Transistor ac output voltage polarity

[amaresh92]

greetings,
when a transistor in common emitter configuration is given a ac signal with base current,the amplified output form from collector of it gives inverted polarity of the input voltage or the output graph of voltage versus time gets inverted of input graph of voltage versus time.how is it so?

thanks.

 

[vk6kro]

First, you need to know about voltage dividers.

If you put a voltage across two (or more) resistors in series, the voltage will divide in proportion to the size of the resistors.
So, if you put 9 volts across a 6 ohm and a 3 ohm resistor in series, there will be 6 volts across the 6 ohm resistor and 3 volts across the 3 ohm resistor.

Now, have a resistor and a transistor in series with a voltage across them.

The transistor can change its resistance depending on its base voltage.

If the base voltage is low, the transistor is like a very large resistor, so the voltage across the transistor is high compared with the voltage across the resistor.

If the base voltage is high, the transistor is like a very small resistor, so the voltage across the transistor is small compared with the voltage across the resistor.

In a common emitter amplifier, we take the output from across the transistor.
So, a large input voltage produces a small output voltage and vice versa.

This is a voltage inversion.

 

[amaresh92]

First, you need to know about voltage dividers.

If you put a voltage across two (or more) resistors in series, the voltage will divide in proportion to the size of the resistors.
So, if you put 9 volts across a 6 ohm and a 3 ohm resistor in series, there will be 6 volts across the 6 ohm resistor and 3 volts across the 3 ohm resistor.

Now, have a resistor and a transistor in series with a voltage across them.

The transistor can change its resistance depending on its base voltage.

If the base voltage is low, the transistor is like a very large resistor, so the voltage across the transistor is high compared with the voltage across the resistor.

If the base voltage is high, the transistor is like a very small resistor, so the voltage across the transistor is small compared with the voltage across the resistor.

In a common emitter amplifier, we take the output from across the transistor.
So, a large input voltage produces a small output voltage and vice versa.

This is a voltage inversion.

the voltage is decreasing then why the AC sin curve is -ve in this case

 

[vk6kro]

On the positive going part of the input sinewave, the collector voltage will be negative going, but still positive. That is, it will be getting less positive.

In the amplifier, the output voltage cannot become negative.

However, there is usually a capacitor to take the signal out of the amplifier and, after this, any DC voltage is removed and the signal swings around zero.

So, the parts of the signal output which were previously only slightly positive will now be negative.