Torque converter for slow speed motor

This is a "torque converter" circuit for a 6V. 2A. Batt. drill motor.
It is designed to run at a extremely slow speed, yet be able to increase it's speed when it is about to stall under a heavy load, thereby increasing torque to the load.

[How it works]:
The motor and it's drive circuitry (Drain and source of mosfet Q6) is on it's own supply
V1. V2 powers the electronics. Both supplies are common grounded.
Under normal running conditions the motor current is driven from V1. through Q6 and R8. With VgQ6 applied by the gate driver Q3. The V.drp..across R8 is at a determined value which keeps the motor running at a constant slow speed. This is acomplished by the value of R7 which helps bias Q3. Now Q1 and the trio of Q2 a,b,c, with it's bias resistors form a typical feedback volt.reg. This VoQ1 is predetermined so as to bias Q3 to the value needed as well as biasing VbQ4 to be at cutoff. Q7 acts as a variable resistance across R23.

With the motor at slow speed transistors Q7,Q8,Q10,Q11, are biased to be at cutoff.
Now when The motor is heavily loaded it begins to slow down more, this decreases it's apparent resistance and more current flows through it at the same time more voltage is dropped across R8 bringing VeQ5 lower until it is low enough for Q4 and Q5 to conduct, it starts conducting the excess current from the motor as it is stalling, and sends it into the voltage Amplifier consisting of Q8,Q9,Q10,Q11 and its bias circuitry, and then this amplified voltage is inputed into the base of Q7 causing it to conduct whereby it begins to bypass current around R23, bringing the VbQ2c, lower making it conduct less, which in turn allows VoQ1 to rise more positive which will increase VbQ3 and ultimately Q6 will rise more positive to drive the motor at a higher voltage.

The Base of Q4 is tied to VoQ1 so that, as long as there is a need to increase motor voltage during heavy loading the VbQ4 rides along with the increase in voltage so as to keep it's base higher than it's emitter thus continuing conducting as long as stall current is flowing.
Once the motor load is released stall current no longer present then the voltage amplifier shuts down which in turn cuts off Q7 then the VoQ1 automatically drops back to its predetermined value which then causes VbQ4 to drop below it's Ve. (cutoff) so as to allow the motor to run at it's normal slow speed again.