Current andTemperature Ratings

Considerations for Current and Temperature Ratings
Introduction

Coilcraft Inductor Current Ratings

This application note describes:

Depending on the type of inductor (chip inductor, powerinductor) we may specify an Isat, Irms, or IDC current.

• How to interpret Coilcraft inductor current and temperature ratings• Our current ratings measurement method and performance limit criteria• Calculations for estimating power performance limits
based on current ratings• How to calculate component temperature from thetemperature rating• How to estimate component DCR at temperaturesother than 25°C.
• How to calculate performance limits in pulsed waveform applications• Detailed rms calculations – Appendix A• Derivation of temperature rise equation – Appendix B• Conversion factors for various waveforms – Appendix C
Electrical ratings are interdependent. The current througha component depends on the applied voltage (waveformand duty cycle) and the component impedance. The impedance of a component depends on the DC resistance(DCR), the applied signal frequency (for AC resistance),
and the component temperature. The temperature of acomponent depends on the thermodynamic (heat transfer) characteristics of the component, the circuit board,
the solder connection, the surrounding environment, theimpedance of the component, and the current through thecomponent. The power dissipation of a component is afunction of all of these variables.
The maximum operating rating of a component must begiven in terms of a specific measurement method andperformance limit. For example, the performance limit
could be defined as exceeding a specified temperaturerise or as a total breakdown of the insulation or wire.
Different measurement methods and performance limit
criteria lead to different conclusions. By establishing themethod of measurement and the performance limit criteria,a baseline is set for evaluating each application.
Ultimately, circuit designers attempt to determine maximum operating limits for temperature, current, voltage,and power for each component. Each of these is specificto the application environment.

• Saturation current (Isat) is the current at which theinductance value drops a specified amount below its
measured value with no DC current. The inductancedrop is attributed to core saturation.
• rms current (Irms) is the root mean square current
that causes the temperature of the part to rise aspecific amount above 25°C ambient. The temperature rise is attributed to I2R losses.
• DC current (IDC) is the current value above which
operation is not recommended without testing thecomponent in its intended application.
For some inductors Isat is lower than Irms. The core
saturates before the component temperature reaches
the performance limit. In this case we may specify only
the Isat as it is the limiting factor. For many inductors, Isatis higher than Irms. In these cases we may specify onlyIrms and the temperature rise above ambient. In manycases, we specify both Irms and Isat current to illustrate
which measurement is more critical. IDC is specified typically when Irms greatly exceeds Isat.

Coilcraft Measurement Method andPerformance Limit Criteria
rms Current – Irms

We determine Irms by measuring the temperature rise
above 25°C ambient caused by the current through a
representative sample inductor. A low DC bias current isapplied to the inductor, and the temperature of the inductor
is allowed to stabilize. This process is repeated until thetemperature rise reaches the rating limit. The measurements are taken with the sample inductor in still air withno heat sinking. The limit is typically a 15°C rise for chip
inductors and a 40°C rise for power inductors.
Therefore, the current rating is based on the data sheet
Irms current and temperature rise. The temperature riseof a component due to current depends on the ambient
temperature. To determine the component temperature
due to rated current at ambient temperatures above
the data sheet ambient temperature, see the “AmbientTemperature Range” section of this application note.
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Saturation Current – Isat

Ambient Temperature Range

Isat ratings are established by measuring the inductanceAmbient temperature is described as a range, such as“– 40°C to +85°C.” The range describes the recommended
ambient (surrounding environment) temperature rangeof operation. It does not describe the temperature of thecomponent (inductor). The component temperature isgiven by Equation 3:

of a representative sample at a specific frequency with
no DC current. The DC current level is then graduallyincreased and the inductance is measured.
The Isat rating is the current level at which the inductancevalue drops a specified amount (in percent) below its
measured value with no DC current.

