Transistor base-emitter voltage pulled down

When the voltage on the base of Q1 goes down its collector goes high. Q1's collector is directly connected to Q2 ,Q3,and Q4 at their bases pulling them up. The collectors are pulled down. The collectors pull the cathodes of LED 1, 2 and 3 causing the current through them to increase making them to glow brighter.

 

With no sounds, C1 has about +4V at the microphone end and about 0.6V at the base of Q1.
Sounds cause the voltage at the microphone to wave up and down. The base-emitter diode of Q1 does not allow the base end of C1 to go up so it is charged more by the mic's resistor R1.. When the microphone signal waves down then the base of Q1 is pulled down by the extra charge on C1.

 

I gurss the result depends somewhat on the nature of that microphone but if it is the source of an ac waveform, impressed on the left of the capacitor, c1, then some proportion of that influence results in a current into the base of Q1. If the base emitter junction of Q1 ultimately looks like a diode, I guess the base will ramp downwards for a time, with each positive cycle.
Have I missed something important?

 

I edited my previous post.
I think C1 charges more only for very loud sounds.
Each positive-going half-wave from the mic that has R1 pulling the capacitor up, C1 charges more because the base-emitter diode of the transistor does not allow the base end of C1 to go up, resulting in C1 charging more. Then when a negative-going wave from the mic goes down, C1's extra charge pulls down the base of Q1.

 

C1 blocks DC and passes the audio signal to modulate the base of Q1 witch amplifies the audio signal
that drives the bases of Q2,3, and Q4. witch amplifies the signal even more and modulates the cathodes of
LEDA 1 2 and 3 causing the brightness of the LEDs to varies in sequence with the audio.