Electronic – npn transistor base current question

I'll start first with definition of amplification. In the most general way amplification is just a ratio between two values. It does not imply that the output value is greater than the input value (although that's the way it's most commonly used). It is also not important if the current change is big or small.

Now let's move to some common amplification values used:

The most important (and the one your question talks about) is \$ \beta\$. It is defined as \$ \beta= \frac {I_c} {I_b} \$, where \$I_c\$ is the current going into the collector and \$I_b\$ is the current into the base. If we rearrange the formula a bit, we'll get \$I_c=\beta I_b\$ which is the most commonly used formula. Because of that formula, some people say that the transistor "amplifies" the base current.

Now how does that relate to the emitter current? Well we also have the formula \$I_c+I_b+I_e=0\$ When we combine that formula with the second formula, we get \$\beta I_b + I_b + I_e=0\$. From that we can get the emitter current as \$-I_e=\beta I_b + I_b= I_b (\beta + 1)\$ (note that \$ I_e\$ is current going into the emitter, so it's negative).

From that you can see that using the \$ \beta \$ as a handy tool in calculations, we can see the relationship between the base current of the transistor and the emitter current of the transistor. Since in practice the \$ \beta \$ is in the hundreds to thousands range, we can say that the "small" base current is "amplified" into "large" collector current (which in turn makes "large" emitter current). Note that I didn't speak about any deltas until now. That's because the transistor as an element does not require current to change. You can simply connect the base to a constant DC current and the transistor will work fine. If the change in current is required, it's not because of the transistor but because of the rest of the circuit which could be blocking the DC part of the input current.

There is another value also used and it's name is \$ \alpha\$. Here's what it is: \$ \alpha = \frac {I_c} {I_e} \$. When we rearrange that, we can see that \$I_c= \alpha I_e\$. So \$ \alpha\$ is the value by which the emitter current is amplified in order to produce collector current. In this case, the amplification actually gives us a smaller output (although in practice \$ \alpha \$ is close to 1, something like 0.98 or higher), because as we know, the emitter current going out of the transistor is the sum of the base current and collector current which are going into the transistor.

Now I'll talk a bit about how transistor amplifies the voltage and current. The secret is: It doesn't. The voltage or current amplifier does! The amplifier itself is a bit more complex circuit which is exploiting properties of a transistor. It also has input node and output node. The voltage amplification is the ratio of voltage between those nodes \$A_v = \frac {V_{out}}{V_{in}}\$. The current amplification is ratio of currents between those two nodes: \$ A_i=\frac {I_{out}}{I_{in}}\$. We also have power amplification which is the product of current and voltage amplification. Do note that the amplification can change depending on the nodes we chose to be input node and output node!

There are few more interesting values related to transistors which you can find here

So to sum this up: We have transistor which is doing something. In order to safely use transistor, we need to be able to represent what transistor is doing. One of the ways of representing processes happening in the transistor is to use the term "amplification". So using amplification, we can avoid actually understanding what is happening in transistor (if you have any semiconductor physics classes, you'll learn that there) and just have few equations which will be useful for a large number of practical problems.

The circuit does not work because you have the bases of the MJ15003's strapped to +5V. Also, the TIP50 can not supply enough current.

Also, you confuse the device's MAXIMUM ratings with the dynamic characteristics. MAXIMUM rates tell you when the device will release the magic smoke.

You do not specify the maximum current you want to draw, I am assuming the maximum for a MJ15003 (about 20A). From the specs, figure 1, I see that the MJ15003 Hfe (current amplification) is about 20 at 20 Amps, meaning the base-emitter current should be about 1A. Figure 3 tells me that the B-E voltage will be about 1.4 Volts.

A TIP50 can supply at most 1A, so it is not powerful enough for this application. It's Hfe is about 30, so to feed 2A as required by the 2 power transistors, it will draw at least 66 mA from the AtMega. This is more than it can source (40mA), so you will need an extra stage.

Have a look at the Darlington configuration for how to make a power amplifier. You will probable need three transistors, not just two, and use one TIP50 for each MJ15003.