Power Limit Calculations
Total or “apparent” power (PA) consists of a combination of the average (real) power (Pavg) and the reactive(imaginary) power (Pvar). While the purely reactive portionof power does not consume energy, there is some loss
associated with its transmission back and forth in the circuit
because transmission lines are not perfect conductors.
Temperature rise due to the heating effect of current
through an inductor is related to the average real power
dissipated by the inductor. The average real power is afunction of the effective series resistance (ESR) of theinductor and the rms current through the inductor, as
shown in Equation 1.
(1)

where:
Pavg = average real power in Watts
Irms = rms current in amps
ESR = effective series resistance in Ohms

The apparent (total) power required by an inductor is afunction of the rms current through the inductor, the rms
voltage across the inductor, and the phase angle difference
between the voltage and current. Equation 2 can be used
to estimate the apparent power required for the inductor.
PA = Irms × Vrms × cos (θ)

Refer to Appendix B for the derivation of this equation.
Example: The ambient temperature range for a power
inductor is stated as –40°C to +85°C, and the Irms is rated
for a 40°C rise above 25°C ambient.
The worst-case component temperature would be (85°C+ 49.4°C) = 134.4°C at the Irms current.

DC Resistance vs. Temperature
Equation 4 can be used to calculate the approximate DC
resistance of a component within the operating temperature range:
DCRT2 = DCR25 × ((1 + 0.00393 (T2 − 25))Equation 1 can be used to estimate a power limit due toreal losses. The real losses include DC and AC losses,
and are described by the ESR of the inductor. The AClosses are frequency-dependent, therefore, we recommend using our simulation models to determine the ESR
for calculating the real power dissipation at your specificapplication frequency.

Since the AC behavior of an inductor is frequencydependent, and Vrms is application-specific, werecommend using our simulation models to determine theapparent power requirements for your specific application
voltage and frequency.

(4)

where:
DCRT2 is the DC resistance at the temperature T2
DCR25 is the data sheet value of DC resistance at 25°C
T2 is the temperature in °C at which the DC resistance isbeing calculated

Pulsed Waveforms
An ideal pulsed waveform is described by the period(T = 1/ frequency), amplitude, pulse width (ton), and duty
cycle, as shown in Figure 1. Ideal pulsed waveforms arerectangular. Real pulsed waveforms have rise time, fall

Pulsed waveform
70% duty cycle(2)

where:
PA = apparent power in Watts
Irms = rms current in amps
Vrms = rms potential across inductor in Voltsθ = phase angle in degrees(3)

where:
Tc is the component temperature
Ta is the ambient (surrounding environment) temperature
Tr is the data sheet temperature rise due to rated current
through the inductor

Amplitude

Pavg = (Irms)2 × ESR

Tc = Ta + Tr × 0.00393 × (229.5 + Ta)

1.2
1.0
0.8
0.6
0.4
0.2
0

ton0T

0.5

1.0

1.5

2.0

Time

Figure 1

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time, overshoot, ringing, sag or droop, jitter, and settling
time characteristics that are not considered in this discussion. We also assume that all pulses in a given application
pulse train have the same amplitude.
The duty cycle (D) is the ratio of the pulse width (time duration) to the pulse period, as shown in Equation 5.
D = ton / T(5)of this discussion, we assume that ESR is frequencydependent, but not waveform-dependent, therefore
ESRpulsed = ESRc. With this assumption, equating theenergy terms of Equation 6 and Equation 7, and canceling the ESR terms:(Irms)2 × T = (Ipulsed)2 × ton(8)(Irms)2 / (I(9))2 = tpulsedon / T

where:
D = duty cycle
ton = pulse width in seconds, T = period in seconds

and since D is ton / T,

Duty cycle is typically given as a percent of the period.
For example, continuous waves have a 100% duty cycle,
since they are “on” for the whole cycle. Square waves
have a 50% duty cycle: “on” for a half cycle and “off” for ahalf cycle. Duty cycle can also be stated as a ratio. A 50%
duty cycle is the same as a 0.50 duty cycle ratio.

