Fairchild datasheets for their LM78XX series of linear regulators contain a circuit for increasing the output using a resistor to hold the reference above ground. I guess how well it works in practice will depend on how stable the quiescent current is over the operating range for the AMS1117.
The above image was from the Fairchild 7805 datasheet that may be found here:
http://www.fairchildsemi.com/ds/LM/LM7805.pdf
First of all you can control the voltage on the inverting input (-) in the range of 9 to 10V.
Opamp will try to keep the voltage on both its inputs the same by varying the output voltage. First assume that opamp is working in its linear region (output voltage is not saturated). This means the voltage on the non-inverting input (+) is exactly the same as the voltage on the inverting input.
If you set the voltage to 10V the voltage difference on the resistor R3 is 0V. Using Ohm's law this yields zero current. This also means there is 0A going through the load.
If you set the voltage to 9V the voltage on the R3 resistor is nor 1V (10V-9V). Using Ohm's law gives 1A. All this current is also going through the load (because opamp's input current is zero).
This way you can control the load current from 0 to 1A.
Now the dynamic behavior.
Assume you set 9.5V with the potentiometer. The voltage on the collector of the transistor is 9.5V. This means R3 voltage is 0.5V and load current is also 0.5A.
Now change the potentiometer to 9.6V. Opamp's inputs are not balanced any more. The inverting input's voltage is higher than the non-inverting input. Therefore opamp will adjust its output by lowering the voltage on the base of the transistor. The collector current will drop and so will the voltage on R3. V(R3) will drop to 0.4V at which point the input voltages will be equal and you will have a steady-state again.
Practical considerations.
Almost any opamp will work correctly in this circuit. You must consider maximal current opamp can give to the gate. If your output current is max. 1A, the gate current has to be 1A/(transformer beta). You must choose an opamp that will provide at least this much current.
You must also be aware that if you want your circuit to work when the output is shorted the voltage on the output has to go down to GND+0.7V. Even if it does not you can very easily correct it by adding a base resistor.
Best Answer
Schematic
The original schematic is the result of experimentation in the simulator, and it makes the circuit looks more difficult. For the purpose of understanding, let me redraw it.
As we see, the circuitry contains three building blocks:
Here's a greatly simplified, step-by-step explanation.
Opamp
The non-inverting amplifier is the heart of the circuit. So first we take a look at the opamp. The opamp has three pins, non-inverting input (V+), inverting input (V-), and output (Vout). Of course, the opamp needs to use power, and we typically use them to handle AC signals (0V centered is the middle), such as audio, it needs not only one, but two DC power rails, let's call it 12v and -12v.
Open Loop
You can think the opamp as a differential amplifier, it only amplifies the voltage difference between (V+ and V-). For example, if V+ is 5 volts , V- is 4 volts (it doesn't matter), V+ is 1 volt greater than V-, the opamp would try to amplify the 1 volt signal at the output side.
But the problem is, an opamp is an amplifier with crazy gain, it will attempt to create a 1,000,000 times greater output than the original difference, so the output voltage starts to rise. Nevertheless, an opamp is not a perpetual motion machine, it cannot output voltage from nowhere. Eventually, it stops at the "maximum positive" voltage, which is the +12v power supply of the opamp in our case. Similarity, if V+ is 1 volt less that V-, the opamp shoots its output straight to the "maximum negative" voltage, the -12v power supply. As the gain of open-loop opamp is so large, it means the slightest imbalance between V+ and V- will make the opamp output swings from +12v to -12, back and forth.
Also, input impedance of V+ and V- is extremely high, it means even the weakest inputs won't be affected if you connect an ideal opamp to them, like an ideal voltmeter. Finally, the ideal opamp has an extremely low output impedance, which means Vout won't drop no matter what is connected to it.
An opamp have enormous gain, amplifies the slightest imbalance of V+ and V- inputs, and attempt to create a 10,000,000 times greater output, but is limited by the power supply +12v and -12v.
V+ and V- have extremely large input impedance, it means the weakest inputs won't be affected by them. On other words, no current flows into them. Just like an ideal voltmeter, the V+ and V- in an ideal opamp behave as if they're not connected.
Vout has extremely low output impedance, it means no matter what is connected, it won't pull the output down.
Unity Gain Buffer
What are the use of such crazy amplifiers? The central idea is to introduce negative feedback.
If we connect the output back to its non-inverting input (V-), something interesting would happen. Imagine, initially, V-, V+ are 0 volts. There is no voltage difference, so Vout is 0 volts. Next, we put +5v to V+, instantaneously, there is a +5 volts difference between the two outputs, the opamp starts attempting to amplify the voltage difference.
If no feedback is connected, Vout will shoot up straight to +12v. However, Vout is connected directly to V-, due to negative feedback, as Vout starts to rise, V- also rises from 0v, at the instant where Vout reaches +5v, V- will also be +5v, and V+ is still +5v. Opamp stops and reaches equilibrium. You can imagine that this process happens so fast, it's almost instantaneous.
