A new method to calculate LED’s thermal impedance and a new voltage-current formula of LED

Abstract

The thermal impedance and voltage-current relation of LEDs can be obtained using a photodetector response to pulsed light emitted by LEDs. In this study, LEDs have been driven by pulsed high currents. Photodetector voltage is proportional to the radiant power of LED. A triangle has been drawn on the top of the photodetector voltage figure, and some geometrical magnitudes have been measured. The junction temperature can be calculated using these magnitudes and the LED’s relative radiant power-junction temperature coefficient. Thermal impedance can be calculated if junction temperature and ambient temperature difference are divided by thermal power dissipated from the LED. In this study, the thermal impedance of LEDs and their error have been calculated and a new voltage-current formula for LEDs has been given. The formulas for LED junction temperature-time and thermal impedance-time variations have been obtained. Besides these, the maximum pulse duration of the current pulse that the LED can work without breaking down has been calculated.

1 Introduction

The voltage-current characteristic of LED has been determined by Shockley diode equation. This very well-known equation is as follows:

$${I}_{D}={I}_{S}\left({e}^{\frac{{V}_{D}}{{nV}_{T}}}-1\right)$$

(1)

In this equation, I_D is the diode current, I_S is the thermal current, V_D is the diode voltage, V_T is the thermal voltage and n is the ideality factor. V_T thermal voltage and I_S thermal current depend on junction temperature. There are no optical parameters in Shockley diode equation that can be used for LED.

There are some other studies about the voltage-current relation of LED. Another LED‘s voltage-current equation is as follows (Ozuturk 2015):

$${V}_{n}\left(t\right)=\frac{\frac{{k}_{pn}{I}_{n}\left(t\right)-{P}_{n}\left(t\right)}{{k}_{pn}{I}_{n}\left(t\right)}}{kZ\left(t\right)a{I}_{n}\left(t\right)}$$

(2)

k and a are constant values in Eq. 2. k is the relative radiant power variation with junction temperature. It can be specified as k = 0.7% / °C. a is a constant and it is equal to a = (0.975 ± 0.025) (Ozuturk 2000, 2013, 2016). Z(t) is the thermal impedance of LED. kpn is electro-optic converter (Ozuturk 2015; Ozuturk 2021 a; Ozuturk 2019; Ozuturk 2016; Ozuturk 2013; Ozuturk 2000). The radiant power-current characteristic of LED can be assumed as linear at low currents where junction temperature can be neglect able (approximately equal to the ambient temperature). kpn has been calculated as Pn(t)/In (t) at low currents where self-heating is neglect able. The dimension of kpn = Pn(t) / In (t) is V/A (photodetector voltage which is proportional to radiant power of LED (Pn(t) / LED’s current (In (t)). kpn × In is equal to the photodetector voltage and its dimension is V for any high currents unless junction temperature does not increase (red linear line in Fig. 1). But the photodetector voltage decreases as kpnIn - Pn(t) because of self-heating. Relative radiant power change can be found if this amount of photodetector voltage variation is divided by kpnIn. This relative radiant power is dimensionless. Junction temperature variation can be found by dividing relative radiant power by k as follows (Ozuturk 2019):

$${T}_{j}\left(t\right)-{T}_{a}=\left(\frac{{k}_{pn}{I}_{n}\left(t\right)-{P}_{n}\left(t\right)}{{k}_{pn}{I}_{n}\left(t\right)}\right)/k$$

(3)

k_pnI_n is photodetector voltage which is proportional to low radiant power, i.e., at a low current of LED. bk_pnI_n is optical magnitude where b is a constant. Also, P_n(t) is proportional to optical power (bP_n(t)). If optical powers are used instead for photodetector voltage, the following equation can be written:

$${T}_{j}\left(t\right)-{T}_{a}=\left(\frac{{bk}_{pn}{I}_{n}\left(t\right)-{bP}_{n}\left(t\right)}{b{k}_{pn}{I}_{n}\left(t\right)}\right)/k$$

(4)

The last two equations are equal. This means photodetector voltages can be used instead of optical radiant powers. The previous equation (which is equal to T_j(t)-T_a) dimension is °C because of the dimension of k.

