The BA6220 is a chip that performs speed stabilization of small motors by means of back EMF. However, the datasheet is very scarce on the details and has no explanation for its workings.

If I understand it correctly, the chip has no PWM and performs linear regulation with a transistor. How would it measure back EMF while driving the motor, then? What RPM stability can be expected wrt. supply voltage and mechanical load? Are there similar contemporary chips? enter image description here


What RPM stability can be expected wrt. supply voltage and mechanical load?

The chip's internal current sink requires some extra power supply voltage (perhaps a few volts). Extra power supply voltage above that required by the current sink is needed to compensate for the motor's internal resistance. A very high power supply voltage allows a larger range of speed control. However, the chip will easily overheat for too large supply voltage input (the thermal dissipation limits are hard to estimate). When the motor is loaded, the extra power supply voltage is used up: more load will simply slow motor RPM after that point.

As far as RPM stability, this speed compensation method can work very well so long as internal motor resistance doesn't change. Motor resistance depends on temperature coefficient of the copper wire wound on the armature. But more important, brush resistance is included in series with copper resistance. Brush resistance might increase with wear, and would require re-calibrating to maintain the constant-speed balance point. At very slow motor speed, brush resistance becomes "jumpy", making speed control difficult. But an uncompensated brush-DC motor has much, much poorer slow-speed RPM stability.
Note also that from an idle start, motor acceleration is improved by this speed-control method.


To understand how this IC works, consider the equivalent circuit of a DC motor:-


simulate this circuit – Schematic created using CircuitLab

Rm is the total series resistance of the windings, commutator and brushes etc. Vm is the voltage the motor generates as it spins. If the motor had zero internal resistance then when power was applied it would instantly speed up until generator voltage (back-emf) equaled the supply voltage, and the speed would stay constant no matter what the load.

However a real motor does have some resistance. The motor has to draw current to produce torque, which causes a voltage drop in the resistance so it slows down. In my example the motor produces 1,000rpm per volt, so to get 5,500rpm it needs 5.5V. However at 0.1A it needs an extra 0.5V across the terminals (6V total) to account for voltage drop in Rm.

This would be fine if the load was constant, but if the load varies then motor speed also varies depending on current draw. One way to improve speed regulation is to produce a 'negative resistance' that cancels out the positive resistance in the motor, leaving only the back-emf (in practice only partial cancellation is possible because if the total resistance becomes even slightly negative the circuit will go unstable).

The BA6220 takes an adjustable proportion of motor voltage as positive feedback pin 4, and negative feedback direct from the motor on pin 8. The positive feedback creates a 'negative resistance' which cancels the internal resistance of the motor, allowing it to 'measure the back-emf'. Motor voltage is then regulated by comparing it to Vref, creating an almost constant internal motor voltage (terminal voltage actually rises as motor current increases, as it attempts to compensate for voltage drop across Rm).

I don't know of any equivalent modern chips, but the TD7274 is another IC that works on the same principle.

  • \$\begingroup\$ Great description, not convinced it matches the schematic though. \$\endgroup\$ – Trevor_G Mar 10 '17 at 3:41
  • \$\begingroup\$ Then again.. they would not be the first IC maker to totally oversimplify their block diagram schematic. \$\endgroup\$ – Trevor_G Mar 10 '17 at 4:13
  • \$\begingroup\$ This is feedforward compensation, not feedback, so motor resistance can be exactly compensated with no stability problems. You can over-compensate too: where motor speed actually increases as you mechanically load it. \$\endgroup\$ – glen_geek Mar 10 '17 at 4:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.