## Inrush current limiters for power supplies.

### Introduction.

ICLs ( Inrush Current Limiters ) are NTCs specially designed for inrush current limiting.
As they can carry the nominal mains current continuously ( if dimensioned properly ), they are inherently safe in this application ( as opposed to fixed resistors ).

### Warning.

ICLs can get very hot ( up to 250 °C ) under fault conditions and must be kept in a safe distance from anything that can melt ( like wire insulation ).
They should be mounted some mm above the PCB.
All though the coating on most ICLs is insulating, it is not considered an insulation barrier by safety standards.
ICLs must be installed with the appropriate distance to secondary circuits and enclosure parts.

### ICLs.

Fig.1: Small power supply rectifier.

Rectifier circuit for a small ( 5..10 W ) universal switch-mode power supply with typical component values.
I1 is a fuse. A Littelfuse 215-series 0.5 A fuse has a cold-resistance of around 1.2 Ω.
Rmains is the series resistance in the mains connection ( 0.25 Ω here ).
R1 is the ICL.
D1..D4 the rectifier.
C1 is the reservoir capacitor. The value shown here will typically have an ESR of around 0.15 Ω.
A simulation suggest that the inrush current at 265 V mains without R1 can be up to 200 A when the power is switched on at the peak of the sine.
I tried to measure it, but found out that my current probe saturates around 35 A.
This is reduced to 12 A with R1.

To select an ICL, you will need its resistance, its energy rating and maximum continuous current rating ( and check that the ICL you select is rated for at least your mains-voltage ).

The resistance is determined from the maximum peak current you will accept. For 265 VAC, 15 A:
R = VAC * √2 / Ipeak = 265 * √2 / 15 = 25 Ω
Pick the nearest higher value ( 33 Ω ).

The maximum energy that can be dissipated in the ICL is the same as the energy stored in the capacitor.
Energy = 0.5 * C * V2 = 0.5 * 22e-6 * 3752 = 1.5 J.
Pick an ICL with a higher energy rating than this.
Some ICLs have their energy rating specified as the capacitance you can discharge into them, for example:
Ctest at 230 VAC: 100 µF
Energy rating = 0.5 * Ctest * ( VAC * √2 )2 = 0.5 * 100e-6 *( 230 * √2 )2 = 5.3 J

The ICLs continuous current rating must be at least double the fuse rating.
The reason for this is that it takes 1.5 to 1.8 times the rated current to blow a fuse.
Note that ICLs must be derated both below and above their operating temperature range. Do read the data-sheet.

The regulator circuit must be designed to operate from a lower supply voltage than it could without the ICL.
For the circuit above with a 10 W load at 85 VAC the regulator must work down to 82 V ( instead of 85 V without the ICL ).

Fig.2: Transformer with rectifier.

The ICL value here is selected for a peak current of 20 A.

The maximum energy that can be dissipated in the ICL is the same as the energy stored in the capacitors and the energy due to the transformer's inrush current.
For the circuit in fig.2:

Energy due to charging the capacitors:
Nominal mains voltage: 230 VAC
Maximum mains voltage: 265 VAC
Nominal voltage on C1: 16 V
Maximum voltage on C1: 16 / 230 * 265 = 18.5 V
Energy stored in one capacitor: 0.5 * 4.0e-2 * 18.5 ^2 = 6.8 J
This is calculated for each output and the results added ( 13.6 J for this example ).

Energy due to transformer inrush current:
Ametherm [1] suggest the following formula for large toroid transformers:
Energy = 30 * steady state current * mains voltage / mains frequency
30 is the factor that the inrush current can exceed the transformer's nominal full-load current.
Dividing by mains frequency means that this can happen for a full mains period.
This is 135 J for the circuit in fig.2, giving a minimum ICL energy rating of 149 J.
You will need a 30 mm ICL or two 20 mm ICLs in series for this.

Assuming that the inrush current will last for a full mains period is definitely erring on the safe side.
Inrush current for a transformer is always uni-polar with the major part in the first or second half-cycle.

I measured the transformer shown i fig.2:
Its primary resistance is 6.3 Ω. This will limit the current at 230 VAC to around 50 A-p.
Its nominal primary full-load current is 1 A.
During a lot of attempts, the highest peak current I could measure was 35 A for 6 ms.
This is an inrush energy of 47 J ( 35 * 230 * 6e-3 ), resulting in a minimum ICL rating of 61 J.
You will need a 22 mm ICL or two 15 mm ICLs in series for this.

I generally use 2 ICLs in series as it gives a lower component height ( but takes more board area ) and it is often cheaper than one large ICL.

If you use this circuit for unregulated power-supplies, you will normally bypass the ICL with a relay to avoid that the ICL modulates the supply due to varying supply-current.
The relay will typically short the ICL after 150..300 ms, but if you want, you can extend this to 1..2 seconds.

### Mounting ICLs.

ICLs must be mounted with some clearance to the PCB as they can get hot ( an 8.5 mm ICL in the example in fig.1 will have a temperature rise of 45 °C at 85 VAC and 10 W load ).
The easiest way to do this is to use ICLs with kinked leads.
If you need small quantities, you will find that most suppliers only stock them with straight leads, so you will have to form the leads yourself.

The leads on the left one was formed with a snipe-nosed plier.
Just make sure that the lead is held between the ICL body and the place you bend it so you avoid mechanical stress on the ICL.
The leads on the right one was formed by a tool designed for this ( I do not know what it is called ).

Fig.4: ICL mounted in PCB.

The leads should be bent on the bottom side of the PCB so the ICL is held mechanically before it is soldered.
Some have suggested that ICLs can actually melt their soldering - I will suggest using a larger ICL.

### Appendix A: Calculating Beta.

In order to simulate ICLs you need four parameters:
The resistance at 25 °C ( R25 in Ω )
The Beta value ( B or β in K )
The dissipation factor ( DF or δth in W / K )
Thermal time constant ( τC in s )
The heat capacity ( Cth in J / K )
That was 5, but as Cth = τC * DF you can find the third if you have the two other.

Few ICL manufacturers specify beta, so you will have to calculate it. For that, you will need the resistance at 2 different temperatures ( RT1 and RT2 at T1 and T2 ).
If you have a resistance versus temperature plot you pick 2 values there ( typically at 25 °C and 75 °C ).

Fig.A1: Typical ICL I, V curve.

If you have an current versus voltage plot, you can find the values there:
Ta is the ambient temperature in °C ( 25 °C here )
DF is the dissipation factor in W/K ( 20 mW/K here )
IT1 = 1.0e-2 ( A )
VT1 = 1.0e-1 ( V )
RT1 = VT1 / IT1 = 1.0e-1 / 1.0e-2 = 1.0e+1 ( Ω )
PT1 = VT1 * IT1 = 1.0e-1 * 1.0e-2 = 1.0e-3 ( W )
T1 = Ta + PT1 / DF = 2.5e+1 + 1.0e-3 / 2.0e-2 = 2.5e+1 ( °C )
IT2 = 2.0e-1 ( A )
VT2 = 1.1e+0 ( V )
RT2 = VT2 / IT2 = 1.1e+0 / 2.0e-1 = 5.5e+0 ( Ω )
PT2 = VT2 * IT2 = 1.1e+0 * 2.0e-1 = 2.2e-1 ( W )
T2 = Ta + PT2 / DF = 2.5e+1 + 2.2e-1 / 2.0e-2 = 3.6e+1 ( °C )

Beta is then:

in this case 5.0e+3 ( K )

References.

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