square wave derivative circuit

[franticEB]

hello to everyone could you help me in the analysis of this circuit?



With a simulation I get the following results:
with a square wave in input 0-5Volts on the node indicated by the red arrow (1) there is the same wave but shifted by 2.5Volt below the 0 level.
The same occurs at the point indicated by the arrow number 2. There is a square wave 0-2V and at the point indicated by the arrow 3 there is a square wave -1V +1 V.

I do not understand this behavior
Why the node indicated by arrow n° 2 is 0-2V and not 0-2.5 V?
thanks

 

[FvM]

I'll start with a request for clarification. Did you draw the battery symbol correctly (minus grounded, plus to resistor)? In this case, the NPN transistor would be operated in common emitter circuit with emitter and collector reversed, resulting in very low current gain.

 

[franticEB]

yes, it is.

 

[FvM]

I see that reverse mode current gain of about 1 will be sufficient to switch the transistor fully on.

But the output voltage is reduced according to the 8k load resistor.

 

[cnktechlabs]

Hi, you are using an NPN transistor. To turn it ON voltage on the base has to be 0.5-0.7V higher then the emitter voltage. Since the voltage swing of the base is -2.5 to +2.5 V the voltage swing on the emitter will be at most 0 to 2.0V. You have a +2.5V voltage source connected to 3.83k resistor and on the other side connected to the emitter of the NPN. In this case emitter becomes a summing point of two currents. One of them is the base current multiplied by the betta of the transistor . The second one is the +2.5 V source connected through 3.83k resistor. In this case emitter voltage will be controlled by the voltage applied to the base: Ve = Vb - 0.5Vdrop. You might get better results if you reverse polarity of the 2.5V battery.