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TC648 Datasheet(PDF) 12 Page - Microchip Technology

Part # TC648
Description  Fan Speed Controller with Auto-Shutdown and Over-Temperature Alert
PDF  28 Pages
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Manufacturer  MICROCHIP [Microchip Technology]
Direct Link  http://www.microchip.com
Logo MICROCHIP - Microchip Technology

TC648 Datasheet(HTML) 12 Page - Microchip Technology

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TC648
DS21448C-page 12
 2002 Microchip Technology Inc.
Per Section 1.0, “Electrical Characteristics”, the leak-
age current at the VAS pin is no more than 1 µA. It is
conservative to design for a divider current, IDIV, of
100 µA. If VDD = 5.0V then…
EQUATION
We can further specify R1 and R2 by the condition that
the divider voltage is equal to our desired VAS. This
yields the following:
EQUATION
Solving for the relationship between R1 and R2 results
in the following equation:
EQUATION
For this example, R1 = (2.27) R2. Substituting this rela-
tionship back into the original equation yields the
resistor values:
R2 = 15.3 kΩ, and R1 = 34.7 kΩ
In this case, the standard values of 34.8 k
Ω and
15.4 k
Ω are very close to the calculated values and
would be more than adequate.
5.4
Output Drive Transistor Selection
The TC648 is designed to drive an external transistor
or MOSFET for modulating power to the fan. This is
shown as Q1 in Figures 5-1, 5-6, 5-7,and 5-8. The
VOUT pin has a minimum source current of 5 mA and a
minimum sink current of 1 mA. Bipolar transistors or
MOSFETs may be used as the power switching ele-
ment, as is shown in Figure 5-6. When high current
gain is needed to drive larger fans, two transistors may
be used in a Darlington configuration. These circuit
topologies are shown in Figure 5-6: (a) shows a single
NPN transistor used as the switching element; (b) illus-
trates the Darlington pair; and (c) shows an N-channel
MOSFET.
One major advantage of the TC648’s PWM control
scheme versus linear speed control is that the power
dissipation in the pass element is kept very low.
Generally, low cost devices in very small packages,
such as TO-92 or SOT, can be used effectively. For
fans with nominal operating currents of no more than
200 mA, a single transistor usually suffices. Above
200 mA, the Darlington or MOSFET solution is
recommended. For the power dissipation to be kept
low, it is imperative that the pass transistor be fully sat-
urated when "on".
Table 5-1 gives examples of some commonly available
transistors and MOSFETs. This table should be used
as a guide only since there are many transistors and
MOSFETs which will work just as well as those listed.
The critical issues when choosing a device to use as
Q1 are: (1) the breakdown voltage (V(BR)CEO or VDS
(MOSFET)) must be large enough to withstand the
highest voltage applied to the fan (Note: This will occur
when the fan is off); (2) 5 mA of base drive current must
be enough to saturate the transistor when conducting
the full fan current (transistor must have sufficient
gain); (3) the VOUT voltage must be high enough to suf-
ficiently drive the gate of the MOSFET to minimize the
RDS(on) of the device; (4) rated fan current draw must
be within the transistor's/MOSFET's current handling
capability; and (5) power dissipation must be kept
within the limits of the chosen device.
A base-current limiting resistor is required with bipolar
transistors. The correct value for this resistor can be
determined as follows:
VOH
=VBE(SAT) + VRBASE
VRBASE =RBASE x IBASE
IBASE
=IFAN / hFE
VOH is specified as 80% of VDD in Section 1.0,
“Electrical Characteristics”; VBE(SAT) is given in the
chosen transistor data sheet. It is now possible to solve
for RBASE.
EQUATION
Some applications benefit from the fan being powered
from a negative supply to keep motor noise out of the
positive supply rails. This can be accomplished by the
method shown in Figure 5-7. Zener diode D1 offsets
the -12V power supply voltage, holding transistor Q1 off
when VOUT is low. When VOUT is high, the voltage at
the anode of D1 increases by VOH, causing Q1 to turn
on. Operation is otherwise the same as in the case of
fan operation from +12V.
R1 + R2
IDIV = 1e
–4A =
, therefore
5.0V
R1 + R2 =
= 50,000
Ω = 50 kΩ
1e–4A
5.0V
VDD x R2
R1 + R2
VAS =
VDD - VAS
VAS
R1 = R2 x
=
R2 x (5 - 1.53)
1.53
VOH - VBE(SAT)
RBASE =
IBASE



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