I am having trouble understanding the three operation modes of a transistor.
When we talk about the modes of operation of a transistor, we're usually talking about cut-off, forward-active, and saturated operation.
The rest of your question seems to be about the different fundamental amplifier configurations, rather than the operating modes, so that's what I'll answer about.
Consider the CE mode. The collector and Emitter both are negative (for npn)
Check your diagrams again. For an NPN CE stage, the base and collector are both biased at higher potential than the emitter.
then how can we reverse bias them?
For a CE stage, the base-emitter junction should be forward biased; the base-collector junction is reverse biased. This is achieved by biasing the collector at a higher potential than the base. This is exactly what's shown in the diagram you posted.
Why is the common pin (the base , the emitter , and the collector respectively) grounded? what is the significance of grounding them?
They aren't necessarily grounded. They are connected to some potential that is equivalent to ground in the ac equivalent circuit. Particularly for common-base stages or PNP common-emitter stages, this is usually not the same as the circuit ground.
The significance is that this node is "common". A node that is used in common between the other nodes to define their potentials. The fact that the emitter is connected to the common node is why we call a common-emitter stage a common emitter stage.
How does a CE mode amplify more than a CB mode ?
A common emitter stage has voltage and current gain. A common base stage is essentially a unity gain current buffer. You need to study the common emitter stage to understand why it has voltage and current gain, and study the common base stage to understand why it is a unity gain buffer. Once you understand those two things, you'll understand why the one stage has more gain than the other.
Kathi, it helps to realize that the signal voltage Vbc (between base and collector) is identical to the signal voltage Vbo (between base and common ground). This is because the internal ac resistance of the DC voltage source (connected to C) can be regarded as zero.
Secondly, in common-collector configuration there is a emitter resitor Re between the emitter node and ground. Hence, the input characteristics of this configuration contains the base-emitter path in series with the resistor Re.
In this context, it is important to realize that the current through Re is beta times larger than the current into the base node. Hence, the voltage drop across this resistor - resulting from the input signal at the base - is correspondingly larger.
This is the well known feedback effect which determines the whole input characteristics. As a result, the input resistance of the whole circuit is r(in)=rbe + beta*Re. (If the current through Re would be identical to the base current Ib the input resitance would be only rbe+Re)
Best Answer
With a PNP transistor, conventional current flows out of the collector so Spice might show negative current to indicate direction.
With NPN, current will flow into the collector. Remember all polarities are reversed between NPN and PNP transistors.
It may be easier to understand if you keep your voltage sources in the same place and flip your transistors upside down depending on polarity.
Hope this helps.