This is what I think it is.
The Q7 and Q8 and the cirucitry around it form a flashing circuit.
How it works:
When voltage is applied a RC timing circuit is set up using R17, R15, C2, R14.
Electron current flows from ground through these components and builds up a neg. voltage at the base of Q7 (NPN) causing Q7 to shut down, for a time denoted by the timing circuit.
That's why R14 is so large to allow a C2 current to build up quickly at the base of Q7. If this resistor was to small then the Q7 would conduct quicker not allowing the capacitor (C2) to fully charge into it's base.
The reasom for that is because now Q8 would be conducting, so current through C2 could be next to nothing, which in turn would stop the flashing process from happening.
Thereby the LED would be dimly on, or fully on depending how much Q8 is conducting.
Another reason for R14 so large is there is no base to ground resistor voltage divider to put at the base, the base is effectively floating.
When the C2 begins to charge to the point where less current flows through it, then the base of Q7, will begin to raise positive, thereby conducting and sending a neg. going signal into the base of Q8, (PNP) which will turn on this transistor causing it to allow current to be dumped quickly through the LED through Q8, the LED flahes on,
But now C2 is being discharged, to the point where current can begin to flow through the RC timing circuiot again shutting down Q7 which in turn shuts down Q8. Now the LED is off again until C2 is charged to the value it was before, starting the whole process over continuously.
If C2 were connected to ground, then there would be no way to discharge the capacitor, but by placing it at the junction of R17 and the LED load then it has a way to discharge for another cycle to begin.
EC2 and R13 is used as a supply decoupling, network, this keeps the oscilator from feeding oscilating currents back into the battery. That's why R13 is so low so as to not have a great effect on the voltage at the base or emitter of Q7 and Q8, but just enough influence to keep isolation between this circuit structure and the rest of the circuitry, which all use the same common battery source.
R17 is mostly low so to be used as a current limiting resistor for the LED.
The LED itself is an IR, so this is probably being used for obsticle avoidance. (I'm just guessing NOW)
R16 is used as a coupling resistor from the Q7 collector to the base of Q8.
Electronic analysis:
Looking at the network (EC2 - R13) this forms a lowpass filter, so any high frequencies from the oscilator circuit will pass through EC2 to ground, thereby keeping the supply free of parasitic oscilations. The cutoff frequency for this network is in the range of 16 HZ. (where F = 1 / 2*pi*R*C)
Now when power is applied to this oscilator, there is a RC time constant also with this network.
Tc = R * C ( where R is in Megohms, and C is in microfarads.)
Tc = 10ms. This is where the EC2 has charged up to 63% of supply voltage.
So in 10ms. the voltage across the EC2 should have reached around 3.78v. which leaves around 2.22v. across the R13. After 5 Tc. the capacitor is considered theoretically fully charged so in 50ms. this network should have close to 6v. dropped across the EC2.
Now positive supply is fed into the oscilator nertwork the base of Q7 and the emitter of Q8.
Since the circuit is powering up there is another Rc network consisting of R13, R14, R15, C2, R17.
After googling Disc capacitor values, found that 104 for C2 would be 0.1uF.
Initial current through this network will be around 3.32uA.
So the voltage drops would be
R17=50uV.
C2=0V.
R15=15.5mV.
R14=5.976V.
R13=332uV.
Therefor the voltage at the base of Q7 ( VBQ7) = 15.65mV.
That would keep Q7 OUT of conduction. Because Vbe of Q7 is around 0.6 to 0.7 V.
Now
When time is greater than zero
A time constant is set up which, This Tc = 180ms. so after 180ms. the voltage across C2 should be around 3.78v.
and now a current flow of approximately 1.23uA.
Voltage drops would be
R17=18.45uV.
C2=3.78V.
R15=5.78mV.
R14=2.214V.
R13=123uV.
So now that puts around 3.78V. at the base of Q7, thereby Q7 is now able to conducting because the base voltage needs to be around 0.7v. to conduct and it is well within that region.
NOW without getting into Transient analysis, it looks like it will take shorter than 1 Tc. for Q7 to begin to conduct. So the frequency is higher than 5.5 Hz. using the 180ms. calculation. (1 Tc)
Just interpolating this, assuming before 1 Tc. the capacitor charges in a linear manner, just for rough calculations, than it would take 30ms. for a VB Q7 to be around 630mV.
So 1/6 of 1 Tc would cause Q7 to start conducting.
This freqency could be around 33 Hz. This is Not using transient analysis, but assuming that before 1 time constant, that the voltage across C2 is very close to linear.
So this rough estimate of freq. for the oscilator is a high enough freq. to be filtered through the EC2 R13 low pass filter, which has a cutoff at around 16 Hz. As stated up above.
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Q1 - Q6 and its circuitry form a motor driving H-Bridge.
How it works:
First motor M1 is continually running, (and running forward) because its connected directly across the supply. I don't know why, but that's what the schem. shows.
Q1 and Q2 are the Base drives to turn on the proper combination of transistor Q3 -Q6, which gives the forward and reverse drive of the motor. (M2).
A positive sig. into Q1 will drive a pos. sig. into Q5. Q5 will drive neg. current through M2 neg. term. and current (electron) will flow the path through M2 and the collector emitter of Q3, to supply voltage.
Now at this point M2 is forward running.
Now if a signal on Q1 is off and a sig. (pos) comes to Q2, then the emitter of Q2 will drive pos. voltage into the base of Q4, so now Q4 turns on and drives electron current through the pos. term. of M2 making it run in reverse, and the electron current continues the path through Q6 collector , emitter, to the supply.
That's an overview.
Electonic component detail:
Pos. voltage signal to the base of Q1,would need to be greater than around 1.4v. Because it's base is 1.4v. above ground. Now its collector would be around 3.8v. above ground due to the Vbe of Q3. So a signal into Q1 base will cause Q1 to conduct, the Base current from Q5 will now be Q1's emitter current, this current will flow into the base of the PNP Q3 causing it to conduct.
So now with Q5, conducting and Q3 conducting there is a completed path for electron current to flow through the motor.
That's about it.
THIS IS MY TAKE ON THIS.
If I'm wrong someone will correct it.