Some parameters used in Eq. 2 can be measured using an appropriate circuit. Such a circuit includes an oscillator and a LED driver (Ozuturk 2021 a; Ozuturk 2021 b; Ozuturk 2021 c; Ozuturk 2019; Ozuturk 2016; Ozuturk 2013; Ozuturk 2000). The oscillator produces periodical pulses. Pulse duration is so short and space time is so long that the pulse can be assumed as a single pulse (not periodically). The junction temperature of the LED has been increased during the pulse and all junction heat has been dissipated during space-time. There is a LED driving circuit at the output of the oscillator. LED is connected in series with current limiter resistance which has been changed to change the amplitude of LED current, 1ohm series current measuring resistor which has been used to calculate LED current (voltage drop of 1 ohm (nominal value) resistance gives current amplitude of LED) and a current switching device power MOSFET. The reason to use power MOSFET as a current-switching device instead of BJT is that the MOSFET saturation voltage is less than the BJT saturation voltage at high currents. It can be assumed that LED has been driving the pulsed current source because the total series resistance is too large compared to LED resistance at pulsed high currents. LED voltage and LED current are constant throughout the pulse duration. A photodetector circuit has been used to measure radiant power of LED (Ozuturk 2015; Ozuturk 2021 a; Ozuturk 2019; Ozuturk 2016; Ozuturk 2013; Ozuturk 2000). The photodetector circuit has been formed by connecting a photodiode and a resistor in photoconductive mode. Photodetector voltage (Pn(t) is proportional to the LED’s radiant power. The position of the LED and photodiode must be fixed during the calculation process.

There is not much research on thermal impedance measurement of LED (Ozuturk 2016). Z(t) depends on junction and ambient temperatures and thermal power dissipated in junction:

$$Z\left(t\right)=\frac{{T}_{j}\left(t\right)-{T}_{a}\left(t\right)}{{P}_{d}\left(t\right)}$$

(5)

Fig. 1

Photodetector voltage versus LED current for two different times (t₂ ˃ t₁)

Z(t) dimension is °C/W. Pd(t) is thermal power and it is equal to the difference of electrical power applied to the junction (Pe = Vn(t) × In) with radiant power (Po optical power) of LED (Pd(t ) = Vn(t)In - Po). The optical efficiency of general-purpose LEDs is maximum 5%. If a constant a = (0.975 ± 0.025) is used, the thermal power can be written approximately as P_d(t) = aVnIn. Using the last two equations, the LED’s thermal impedance can be found as follows (Ozuturk 2016):

$$Z\left(t\right)=\left(\frac{{k}_{pn}{I}_{n}-{P}_{n}\left(t\right)}{{k}_{pn}{I}_{n}k}\right)/\left(a{V}_{n}{I}_{n}\right)$$

(6)

Using the last equation, the voltage-current characteristic of LEDs can be obtained as written Eq. 2.

2 Material and method

It is possible to calculate the junction temperature and thermal impedance of LED using the light pulse response of the photodetector (Ozuturk 2013). Photodetector voltage is proportional to the LED’s radiant power. Photodetector voltage decreases as junction temperature increases because of the self-heating of LED in pulse duration. The junction temperature of the LED can be calculated by drawing a triangle on the top of the photodetector voltage pulse (Fig. 2) (Ozuturk 2021 a; Ozuturk 2000). There are some relationships between Figs. 1 and 2. IOAI is equal to k_pnI_n and junction temperature at this time is equal to ambient temperature (T_a). The top of the photodetector voltage pulse decreases as the junction temperature of the LED increases (Fig. 2). The relative change of photodetector voltage is (IOAI - P_n(t₂) / IOAI = IBCI / IOAI. If this relative amount is divided by k, T_j(t₂) - T_a can be found as follows (Ozuturk 2021 a):

