FAN8902 데이터 시트보기 (PDF) - Fairchild Semiconductor
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FAN8902 Datasheet PDF : 14 Pages
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FAN8902(KA3902)
In the steady state, the terminal voltage on a motor is consisted of a back EMF and the voltage drop on the armarture resistors.
When the motor happens to be stalled, the back EMF becomes zero, and the motor current (IM) is quickly increased until a
maximum values.
Therefore the duty of the pin #12 output becomes lower because of the increase of the sense voltage (VRS). Also it makes the
voltage (VLa) be lowered, then it makes the duty become lower again.
This mechanism makes the motor current hold very low value in the stalled motor state.
The voltage on pin #14 (VLa) ia calculated as follows:
VLa
=
VB
A
T
×
D
×
--------R----3---------
R2 + R3
Assumed the saturation voltage of Q1 is zero.
We can choose the ratio of the resistors, R2 and R3, as follows:
- Applied the rated voltage on motor, and then measured the current IRAT
- Matched the maximum command current, VCMD,MAX to IRAT.
VCMD,MAX = VLa,MAX = RS × IRAT × 5 × 5
for example, if RS = 10mΩ and IRAT = 20[A] at VBAT = 13[V],
VCMD,MAX = VLa,MAX = 10mΩ × 20 × 25 = 5V
- VLa,MAX = 5V = VBAT × 1 × R3 / (R2 + R3)
Ratio = R3 / (R2 + R3) = VCMD,MAX / VBAT = 5 / 13
Therefore, R2 : R3 = 8 : 5
VBAT
VMOTOR
Operating curve
A
B
VLa curve × (1 / R)
B’
VMOTOR,STALL
B”
VLA VLA
The voltage drop
on the amarture resistance
VCMD VLA
IRAT
The buffer OP-amp selects the lower command between VCMD and VLa so as to limit the stalled motor current to very low in
the above figure. Because of much larger VLa than VCMD, the motor operating point stays at A. But the point gradually moves
toward B’ and then B” through the curve from the instance of stall as the below figure.
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