7

LT1640L/LT1640H

Hot Circuit Insertion

When circuit boards are inserted into a live –48V backplane,

the bypass capacitors at the input of the board’s power

module or switching power supply can draw huge tran-

sient currents as they charge up. The transient currents

can cause permanent damage to the board’s components

and cause glitches on the system power supply.

The LT1640 is designed to turn on a board’s supply

voltage in a controlled manner, allowing the board to be

safely inserted or removed from a live backplane. The chip

also provides undervoltage, overvoltage and overcurrent

protection while keeping the power module off until its

input voltage is stable and within tolerance.

Power Supply Ramping

The input to the power module on a board is controlled by

placing an external N-channel pass transistor (Q1) in the

power path (Figure 6a, all waveforms are with respect to

detection and R2 prevents high frequency oscillations.

Resistors R4, R5 and R6 provide undervoltage and over-

voltage sensing. By ramping the gate of Q1 up at a slow

rate, the surge current charging load capacitors C3 and C4

can be limited to a safe value when the board makes

connection.

Resistor R3 and capacitor C2 act as a feedback network to

accurately control the inrush current. The inrush current

can be calculated with the following equation:

+

LT1640H

PWRGD

UV = 37V

OV = 71V

SENSE

C1

0.033

µF

24V

C3

0.1

µF

100V

C4

100

µF

100V

C5

100

µF

16V

Q1

IRF530

R2

10

Ω

5%

R3

10k

5%

C2

3.3nF

100V

R4

562k

1%

R5

9.09k

1%

R6

10k

1%

R1

0.02

Ω

5%

4

3

2

OV

GND

– 48V

UV

56

8

7

1

GATE

DRAIN

VICOR

VI-J3D-CY

+

–

+

5V

1640 F06a

GATE IN

–

+

CONTACT

BOUNCE

Capacitor C1 and resistor R3 prevent Q1 from momen-

tarily turning on when the power pins first make contact.

Without C1 and R3, capacitor C2 would pull the gate of Q1

LT1640 could power up and actively pull the gate low. By

placing capacitor C1 parallel with the gate capacitance of

Q1 and isolating them from C2 using resistor R3 the

problem is solved. The value of C1 should be 10 times the

value of C and R3 • C1

≈ 300µs.

Figure 6b. Inrush Control Waveforms

1640 F07b

Figure 6a. Inrush Control Circuitry