Hi guys

I designed the attached transistor switch, the problem with it is that the negative part of the output does not reach the set amplitude(4.5v) unlike the positive part.
How the circuit works: the Vcc switches the Vdc on and off. In on position, Vcc allows two cycles(two oscillations of Vdc) to pass.

Anything im missing here?

 

An NPN transistor can only have positive voltage on the collector. You are applying plus and minus voltages. You must add a +4.5V offset to the AC voltage so it goes between 0V and 9V positive voltage.

 

thanks for your reply. But how do I offset the voltage?

 

You can see the following info, it will help you more understand about the double power used in Amplifier.

Class B audio amplifier.
http://www.allaboutcircuits.com/vol_6/chpt_6/10.html

Also you can use the search engine of All About Circuits to searching by : Class B audio amplifier.

 

See attached image below.

 

I still cant get anything ut if this. It seems like is quite difficult to do this offset thing, since nobody seems to be coming with a precise solution.

How do i go about doing it?

 

To offset the sine wave by 4.5V on the simulator is easy. Just change the VDC "sine (0 4.5 40k 1n 1n)" to "sine (4.5 4.5 40k 1n 1n)".

In real circuit, you need a DC of 4.5V super-imposed with sine wave of 4.5V amplitude on the VDC.

Allen

 

yes, coz i have designed a wein bridge oscillator that gives around 4.5 Vp. so what circuit would enable me to shift it to 0-9v?

 

Use a series dc-block capacitor to couple the signal into a resistor divider to bias the transistor. See this.

 

What you need is a linear gate such as an FET or CD4016/4066.

 

What you need is a linear gate just as an FET or CD4016/4066.

Would the circuit be something like attached circuit?
found at http://www.circuitsgallery.com/2012/05/binary-amplitude-shift-keying-bask-or.html