Fig. 2

Photodetector voltage versus time for one LED current (In)

$${T}_{j}\left({t}_{2}\right)-{T}_{a}=\frac{\frac{{k}_{pn}{I}_{n}-{P}_{n}\left({t}_{2}\right)}{{k}_{pn}{I}_{n}}}{k}=\frac{b\left|BC\right|}{b\left|OA\right|k}=\frac{\left|BC\right|}{\left|OA\right|k}$$

(7)

The last equation shows how to calculate junction temperature using photodetector voltages instead of the radiant power of LED. bIBCI and bIOAI are optical magnitudes. Thermal impedance can be calculated as follows:

$$Z\left({t}_{2}\right)=\frac{\left|BC\right|}{\left|OA\right|ka{V}_{n}{I}_{n}}$$

(8)

Another new LED’s voltage-current formula can be obtained using the last equation as follows:

$${V}_{n}\left(t\right)=\frac{\left|BC\right|}{\left|OA\right|kaZ\left(t\right){I}_{n}\left(t\right)}$$

(9)

A photodetector voltage pulse (Ozuturk 2000) and a triangle at the top of it have been shown in Fig. 3 (Ozuturk 2021 a). The LED identity is TLHRS 101 (TFK (ITT). The junction temperature is equal to the ambient temperature at the beginning of time (t = 0). The current amplitude of LED is I_n=10.92 A and the voltage drop of LED for this current pulse is V_n=9.8 V and these values are constant. The decreasing photodetector voltage amount for t₂ = 18µs is IBCI = 1.5–1.15 = 0.35 V. The increasing amount of junction temperature at this time is T_j(t₂) - T_a = (I BC I / I OA I) / k = (0.35 / 1.5) / 0.007 = 33 °C (Ozuturk 2021 a). The thermal power is approximately P_d = aV_n I_n = 0.975 × 9.8 × 10.92 = 104.34 W. Thermal impedance can be calculated as (T_j(t₂) -T_a) / P_d = 33 °C / 104.34 W = 0.316 °C/W. Z(t₁) (t₁ = 12µs) can be calculated as follows:

$$Z\left({t}_{1}\right)=\frac{\left|DE\right|}{\left|OA\right|ka{V}_{n}{I}_{n}}=\frac{(1.5-1.25)}{1.5\times 0.007\times 0.975\times 9.8\times 10.92}=0.228 ^\circ \text{C}/\text{W}$$

(10)

Fig. 3

A photodetector voltage pulse (Ozuturk 2000) and a triangle at the top of it (Ozuturk 2021 a). The current amplitude of LED is I_n=10.92 A and the voltage drop of LED for this current pulse is V_n=9.8 V

Another example has been given in Fig. 4 (Ozuturk 2021 a; Ozuturk 2000). This graph has been obtained for a different TLHRS 101 LED. This photodetector voltage pulse has been obtained for 12.5 A LED current and 10.5 V LED voltage and they are constant. In this figure, I BC I is equal to 4.8–2.2 = 2.6 V (t₂ = 18µs). The increasing amount of junction temperature at this time is (I BC I / I OA I) / k = (2.6 / 4.8) / 0.007 = 77.4 °C (Ozuturk 2021 a). Thermal impedance is Z(t₂) = (T_j(t₂) - T_a) / aV_nI_n = 77.4 / (0.975 × 10.5 × 12.5) = 77.4 / 127.97 = 0.605 °C/W. For t₁ = 10µs, the junction temperature is {(IDEI / IOAI) / k} = {[ (4.8–3.3) / 4.8 ] / 0.007} = 44.6 °C (Ozuturk 2021 a). Thermal impedance for 10µs is Z(t₁) = ( T_j(t₁) - T_a ) / aV_nI_n = 44.6 / (0.975 × 10.5 × 12.5) = 44.6/127.97 = 0.348 °C/W.