In all negative feedback configuration, we wave our hands and assume the equilibrium where V+ = V- is reached instantaneously. Now we come to an important conclusion.
As a result, Vout of the opamp always follows V+: it takes V+, and uses its own power supply to create a replica.
Is it useful? Yes, because the opamp works like a repeater, it can receive a weak signal (like a 5 volts voltage source with a 1 megaohm resistor in series, which is still 5 volts, but the maximum current is less than 1 mA), and drive a powerful replica of that signal using its own power supply. The voltage is the same, but the output resistance is now near-zero, with nearly unlimited current. We call it a buffer.
Non-inverting Amplifier
This time, instead of connecting a wire from Vout to V-, we use R1 and R2.
This is called a voltage divider, where
$$ V_{out} = V_{in} \times \frac{R_2}{R_1 + R_2}$$
And in our non-inverting amplifier, assume R2 = 1000 ohms, R1 = 2000 ohms
$$ V_{-} = V_{out} \times \frac{R_2}{R_1 + R_2}$$
$$ V_{-} = V_{out} \times \frac{1000}{3000}$$
$$ V_{-} = V_{out} \times \frac{1}{3} $$
$$ V_{out} = 3 V_{-} $$
Assume Vin is 5v. Again, the opamp output would attempt eliminate the voltage difference at V-. But this time, it has a voltage divider, when Vout = 5v, V- only gets 1/3 of it, so the opamp continue to rise, until Vout is 3x larger than Vin.
It's a non-inverting amplifier.
Now analyze this circuit,
What is Vout? It's 3 x V+. If you have a unchanging reference voltage of V+ = 3.0, by changing the radio by R2 and R3, you can set the output of the opamp to an arbitrary voltage.
If you've read to this point and understood all my previous ideas, congrats, you've already understood the principle of operation of your circuit.
The output of your circuit is simply,
$$ V_{out} = 3.0 \times (\frac{1}{\frac{1000}{2000+1000}}) $$
$$ V_{out} = 3.0 \times \frac{2000+1000}{1000} $$
$$ V_{out} = 3.0 \times (1 + \frac{2000}{1000}) = 9.0 $$
Furthermore, if Vout ever drops, the opamp will "notice" because V+, the reference voltage is no longer equals to V-, the "sampled" output voltage, and starts to output a higher output to correct it. Likewise, if Vout is too high, the opamp will output a lower voltage.
The opamp can notice the the slightest difference, between the actual output voltage and the intended voltage, even a 0.05 volt error, because the opamp has enormous gain. Therefore, when an opamp non-inverting amplifier is used for this purpose, we also call it an error amplifier.
If the components are ideal, all of it occurs nearly instantaneously.
Zener Diode
Now, where can we obtain a stable 3.0v voltage reference? Using a Zener diode.
A silicon diode only allows the current to flow to one direction, with a constant 0.6 volts voltage drop across the diode. This voltage drop represents power loss, but it's not always a nuisance, because it enables you to create a 0.6 volts constant voltage reference independent from the power supply voltage. But this voltage is always close to 0.6 volts and inflexible.
A Zener diode is special diode designed to work in backwards. If you connect a Zener diode backwards, it will breakdown at a low breakdown voltage (A normal silicon power diode has a breakdown voltage of hundreds of volts, and it's not useful). Across the diode, there will be a constant voltage drop as well. We can use this voltage to create a reference voltage.
A ZPD3.0 diode has a 3.0 volts breakdown voltage. The resistor is ensure there won't be a short circuit across the diode. The proper current flows through the Zener diode can be found from the datasheet, typically ~10 mA.
Emitter Follower
Like I mentioned before, the ideal opamp has an extremely low output impedance, which means Vout won't drop no matter what is connected to it, and can output unlimited current, so an opamp itself can be used as a buffer.
But in reality, an opamp is typically used for instrumentation and low-power applications. If you use it as a power source, it will quickly overheat and may be damaged. So we won't use the opamp itself as the output stage, but use it only as an error amplifier.
A transistor in this configuration is called an emitter follower.
Just like an opamp buffer,
Vout is a replica of Vin, but is replicated and driven by the transistor using its own power source.
A power transistor allows higher current output, and may come with a heatsink. This is the final step, instead of driving the output directly, we use the opamp to drive the power transistor, which then drives the output.
Putting it all together
A 3.0 volts voltage reference is generated by R1 and D1.
A non-inverting amplifier is with a gain of 3, set by R2 and R3, determines the output voltage of the regulator as 9 volts.
A BJT power transistor, as emitter follower, buffers the opamp and drive the output power rail.
All linear voltage regulators, like a LM317, works by following the same principle of operation.
In integrated circuits, a 1.25 volts bandgap voltage reference is often used as a reference voltage and has much higher performance than a Zener diode, and allows one to build a very precise voltage regulator.
Which is why the formula for almost all linear voltage regulator is,