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PDF BA6209 Data sheet ( Hoja de datos )

Número de pieza BA6209
Descripción Reversible motor driver
Fabricantes ROHM Semiconductor 
Logotipo ROHM Semiconductor Logotipo



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No Preview Available ! BA6209 Hoja de datos, Descripción, Manual

Motor driver ICs
Reversible motor driver
BA6209 / BA6209N
The BA6209 and BA6209N are reversible-motor drivers suitable for brush motors. Two logic inputs allow three output
modes : forward, reverse, and braking. The motor revolving speed can be set arbitrarily by controlling the voltage applied
to the motor with the control pin voltage VR.
FApplications
VCRs and cassette tape recorders
FFeatures
1) Power transistors can handle a large current
(1.6A maximally).
2) Brake is applied when stopping the motor.
3) Built-in function to absorb rush currents generated by
reversing and braking.
4) Motor speed controlling pin.
5) Small standby current.
(VCC = 12V, IO = 5.5mA typically)
6) Stable operation during mode changes either from
forward to reverse or vice versa.
7) Interface with CMOS devices.
FBlock diagram
386

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BA6209 pdf
Motor driver ICs
FCircuit operation
Input / output truth table
Forward / reverse control, forced stop, and rush current
absorption are controlled by the combination of FIN and
RIN input states.
(1) Forward / reverse control circuit
When FIN is HIGH and RIN is LOW, current flows from
OUT1 to OUT2. When FIN is LOW and RIN is HIGH, cur-
rent flows from OUT2 to OUT1 (refer to the truth table).
(2) Forced stop circuit
By setting RIN and FIN both HIGH or both LOW, power
supply to the motor is shut down and a brake is applied
by absorbing the motor counter-electromotive force.
FPin descriptions
BA6209 / BA6209N
(3) Rush current absorption circuit
When a high voltage (caused by such as a motor rever-
sal) is generated on OUT1 and OUT2, an internal
comparator detects the high voltage and turns on an in-
ternal circuit that absorbs rush currents.
(4) Drive circuit
The forward direction of the motor connected between
OUT1 and OUT2 corresponds to the current flow from
OUT1 to OUT2, and the reverse direction corresponds to
the current flow from OUT2 to OUT1. The output voltage
(VOUT) applied to the motor is given by the equation :
VOUT (V) = VZD * VCE (sat.) = VZD * 0.2 (IOUT = 100mA)
where VZD is the zener voltage of the constant voltage
diode (ZD) connected to pin 4.
If Vref is left OPEN, the output voltage (VOUT) is given by
the equation :
VOUT (V) = VCC1 * VCE (sat.) (PNP) * 2VF * VCE (sat.)
= VCC1 * 1.8 (IOUT = 100mA)
390

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