Fig. 4

A photodetector voltage pulse (Ozuturk 2000) and a triangle at the top of it (Ozuturk 2021 a). This photodetector voltage pulse has been obtained for 12.5 A LED current and 10.5 V LED voltage

2.1 Junction temperature-time, thermal impedance-time variations and maximum pulse duration

ABC and ADE triangles are similar triangles which yield in Figs. 3 and 4. Because of that, using the triangles that yield in Fig. 3 [BC]/t₂ = 0.35 V/18µs ≅ 0.02 × 10⁶ can be found. [BC]/t₂ is equal to IDEI/t₁ = 0.25 V/12µs = 0.02 × 10⁶. So, for any given time (t) in pulse duration, the voltage reduction from IOAI can be obtained as (IBCI/t₂)×t. For exemple, for t = t₁ = 12µs, IDEI=(IBCI/t₂)×t₁=(0.35 V/18 µs)×12 µs = 0.233. The difference between the two IDEI values has been originated from the reading error of Fig. 3. The pulse in Fig. 3 has been obtained for the current amplitude I_n=10.92 A. The voltage drop of LED for this current pulse is V_n=9.8 V.

The junction temperature formula can be written as follows;

$${ \text{T}}_{\text{j}}\left(\text{t}\right)=\left(\frac{\text{I}\text{B}\text{C}\text{I}}{{\text{t}}_{2}}\text{t}\right)\frac{1}{\text{I}\text{O}\text{A}\text{I}}\frac{1}{\text{k}}+{T}_{a}=\text{0,02}\times {10}^{6}t\frac{1}{1.5}\frac{1}{0.007}+25=1904762t+25$$

(11)

Similarly, the thermal impedance formula can be written as follows;

$$\text{Z}\left(\text{t}\right)=\left[\left(\frac{\text{I}\text{B}\text{C}\text{I}}{{\text{t}}_{2}}\text{t}\right)\frac{1}{\text{I}\text{O}\text{A}\text{I}}\frac{1}{\text{k}}\right]\frac{1}{a{V}_{n}{I}_{n}}$$

(12)

According to Fig. 4, IBCI is equal to 2.6 V and the time is 18µs. IOAI is 4.8 V. Thus, the junction temperature equation is as follows;

$${ \text{T}}_{\text{j}}\left(\text{t}\right)=\left(\frac{\text{I}\text{B}\text{C}\text{I}}{{\text{t}}_{2}}\text{t}\right)\frac{1}{\text{I}\text{O}\text{A}\text{I}}\frac{1}{\text{k}}+{T}_{a}=\frac{2.6}{18\times {18}^{-6}}\frac{1}{4.8}\frac{1}{0.007}+25=4298941.7t+25$$

(13)

Equation 11 and Eq. 12 are line equations. Figure 5 shows junction temperature variation according to time in pulse duration. There are two line variations in this figure. The change in line 1 depends on the pulse change in Fig. 4. The LED’s voltage drop is V_n=10.5 V and the current amplitude is I_n=12.5 A and they are constant while getting line 1. The change in line 2 depends on the pulse change in Fig. 3. Similarly, the LED’s voltage drop is V_n=9.8 V and the current amplitude is I_n=10.92 A while getting line 2 and they are constant.

Fig. 5

Junction temperature-time variations. The LED’s voltage drop is V_n=10.5 V and the current amplitude is I_n=12.5 A for line 1. The LED’s voltage drop is V_n=9.8 V and the current amplitude is I_n=10.92 A for line 2

Thermal impedance-time variations can be drawn using Eq. 11 and Eq. 13. The following equation can be written using Eq. 11:

$$Z\left(t\right)=\frac{{T}_{j}\left(t\right)-{T}_{a}}{a{V}_{n}{I}_{n}}=\frac{1,904,762}{0.975\times 9.8\times 10.92}t=18255.2\times t$$

(14)

This variation has been shown as line 2 in Fig. 6. Similarly, the following equation can be written using Eq. 13:

$$Z\left(t\right)=\frac{{T}_{j}\left(t\right)-{T}_{a}}{a{V}_{n}{I}_{n}}=\frac{4,298,941}{0.975\times 10.5\times 12.5}t=33585.5\times t$$

(15)

The variation determined by Eq. 15 has been shown in Fig. 6 as line 1.

