AS Series

INRUSH CURRENT LIMITERS

When a battery is connected to a load with capacitive input, there is an Inrush current

surge as the capacitance is being charged to the battery voltage. The Input current

depends on the input capacitance; the larger the batteries and the more powerful the

load, the larger the input capacitance. A large Inrush current (in the precharge circuit,

without protection) can cause the following:

• Damage to input filter capacitors

• Blowing of the main fuse if asked to carry the inrush current without protection

• Contact failure (as well as reduction in current carrying capacity)

due to arcing and pitting that results from high inrush current

• Damage to the battery cell, which is not rated for inrush current

A typical precharge circuitry for battery operation is below with the timing diagram,

showing how the circuit operates. (Courtesy of Lithium -ION BMS)

Thermistor Protection for Precharge Circuit on Lithium Ion Batteries

+

–

R1

K1

10

BATTERY

CONTROLLER

PRECHARGE

HV OUT

K2

B=

K3

B-

K1+

K1+

K1–

K1+

K1+

K1–

+

–

The minimum resistance of the thermistor is

determined by the following:

1. Ambient temperature

2. Input capacitance value

(of the precharge circuit)

3. Battery voltage

The precharge surge current reaches 63.2% (1/e)

of its initial value after a time τ = RC.

In the selection of the thermistor, we consider a

time value of “five time-constant” when the

capacitances are fully charged and the surge

current reaches the normal operating current.

For the purpose of our design, let us assume the

following quantitative values:

Precharge time: 20 millisecond

Ambient operating temperature: Varies between

10°C to 50°C.

Battery voltage: 100 volt

Capacitor bank : 50,000 µF

5τ = RC

R = 5τ / C = 5 (0.02 sec) / 0.05F = 2.0 .

Now, look at the at R-T curves for Ametherm

thermistor at ambient of 50°C. The material

“C”exhibits

R @ 50°C/ R @ 25°C = 0.412 @ R @ 10°C / R

@ 25°C = 1.70

Therefore, minimum resistance @ 25°C = 2.0 /

0.454 = 4.40 , so our standard part has 5.0

ohm nominal resistance

At 10°C, the standard part will have a resistance

of 5.0 x 1.70 = 8.50 , which will meet our

minimum resistance.

Determine the energy the thermistor needs to

handle with out self-destruction,

E = ½ C V2 = ½ (0.05) (100) 2 = 250 Joules.

The steady state current is not calculated because

in most precharge circuits the steady state current

goes through the contactor.

The part, which would meet your specification, is

AS32 5R020.

In its most basic form, the Precharge circuit operates as follows:

• OFF: When the system is OFF all relays / contactors are off.

• Precharge: When the system is first turned on, K1 and K3 are turned on to

Precharge the load, until the Inrush current has subsided. R1 shows the location

of Thermistor in the Precharge circuit.

• ON: After Precharge, contactorK2 is turned on (relay K1, must be off to save

coil power)

V

Precharge Surge

Normal Operating Current

Load

Voltage

t

A

Battery

Current

t

Off

On

Pre

charge

K1

K1

K1

t

Off

SELECTION OF

THE THERMISTOR

For this application note, let us limit our discussion

to the selection of the Thermistor