How can i configure this Circuit to achieve open gain without DC offset ?

How can i configure this Circuit to achieve open gain without DC offset ?

Hi Nyame-Perhaps I'm misunderstanding the question - but since the output is capacitor coupled there should be no DC offset. 
-Rick

but there is DC offset in the mention Circuit kindly go via carefully please 

Nyame, I'd agree with Rick.  It looks like you are measuring at the junction between the collector of the transistor and the bottom of the 8.3K resistor.  If you measure after the 1000nF capacitor (ie where the right hand meter labelled "V" is" you should get a signal that's centred around ground.If you don't, it's possible the 1000nf capacitor has a short circuit, and is allowing the DC through to the output.A couple of other points....1. It's better to say "1uF" than "1000 nF" - you should try and keep values between 0.1 and 100.  This may be something to do with your circuit drawing program, though.2. A circuit like this, while it works ok, may give problems due to transistors with different gains or changes in supply voltage.  I'd suggest the following...  change the 30K resistor to 33k (preferred value), change the 1.6K resistor to 3.3K, and put a resistor between the transistor emitter and ground of 1K,  Also change the 8.3K resistor to 10K.  Using preferred values will  make it easier and cheaper to get your resistors.  the gain of the circuit will now be around 10.  If you want  higher gain, put a 47uF capacitor across the 1K resistor.This will result in a much more stable circuit which will give repeatable results for different supply voltages and transistors with different gains.  If a transistor has higher gain, it will tend to pass more current, which will raise the voltage on the new 1K resistor, which will lower the base-emitter voltage on the transistor and tend to reduce the current.  Finally, you posted your question twice.  This does happen, but if you look on the right or your comment you will see 3 dots.  Click on them and you will be offered the opportunity to edit or delete the comment - you can use this to delete the second one.Good luck with your endeavours!  Cheers // David

There will be no DC offset across the output  "V".  I think something is wrong with your simulation.
I have to admit- my favorite simulator is hardware! Have you built the circuit?
-Rick 

No capacitor is perfect and in practice there may be a DC voltage on the output if the measuring instrument is very high impedance.  However I wouldn't expect this on a simulation.  You could try putting a 100K resistor across the output and see what happens....

This is a simulation, so your  voltmeter may be an ideal one. If so, then it will have infinite resistance and that end of the capacitor will effectively be unconnected. Depending on the fidelity of the simulation of the capacitor itself, the DC offset could easily approach the DC offset of the other end.
If you add a load resistor in parallel with the voltmeter, this will give the capacitor a small grounded load, and the simulation may converge to reality a bit better. As an educated first guess, a 220k resistor sounds reasonable to confirm this hypothesis.

David,
I agree, this will likely  fix the problem. A few weeks ago, I was simulating a bridge rectifier circuit and got some similarly strange results  until I put a load on the output.
In this case the result he got makes sense when you think about it. The capacitor starts with zero volts across it and has no reason to change if the output is floating.

Spot on Elizabeth.  The simulator is treating the capacitor as a perfect capacitor, and the voltmeter as a perfect (ie infinite resistance) voltmeter.  Makes perfect sense!  Nyame, can you put a 100K resistor on the output (ie across the "V" meter) and see if you still get the DC offset?