts out discharged, current will flow through R8 and turn on Q3 and Q2. When Q3 turns on, it feeds back into the base of Q6. This allows Q2, Q3, and Q6 to remain turned on for a longer period of time. Q2 supplies power to the control circuit in the inverter, turning it on.

Note that when Q6 turns on, the base of Q4 is pulled down. This is necessary to prevent damage to R6 and Q4 when the inverter connects a high voltage to the N_IN terminal. D1 also prevents the high voltage on the L_IN terminal from damaging Q1.

As C2 charges up, less and less current will flow through the bases of Q3 and Q2. Eventually, Q3 begins to turn off. Once the collector current of Q3 is low enough, Q6 turns off as well. Furthermore, Q2 also switches off, and power to the inverter is cut. However, we want the inverter to remain on not just for a few seconds, but as long as a load is connected. To do this, C2 must be periodically discharged so that enough current can flow through the bases of Q3 and Q2 to keep them turned on.

The circuit senses currents in the load with the current transformer T1. The current transformer is made of a 6V mains transformer with a single turn of thick wire between the original windings and the core. When a large current flows between pins 1 and 2 of the current transformer, a smaller current is induced in the winding between pins 3 and 4. This current flows either through D2 or through the base-emitter junction of Q5, depending on the direction of the current. When current flows through the base-emitter junction of Q5, Q5 turns on, discharging C2 and keeping the inverter turned on.

Pictures

Picture of the inverter with the current transformer and circuit board visible.
Closeup image of the circuit board:

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