Fig. 6

Thermal impedance-time variations. The LED’s voltage drop is V_n=10.5 V and the current amplitude is I_n=12.5 A for line 1. The LED’s voltage drop is V_n=9.8 V and the current amplitude is I_n=10.92 A for line 2

The maximum pulse duration can be calculated using these equations. For example, the maximum junction temperature can be accepted as approximately 120 °C. New carriers have been generated in the depletion region of LED if the junction temperature is bigger than 120 °C. Maximum pulse duration t_d can be calculated for Fig. 4 and Eq. 13 as follows;

$${T}_{j}\left(t\right)-{T}_{a}=120-25=95=4298941t$$

$$t={t}_{d}=\frac{95}{4298941.7}=22.1\mu s$$

This is the maximum pulse duration. LED has been broken if pulse duration exceeds this time. The maximum pulse duration also depends on ambient temperature and current amplitude. If the current amplitude increases, the pulse duration becomes shorter. Maximum pulse duration decreases as ambient temperature increases. The pulse in Fig. 4 has been obtained for the current amplitude I_n=12.5 A. The voltage drop of LED for this current pulse is V_n=10.5 V. The current amplitudes and voltage drops on LED are constant.

2.2 Error calculation

Relative error of thermal impedance can be calculated as follows:

$$Z\left({t}_{1}\right)=\frac{\left|DE\right|}{\left|OA\right|ka{V}_{n}{I}_{n}}$$

$$\text{l}\text{n}\left(Z\left({t}_{1}\right)\right)=\text{l}\text{n}\left(\frac{\left|DE\right|}{\left|OA\right|ka{V}_{n}{I}_{n}}\right)$$

$$\text{ln}\left(Z\left({t}_{1}\right)\right)=ln\left|DE\right|-\text{l}\text{n}\left(\left|OA\right|ka{V}_{n}{I}_{n}\right)$$

$$\text{d}(\text{ln}\left(Z\left({t}_{1}\right)\right))=d(ln\left|DE\right|+ln\left|OA\right|+lnk+lna+ln{V}_{n}+ln{I}_{n})$$

$$\frac{\text{d}\left(Z\left({t}_{1}\right)\right)}{{Z(t}_{1})}=\frac{d\left(\left|DE\right|\right)}{\left|DE\right|}+\frac{d\left(\left|OA\right|\right)}{\left|OA\right|}+\frac{dk}{k}+\frac{da}{a}+\frac{d{V}_{n}}{{V}_{n}}+\frac{{dI}_{n}}{{I}_{n}}$$

$$\frac{{\Delta }\left(Z\left({t}_{1}\right)\right)}{{Z(t}_{1})}\cong \frac{\varDelta \left(\left|DE\right|\right)}{\left|DE\right|}+\frac{\varDelta \left(\left|OA\right|\right)}{\left|OA\right|}+\frac{\varDelta k}{k}+\frac{\varDelta a}{a}+\frac{\varDelta {V}_{n}}{{V}_{n}}+\frac{{\varDelta I}_{n}}{{I}_{n}}$$

According to Fig. 3, the following values can be written:

IDEI=(0.25 ± 0.01)V, IOAI=(1.5 ± 0.01)V, k=(0.007 ± 0.001)/°C, a = 0.975 ± 0.025, V_n=(9.8 ± 0.1)V, I_n=(10.92 ± 0.01)A, Z(t₁) = 0.228 °C/W

$$\frac{{\Delta }\left(Z\left({t}_{1}\right)\right)}{{Z(t}_{1})}\cong \frac{0.01}{0.25}+\frac{0.01}{1.5}+\frac{0.001}{0.007}+\frac{0.025}{0.975}+\frac{0.1}{9.8}+\frac{0.01}{10.92}=0.226=22.6\%$$

