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This circuit is supposed to be a 12v/120v DC-AC inverter for a project I'm working on. It seems that if I set the turns ratio of the transformer to 0.2 the circuit will (sometimes) simulate properly. If I set the turns ratio to 5, the simulation shows 0v or -22.4mV on the secondary side of the transformer.

If anyone could help me get this circuit working, I would be greatly appreciative.

Thanks! :) -Nick

A 120 volt inverter with a 2 ohm load would be generating e squared over r, or 7,200 watts. A bit of a stretch for a 2N3055 or two.

You need to up the load resistance to 200 ohms or so.

Also this is a great example of a circuit that might work in simulation but blow up in many different ways in practice. Think hard about the signals at the transistor bases during operation and when you shut off the juice.

Also the diodes in the collector circuit are not going to do what you think they will do.

I concur - when Q5 is on, the collector of Q6 will be about 25V, so there will be nearly 1.5 Amp flowing in R7. This will require 1.5A to flow in Q5 as well, in addition to any loading on the secondary. This serves no obvious purpose.

The huge caps to the bases of the transistors provide a bigger problem, but one that will not be evident in a sim. Both caps will charge to about 25V, so when Q5 turns on, it will pull the base of Q6 to -24V. The breakdown of the emitter-base junction is about 7V; the transistors would fry. (The simulator will not show this breakdown, but a breadboard would likely die a quick death).

Also it's wildly inefficient to derive the base drive current from the collector swing. The collectors swing about 25 volts but the bases only swing under a volt, so the impedance match is howibble. Most real inverters use a separate feedback winding to supply base drive at the right voltage level.

Not your fault, the textbooks and the web are full of shaky circuits like this.