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Fabricante

43631

High Voltage Surge Stopper with Current Limit

Linear Technology 

LT4363

Applications Information

significantly by locating resistive dividers close to the pins

with short VCC and GND traces.

Design Example

As a design example, take an application with the follow-

ing specifications: VCC = 8V to 14V DC with a transient of

150V and decay time constant (τ) of 400ms, VOUT ≤ 27V,

current limit (ILIM) at 5A, low battery detection of 6V, input

overvoltage level at 60V, and 1ms of overvoltage early

warning (Figure 8).

Selection of SMAJ58A for D1 will limit the voltage at the

VCC pin to less than 71V during 150V surge. The minimum

required voltage at the VCC pin is 4V when VIN is at 8V;

the supply current for LT4363 is 1.5mA. The maximum

value for R7 to ensure proper operation is:

R7 = 8V – 4V = 2.67kΩ

1.5mA

Select 1kΩ for R7 to accommodate all conditions.

The maximum current through R7 into D1 is then calcu-

lated as:

ID1

=

150V – 64V

1kΩ

=

86mA

which is easily handled by the SMAJ58A for more than

500ms.

With 0.1µF of bypass capacitance, C2, along with 1k of

R7, high voltage transients up to 200V with a pulse width

less than 10µs are filtered out at the VCC pin.

Next, calculate the resistive divider value to limit VOUT to

27V during an overvoltage event:

VREG

=

1.275V •(R1+R2)

R2

=

27V

Set the current through R1 and R2 during the overvoltage

condition to 250µA.

R2 = 1.275V = 5kΩ

250µA

Choose 4.99kΩ for R2.

R1= (27V – 1.275V) •R2 = 100.7kΩ

1.275V

The nearest standard value for R1 is 100kΩ.

Next calculate the sense resistor, RSNS, value:

RSNS

=

50mV

ILIM

=

50mV

5A

=

10mΩ

CTMR is then chosen for 1ms of early warning time:

CTMR

=

1ms • 6µA

100mV

=

60nF

The nearest standard value for CTMR is 47nF.

Finally, calculate R4, R5, and R6 for 6V low battery detec-

tion and 60V input overvoltage level:

6V • R5+R6 = 1.275V

R4+R5+R6

60V • R6 = 1.275V

R4+R5+R6

Choose 10kΩ for R6.

R4+R5 = 60V •10kΩ – 10kΩ = 460.6kΩ

1.275V

R5 = 1.275V • 460.6kΩ +10kΩ – 10kΩ = 90kΩ

6V

R4 = 460.6kΩ – 90kΩ = 370.6kΩ

Select 90.9kΩ for R5 and 374kΩ for R4.

The pass transistor, Q1, should be chosen to withstand a

short-circuit with VCC = 14V. In the case of a severe output

short where VOUT = 0V, the total overcurrent fault time is:

tOC

=

47nF • 0.875V

45.5µA

=

0.904ms

For more information www.linear.com/LT4363

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