1) You're operating the transformer in recerse mode 9 V : 230 V, right? The former secondary (now primary) 9 V winding will have a rather low impedance (it once was menat to delver power with as lottle loss as possible). So when you turn on the transistor, the resulting current may be too much for the 6 V power supply and the voltage may break down. Have you measured this?
2) Since you put the transformer in the emitter loop of the transistor, you cannot use the full 6 V power supply voltage across the windings. The voltage on the transformer will be max. Vout(Wien)-0.6 V
3) Your use of a Wien bridge oscillator suggest you are using sine waves. However, this simple schematic looks like only one half wave is active at the transformer (for positive Vbe). During the negative half of the sine wave (Vbe<0) the transistor will be closed and no current flows into the transformer. So you're operating the transformer with an pulsating DC current. The transformer may not like this and go into saturation, thus the transformer ratio is reduced markedly.
To remedy 2:
Put the transformer in the collector loop of the transistor. Remember to add a base resistor from the Wien bridge to the base to limit base current.
For 3 you could do several things:
3.1) use an amplifier with true bipolar output OR
3.2) add a series capacitor between the collector (see #2) and the transformer
3.3) add a freewheling diode across the 9 V winding of the transformer (revers epolarity: cthode to plus 6 V)
Finally you could forego all the effort of using a Wien bridge oscillator and just use square waves. Your voltage multiplier circuit will destroy any similarity of the output voltage with a sine wave anyway.
Harald
