Project: Torque converter for slow speed motor

Discussion in 'The Completed Projects Collection' started by hobbyist, Oct 15, 2008.

  1. hobbyist

    Thread Starter Distinguished Member

    Aug 10, 2008
    Praise God!. It works, I've been designing this circuit for the last 3 weeks, ran into a lot of snags but once I put the motor on its own supply, the design went smoothly.

    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.

    [Equipment Used in this design]:
    1. Breadboard and wire kit
    2. Digital Multimeter
    3. Bench top variable supply.
    4. Resistor Substitution Box (RSB)..excellent design tool..
    5. plenty jumper wires (with aligator clips)
    6. 6V. 2A. bat. drill. motor.
    7. Design Notebook (for recording voltages and all kinds of data)
    8. A good scientific calculator.

    [Knowledge needed]:
    1. Thourogh understanding of "OHMS LAW"
    2. advanced basic understanding of BJT. NPN and PNP.
    3. Limited understanding of POWER MOSFETS.
    4. Some algebra (rearrange Volt div. equ., and such)
    5. Thourogh understanding of what I wanted the circuit to do.
    So as to make design choices of voltages, currents, resitances,
    and circuit configurations.

    [Design Procedures]:
    This is a very, brief topical explanation of the procedures involved.
    Using my Resistor sub box. (RSB), I biased Q3 to produce both min. and max. speed of motor.
    Then I recorded the VbQ3 for both speeds.
    Then I designed a typical feedback volt. reg. to accomodate the required voltages for Q3 bias.
    I then added Q7 and with RSB. I biased it to cause VoQ1 to go max.output.
    This resistance is temporary just for calculating voltages and currents needed for the next stage. With the VbQ7 recorded I then proceded to bias Q8 stage to take the place of the temp. resistor. This stage brings Q7 back to cutoff.

    Now I worked on darlington pair Q4,Q5, and biased this pair to have a VbQ4 lower than VeQ5 putting this pair at cutoff.
    Now with my RSB. I temporarily biased Q8 to cutoff and recorded VbQ8. Then I calculated the current through this resistance and then proceded to design stage Q9. with a Vc = to the VbQ8 and Ic = to current calculated. Then I again biased Q9 to be cutoff recorded volts. and amps. and designed Q11 stage to do this same thing, and then designed Q10 stage to take the place of the resistor bias for Q11.

    [Test Voltages]:
    See schematic.
    Motor no load. (approx. 50 rpm's)
    VTP1= 0.68V
    VTP2= 1.38V
    VTP3= 9.55V
    VTP4= 1.15V

    Motor heavily loaded near stalling.
    VTP1= 1.00V
    VTP2= 1.93V
    VTP3= 13.55V
    VTP4= 2.34V

    Delta (heavy load - no load)
    Delta VTP1= 0.32V
    Delta VTP2= 0.55V
    Delta VTP3= 4.00V
    Delta VTP4= 1.19V

    As the motor is heavily loaded the voltage across the motor increases around 300mv.
    When the motor is at a complete stall for a few seconds then when the load is released the motor voltage will then increase as high as 2.5V across it for a few seconds then drop back to its original no load speed and remain there until another load is put on it.
    Last edited: Oct 16, 2008
  2. scubasteve_911

    Senior Member

    Dec 27, 2007
    Really impressive stuff! Thanks for the contribution!

  3. Capt-Killjoy

    Active Member

    Sep 29, 2008
    Very nice build and schematic design. Nice of you to go through the actual workings of the circuit as well. Good job all the way around