Solving for the equivalent pulsed current,(Irms)2 / (Ipulsed)2 = D

To convert to pulsed power values from the equivalentcontinuous (100% D) waveform estimated power, equate
the calculated energy values for a cycle, and solve theequation for the pulsed power. The calculation uses thedefinition: energy = power × time.
Energy = Pavg × T(6)For one cycle of a continuous (100% duty cycle) waveform,
and invoking Equation 1:
Energy = (Irms)2 × ESRc × T
where:
ESRc = effective series resistance to continuous current

The ESR is frequency-dependent, and can be obtained
from our simulation model of the chosen inductor.
For one cycle of a pulsed waveform,
Energy = Ppulsed × ton(7)

Energy = (Ipulsed)2 × ESRpulsed × ton
where:
Ppulsed = equivalent pulsed power
ton = pulse width in seconds
ESRpulsed = effective series resistance to the pulsed current

While the pulse width (ton) and the pulsed current amplitude (Ipulsed) are defined for a specific application, theabove logic requires either knowledge of ESRpulsed or theassumption that it is equivalent to ESRc. For the purposes

Ipulsed = ((Irms)2 / D)0.5(10)

Example: A chip inductor has an Irms rating of 100 mA
for a 15°C rise above 25°C ambient. The duty cycle fora particular application is calculated using Equation 5
to be 30%.
Using Equation 10,
Ipulsed = ((0.1 A)2 / 0.30)0.5
Ipulsed = 0.183 A or 183 mA

This calculation predicts an equivalent 15°C rise above 25°C
ambient for the 183 mA current pulse at 30% duty cycle.
The previous calculations assume that, regardless of thepulse amplitude, width, and duration, if the same amount ofenergy calculated from the power rating is delivered to thecomponent over a cycle, the component will dissipate theenergy safely. There may be physical conditions that limit
this assumption due to the heat dissipation characteristicsof the component, the solder connection, the circuit board,and the environment.
In the previous example, if the duty cycle is reduced to10%, the calculated pulsed current would be 316 mA,
which is more than three times the rated rms current, although only applied for 1/10 of a cycle. Our current ratings
are based on steady-state measurements, not on pulsed
current waveforms. While the duty cycle assumption isvalid in many cases, we have not verified our ratings forpulsed waveforms. Therefore, we recommend testing
your specific application to determine if the calculation
assumptions are valid.

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Appendix A – rms Calculations

Rectified sinusoidal waveform

Figure 2 shows a typical sinusoidal waveform of alternating
current (AC) illustrating the peak and peak-to-peak values.
The horizontal axis is the phase angle in degrees. The
vertical axis is the amplitude. Note that the average valueof a sinusoidal waveform over one full 360° cycle is zero.

Full sinusoidal waveform
Peak

1.0

Peak-to-peak

1.00.50
rms = 0.707

Average = 0.636-0.5-1.00.50100

2003000

Figure 3-0.5

value to obtain all positive values; take the mean value;
and then take the square root.

-1.00100

200

300

Figure 2
Figure 3 shows the same full sinusoidal waveform ofFigure 2, full-wave rectified. The average rectified andrms values are illustrated for comparison.
Calculate the root-mean-square (rms) value by the following sequence of calculations: square each amplitude

The rms value, sometimes called the “effective” value,is the value that results in the same power dissipation(heating) effect as a comparable DC value. This is true of
any rms value, including square, triangular, and sawtooth
waveforms.
Some AC meters read average rectified values, and others
read “true” rms values. As seen from the conversion equation in Appendix B, the rms value is approximately 11%
higher than the average value for a sinusoidal waveform.

