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How to increase amplitude of output voltage from 100 m V to 1V

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Or another attempt for explanation:
Each transistor (BJT as well as FET) has a non-linear transfer function Ic=f(Vbe) starting at the origin. This curve has an exponential shape like a pn diode.
Therefore, to operate the device in a small and quasi-linear region along this function (with the aim not to distort a sinusoidal input signal) we have to "bias" the device with a suitable dc voltage Vbe,dc and a corresponding dc current Idc. This gives the often mentioned "bias point" or Q-point.
 
The coil at the base of your second transistor caused its average voltage to be 0V and it swung up to +0.05V and down to -0.05V. But the transistor begins to turn on when its base voltage is about +0.6V and is fully turned on when it is +0.7V. Therefore your second transistor was turned off all the time.

When the base is capacitor-coupled and is biased at about +0.65V by a resistor or two then the transistor is turned on half-way and can amplify a low level signal.
 
Audioguru said:
When the base is capacitor-coupled and is biased at about +0.65V by a resistor or two then the transistor is turned on half-way and can amplify a low level signal.
Click to expand...

Why capacitor coupled? Can I simply have a voltage divided that sets my base voltage ( of 2nd npn) at about +0.7V ?
 

Neyolight said:
Why capacitor coupled? Can I simply have a voltage divided that sets my base voltage ( of 2nd npn) at about +0.7V ?
Click to expand...
The coil will short the base of the second transistor to ground. You need a coupling capacitor so that the base can be at about +0.65VDC.
As I showed, with a voltage divider then you also need an emitter resistor (that has a bypass capacitor) so that the bias voltage will still work if the transistor actually needs 0.6V or 0.7V and so that temperature change does not cause malfunction when it is hot and when it is cold. The emitter resistor will have DC current in it then the emitter and base voltages will be higher.
 

You have to use transistors amplifier for increasing that output voltage.
 

There is just no way you can get a clean and sharp square wave with just 1 transistor, the first needs to amplify, clips the heck out of the low signal and the second is needed for fast switching to get a sharp and clean waveform, that is why the second transistor in post #10 is bias differently.

Here is the output I should have included when I posted earlier. You should download LTSpice simulator, its free and it makes things much easier to understand and learn
 

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Iimagine,
Are you left-handed or something? Most of us read words and schematics from left to right so our input is on the left side and the output is on the right side.
Yours is backwards.
Oh, are you in Down Under??:wink:
 

Audioguru said:
Iimagine,
Are you left-handed or something? Most of us read words and schematics from left to right so our input is on the left side and the output is on the right side.
Yours is backwards.
Oh, are you in Down Under??:wink:
Click to expand...

That circuit was a part of the project I'm working on, that's why its drew like that :cool:
 

Breaking news: Circuit in post #10 and #26 does not work on breadboard
 

a pic of your output would help a lot to identify the problems
 

Neyolight said:
Its a straight line at 600mV
Click to expand...
I simulated the circuit and it works fine.
1) The input must have a low impedance source of about 1k ohms maximum.
2) There must be no load but a high resistance load (1M) is fine.
3) The circuit worked fine when the input was 10mV peak to 200mV peak.
 
It just didnt work on breadboard. I showed it to a couple of other student to check my connection - but no one found any mistake.
 
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