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.
BJTs tend towards either amplification (Ic increases with Ib) or switching (Ic is on, or off). Amplifier transistors have good gain linearity, but DC gain is moderate, and switching time and Vce(sat) may not be very low (e.g. the 2n5550). Switching transistors have very fast switching time, high DC gain and low Vce(sat) but gain linearity is not much of a concern (e.g. MMBT4401. Then there are general purpose transistors that are middle of the road for both types of operation (e.g. 2n3904).
Unfortunately none of these three example BJTs are suitable for the purpose of the specified circuit, for a multitude of reasons:
- For driving the load represented by the motor specified, a sustained current rating of 240 mA (
12 V / 50 Ohm DC resistance
) or 259 mA (from datasheet) is needed. The 2n5550 will handle about 50 mA sustained current with ease, with Vce(sat) hitting 0.25 Volts. At higher continuous currents, heating becomes an issue: Note the derating values for device dissipation above 25 degrees C.
- Even if we were to drive 259 mA, the transistor would need heaps of base drive current, depending on the worst-case DC gain characteristic of the transistor of choice. More than 20 mA typically for the examples mentioned above, to drive 250 mA Ic. The anemic outputs of the Raspberry Pi are simply not designed for that kind of abuse. One would hesitate to draw more than single-digit current from its GPIO, if that.
- A high-gain moderate current switching transistor such as NTE2503 might serve the purpose, with its 800 minimum DC gain, and 700 mA continuous current rating.
- A Darlington pair might sound like another good alternative, but Darlington pairs have a larger voltage drop between collector and emitter, as they have two sets of NPN (or PNP) potentials in the current path. This means more heat generated when current is being passed, therefore need for larger packages and / or heat sinks and cooling.
That leaves a MOSFET as an excellent alternative. For instance, the inexpensive IRLML2502 enhancement mode N-type MOSFET will very comfortably drive over 2 Amperes with a gate voltage of 2.8 Volts, and will not feel warm to the touch at 259 mA Id. Negligible gate drive current is needed except at switching, and this gate current can be limited to a couple of mA using a gate resistor, if very fast switching is not required.
If the motor is to be driven with PWM, then a gate driver could be used as an interim stage to drive the MOSFET gate, either a discrete solution using one of those 2n5550 BJTs that you already have and won't be needing, or using a gate driver IC if truly stupendous PWM frequencies are intended.
Best Answer
Your question seems to be about beta or hFE. Yes, this can vary significantly between parts, even from the same production batch. It also varies somewhat with collector current and collector voltage (using emitter as the 0 V reference). However, for any one transistor, its gain actually varies rather little as a function of collector current across a reasonable range, and assuming the collector voltage is held high enough.
The big point you seem to be missing is that you shouldn't be worrying about the exact gain. A good circuit with bipolar transistors works with the minimum guaranteed gain over the intended operating region, but otherwise works fine with the gain being anywhere from there to infinite. It's not out of line for any one transistor at a particular operating point to have 10x more gain than the minimum guaranteed by the datasheet. After taking that into account in the circuit design, it's really just a minor step to make sure the circuit works with the transistor's gain all the way to infinity.
Designing for such a wide range of gain may sound difficult, but it's actually not. There are basically two cases. When the transistor is used as a switch, then some minimum base current, computed from the minimum guaranteed gain, will drive it into saturation. If the gain is higher, then the transistor will just be more into saturation at the same base current, but all the voltages across it and currents through it will still be pretty much the same. Put another way, the rest of the circuit (except for unusual cases) won't be able to tell the difference between the transistor driven 2x or 20x into saturation.
When the transistor is used in it's "linear" region, then negative feedback is used to convert the large and unpredictable gain to a smaller but well controlled gain. This is the same principle used with opamps. The DC and AC feedback may be different, with the first setting the operating point, sometimes referred to as biasing the transistor, and the second controlling what happens when the desired signal is passed through the circuit.
Added:
Here is a example circuit that is tolerant of a wide range of transistor gain. It will amplify small audio signals by about 10x, and the output will be around 6 V.
To solve this manually, it's probably easiest to do it iteratively. Start by assuming OUT is 6V, and work from there. Since the gain is infinite, there is no base current, and the base voltage is set directly by the R1-R2 divider from whatever OUT is. The divider has a gain of 1/6, so the base is at 1.00 V. Minus the B-E drop of 600 mV, that puts the emitter at 400 mV, and the emitter and collector currents at 400 µA. The R1-R2 path draws 50 µA, so the total drawn from OUT it 450 µA, so the drop across R3 is 4.5 V, so OUT is at 7.5 V. Now go through the above calculations again assuming OUT is 7.5 V, and maybe one more time after that. You will see the results converge rapidly.
This is actually one of the few cases a simulator is useful. The main problem with simulators is that they give you very accurate and authoritative looking answers despite the input parameters being vague. However, in this case we want to see the affect of changing just the transistor gain, so a simulator can take care of all the drudge work for us, as performed above. It's still useful to go through the process in the previous paragraph once to get a feel for what is going on, as apposed to just looking at the results of a simulation to 4 decimal places.
In any case, you can come up with the DC bias point for the circuit above assuming infinite gain. Now assume a gain of 50 for the transistor and repeat. You will see that the DC level of OUT only changes a little.
Another thing to note is that there are two forms of DC feedback, but only one for the AC audio signals.
Since the top of R1 is connected to OUT, it provides some DC feedback that makes the operating point more stable and less sensitive to the exact transistor characteristics. If OUT goes up, the current into the base of Q1 goes up, which makes more collector current, which makes OUT go down. However, this feedback path does not apply to the audio signal. The impedance looking into the R1-R2 divider is R1//R2 = 17 kΩ. The high pass filter rolloff frequency formed by C1 and this 17 kΩ is 9.5 Hz. Even at 20 Hz, R1//R2 isn't much of a load on the signal coming through C1, and it gets more irrelevant proportional to frequency. Put another way, R1 and R2 help set the DC bias point, but don't get in the way of the intended audio signal.
In contrast, R4 provides negative feedback for both DC and AC. As long as the gain of the transistor is "large", then the emitter current is close enough to the same as the collector current. This means whatever voltage is across R4 will appear across R3 in proportion to their resistances. Since R3 is 10x R4, the signal across R3 will be 10x the signal across R4. Since the top of R4 is at 12 V, OUT is 12 V minus the signal across R3, which is 12 V minus 10x the signal across R4. This is how this circuit achieves a fairly fixed AC gain of 10 as long as the transistor gain is significantly larger than that, like 50 or higher.
Go ahead and simulate this circuit while varying parameters of the transistor. Look at both the DC operating point and what the overall transfer function from IN to OUT of a audio signal is.