As far as I understood, you are trying to make some kind of a sound level detector, which will let you detect if there is a sound with a certain volume or not. You can do this with minor changes to the schematic you have. But before that, you should understand the circuit.
Let's break that circuit down. First of all the part with the microphone.
R1 is for supplying power that is needed by the microphone and this is called biasing the microphone. A microphone generates an AC voltage, which is sometimes negative and sometimes positive and it changes most of the time. Think of a sine wave. But remember, we had some biasing to it which is a DC voltage. We have to take that out and give only the AC voltage to the amplifier. And doing this is easy with a simple, single capacitor. A capacitor does not let the DC to pass, but lets AC pass easily. We have blocked the DC portion of the voltage on the electret microphone.
Now, let's look at the amplifier itself. Imagine that there is nothing else but the below schematic:
In this configuration, the transistor is biased to be in the linear region. It is in the edge of being turned ON or turned OFF, but it is neither of it. If it was fully ON, it would be saturated. If it was fully OFF, it would be not conducting at all. But it is in the middle, which is called the linear region.
When it is configured like that, if you touch (not literally) to the base of it, creating a small change, the output will be changing largely. This is what amplification called. You can beg Google for more detailed information.
What if we combine the two circuits mentioned above. A biased electret microphone with a capacitor will output small changes with respect to sound. The transistor will amplify these small changes so they can be viewed easily:
Notice that I have changed C1 to 1uF. You can use values up to 100uF. You will probably need electrolytic capacitors. Also, notice that there is no more an output capacitor. This means that you will have an output voltage somewhere between 0 and 5 V, depending on the sound level. If you have an oscilloscope, view the waveform on the output. If you do not, try lighting an LED if the analog read is higher than, for example, 750. Experiment with different values than 750, then report me the results.
Best Answer
Remove R5 and you will have what you describe. The configuration of Q5 is called common collector or emitter follower. Essentially, the voltage at the emitter is the voltage at the base minus 0.6V, but the emitter current can be much more than the base current, because the gain of the transistor will draw more current from the collector. Thus, it's a current amplifier.
Remember, the base-emitter junction is a diode. So, the emitter will be about 0.6V below the base if you forward bias it. With R5 removed, you can pull the emitter up to \$V_{cc} - 0.6V\$. With R5 present, you won't get it as high, since some voltage will be dropped when current flows in R5.
Since there are things that will limit the current in the emitter leg of Q5, you don't need R5 to limit the base current, which isn't true of Q2 or Q4, which have their emitters shorted to ground, or Q1, with its emitter shorted to \$V_{cc}\$.
See Why would one drive LEDs with a common emitter?
There isn't much difference in performance. In circuit 1, the anode of D1 will be at \$V_{cc} - 0.2V\$, whereas in circuit 2, it will be at \$V_{cc} - 0.6V\$, so the LED current is a bit higher in circuit 1, assuming R1 and R4 are the same value.
Circuit 2 has the advantage that the base current goes towards powering the LED, but since the base current is small, this isn't a big effect.
The last subtle difference is that in circuit 1, Q1 enters saturation, which will charge the base-emitter capacitance. When you then turn it off, this capacitance has to discharge before Q1 really goes off, adding a bit of delay from when your MCU output goes low to when the diode gets switched off by Q1. Q5 never enters saturation, because the emitter voltage is brought up to just the point where the transistor enters saturation, but not more. So, no turn-off delay. The delay is very short, and probably not significant until you are switching at least 50kHz.