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Appendix B – Derivation of Temperature Rise Equation
Our current ratings are typically based on a specifictemperature rise (Tr) above 25°C ambient, resulting fromthe rated current. When a component is used in a higher
ambient temperature environment, the resistance of thecomponent wire is higher in proportion to the temperature
difference between the higher ambient (Ta) and 25°C.
The increase in resistance depends on the thermal coefficient of resistance (TCR) of the wire (α). For copperwire, α ≈ 0.00393. When full rated current is applied tothe component at a higher ambient temperature, the increased wire resistance results in increased I2R losses.
The increased losses are assumed to be converted toheat, resulting in a temperature rise that is proportional
to the increase in resistance of the wire.
The derivation of the equation for determining the component temperature (Tc) when operating at a higher ambienttemperature (Ta) follows. We begin with the definition of
the temperature coefficient of resistance, using 25°C as
our reference temperature.

α=R – R25∆R25R – R25∆R/ (Taa – 25) = aa/ (Taa –α25)/ (Ta – 25) = a=/ (TaR25
R252525
RR∆R= α(Taa – 25)
R252525∆R= α(Ta – 25)
R25

25∆R increase in temperature
The∆R× Trr
× Tr
due
to theincreased resistance =R25
25
R25

∆R
∆R× Trr
Tcc = Taa + Trr +Tc = Ta + Tr +× Tr
R25
25
R25
∆R
∆R
Tcc = Taa + Trr (1 +)Tc = Ta + Tr (1 +)
R25
25
R25
Tcc = Taa + Trr (1 + α (Taa –25))
Tc = Ta + Tr (1 + α (Ta –25))
1Tcc = Taa + αTrr ( α + (T aa –25))
Tc = Ta + αTr ( 1 + (T a –25))
αFor α = 0.00393
For α = 0.00393
Tcc = Taa + 0.00393 × Trr (229.5 + Taa ) T = T + 0.00393 × T (229.5 + T )caraDefinitions:∆R = Increase in resistance due to higher ambienttemperature Ta
Ta = Ambient temperature (assumed to be >25°C)
Ra = Resistance of copper wire at ambient temperature (Ta)R25 = Resistance of copper wire at 25°C ambient
Tr = Data sheet temperature rise due to rated current
Tc = Component temperature due to rated current at
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Appendix C – Conversion Factors for Various Waveforms
Use the following equations to convert between average, rms, peak, and peak-to-peak values of various waveformsof current or voltage.

Sinusoidal Waveforms
Given an Average Value:rms = 1.112 × Average

Given an rms Value:
Average = 0.899 × rms

Peak = 1.572 × Average

Peak = 2 × rms (Q1.414 × rms)
Peak-to-Peak = 2 × 2 × rms (Q2.828 × rms)

Peak-to-Peak = 3.144 × Average
Given a Peak Value:
Average = 0.636 × Peakrms = 1/ 2 × Peak (Q0.707 × Peak)
Peak-to-Peak = 2 × Peak

Given a Peak-to-Peak Value:
Average = 0.318 × Peak-to-Peakrms = 1/(2 × 2) × Peak-to-Peak (Q0.354 × Peak-to-Peak)
Peak = 0.5 × Peak-to-Peak

Squarewave Waveforms
Average = rms = Peak

Peak-to-Peak = 2 × Peak

Triangular or Sawtooth Waveforms
Given an Average Value:rms = 1.15 × Average
Peak = 2 × Average
Peak-to-Peak = 4 × Average

Given an rms Value:
Average = 0.87 × rms
Peak = 3 × rms (Q1.73 × rms)
Peak-to-Peak = 2 × 3 × rms (Q3.46 × rms)

Given a Peak Value:
Average = 0.5 × Peakrms = 1/ 3 × Peak (Q0.578 × Peak)
Peak-to-Peak = 2 × Peak

Given a Peak-to-Peak Value:
Average = 0.25 × Peak-to-Peakrms = 1/(2 × 3 ) × Peak-to-Peak (Q0.289 × Peak-to-Peak)
Peak = 0.5 × Peak-to-Peak

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