This is again a compensated circuit (now there is monomania) – it must allow the internal resistance of the motor to run at a constant speed – or close to it – regardless of speed.

However, despite the many scratches and many sheets of paper, I can’t output a transfer function to prove to me that V_{emf} can only be proportional to V_{in} (actually -V_{in} since the control voltage must be negative) regardless of the motor current – not to mention the choice of the ratio of the resistor that will ensure this connection.

Furthermore, I think the text tells me that the replacement of R_{3} with a variable resistor will allow me to adjust the situation while the motor warms up and R_{m} changes, but it seems to me that it will adjust the gain at the same time.

According to the text R_{s} must be less than R_{m}.R_{1}/ Р_{2} (which seems to make sense).

And R3 must be equal to R_{2}/ ((R_{m}/ Р_{s}) – (Р_{2}/ Р_{1})), which seems to be an open circuit if R_{s} was the same as P_{m}.R_{1}/ Р_{2} instead of less? – but even this simple scheme defeats me.

It will help!

**UPDATE**:

Someone (I don’t know who) from Texas Instruments noticed this blog and helpfully arranged an explanation:

The core circuit claims to be a negative resistance generator (represented by the engine load). If so, then the IR losses in the motor will be canceled. Only EMF remains in the simplified engine model.

The new task is to calculate how this is negative resistance. Is it really -12 ohms?

Yes it is.

Since this is a linear circuit, the easy method is to find Vth and In (Thevenin voltage and Norton current) – open / short test.

The V is simple, with an open load [think ideal lossless motor] Vth is VIN * -R2 / R1 because there is 0V as common mode voltage (Vcm) and 0V on R3 – so there is no current on R3; as if he were not there. Vo = Vin * -10

“In” is a little harder; short load makes Vo = Vcm = Rs * In; The input current of the operational amplifier is assumed to be zero.

C Vcm = Vo; no R2 current. So R1 and R3 make a simple voltage divider. Vcm = Vo = VIN * R3 / (R1 + R3)

Put them together; Vo = Vcm = In * Rs = VIN * R3 / (R1 + R3)

In * Rs = VIN * R3 / (R1 + R3)

In = VIN * R3 / (R1 + R3) / Rs

B = VIN * 5/6

With VIN = 0; both sets of equations are obtained; 0V and 0A; so the end result will be resistive (zero offsets)

Vth / In = VIN * -R2 / R1 / (VIN * R3 / (R1 + R3) / Rs)

Vth / In = VIN * -R2 / R1 * (R1 + R3) * Rs / (VIN * R3)

Vth / In = -R2 / R1 * (R1 + R3) * Rs / R3)

Rout = -R2 / R1 * (R1 + R3) * Rs / R3)

Route = -10 * 6k * 1 ohm / 5k = -60k / 5k * [1 ohm]

Route = -12 ohms

The purpose of the R3 is to make the engine voltage higher than the ideal Vemf. Placing R3 gives an increase in motor voltage beyond the EMF of the motor.

Looking at this step by step:

1) The voltage in general mode is I[motor] * Rs

2) This voltage in common mode creates a current of resistor R3; I[motor] * Rs / R3

3) This current R3 also passes through R2 to make the motor voltage higher

4) The increased motor voltage is I[motor] * Rs / R3 * R2

Remember that this open loop solution is open; So experimental compensation will probably be needed.

Many thanks for this, it is the most comprehensive in terms of negative resistance – I had completely forgotten about the analysis of short circuit and open circuit.

I wonder if some good soul in TI could derive a common transfer function for Vemf with respect to Vin?