$$\Delta Z\left( {{t_1}} \right) = 0.226 \times 0.228 = 0.051^\circ C/W$$

$$Z\left( {{t_1}} \right) = \left( {0.228 \pm 0.051} \right){\rm{ }}^\circ C/W$$

Similarly Z(t₂)±ΔZ(t₂) can be calculated as follows:

$$\frac{{\Delta }\left(Z\left({t}_{2}\right)\right)}{{Z(t}_{2})}\cong \frac{\varDelta \left(\left|BC\right|\right)}{\left|BC\right|}+\frac{\varDelta \left(\left|OA\right|\right)}{\left|OA\right|}+\frac{\varDelta k}{k}+\frac{\varDelta a}{a}+\frac{\varDelta {V}_{n}}{{V}_{n}}+\frac{{\varDelta I}_{n}}{{I}_{n}}$$

$$\frac{{\Delta }\left(Z\left({t}_{2}\right)\right)}{{Z(t}_{2})}\cong \frac{0.01}{0.35}+\frac{0.01}{1.5}+\frac{0.001}{0.007}+\frac{0.025}{0.975}+\frac{0.1}{9.8}+\frac{0.01}{10.92}=0.215=21.5\%$$

$$\Delta Z\left( {{t_2}} \right) = 0.215{\rm{ }} \times 0.316 = 0.068^\circ C/W$$

$$Z\left( {{t_2}} \right) = \left( {0.316 \pm 0.068} \right){\rm{ }}^\circ C/W$$

According to Fig. 4, the following values can be written:

IDEI=(1.5 ± 0.01)V, IOAI=(4.8 ± 0.01)V, k=(0.007 ± 0.001)/°C, a = 0.975 ± 0.025, V_n=(10.5 ± 0.1)V, I_n=(12.5 ± 0.01)A, Z(t₁) = 0.348 °C/W

$$\frac{{\Delta }\left(Z\left({t}_{1}\right)\right)}{{Z(t}_{1})}\cong \frac{\varDelta \left(\left|DE\right|\right)}{\left|DE\right|}+\frac{\varDelta \left(\left|OA\right|\right)}{\left|OA\right|}+\frac{\varDelta k}{k}+\frac{\varDelta a}{a}+\frac{\varDelta {V}_{n}}{{V}_{n}}+\frac{{\varDelta I}_{n}}{{I}_{n}}$$

$$\frac{{\Delta }\left(Z\left({t}_{1}\right)\right)}{{Z(t}_{1})}\cong \frac{0.01}{1.5}+\frac{0.01}{4.8}+\frac{0.001}{0.007}+\frac{0.025}{0.975}+\frac{0.1}{10.5}+\frac{0.01}{12.5}=0.188=18.8\%$$

$$\Delta Z\left( {{t_1}} \right) = 0.348 \times 0.188 = 0.065^\circ C/W$$

$$Z\left( {{t_1}} \right) = \left( {0.348 \pm 0.065} \right){\rm{ }}^\circ C/W$$

Similarly Z(t₂)±ΔZ(t₂) can be calculated as follows:

$$\frac{{\Delta }\left(Z\left({t}_{2}\right)\right)}{{Z(t}_{2})}\cong \frac{\varDelta \left(\left|BC\right|\right)}{\left|BC\right|}+\frac{\varDelta \left(\left|OA\right|\right)}{\left|OA\right|}+\frac{\varDelta k}{k}+\frac{\varDelta a}{a}+\frac{\varDelta {V}_{n}}{{V}_{n}}+\frac{{\varDelta I}_{n}}{{I}_{n}}$$

$$\frac{{\Delta }\left(Z\left({t}_{2}\right)\right)}{{Z(t}_{2})}\cong \frac{0.01}{2.6}+\frac{0.01}{4.8}+\frac{0.001}{0.007}+\frac{0.025}{0.975}+\frac{0.1}{10.5}+\frac{0.01}{12.5}=0.185=18.5\%$$

$$\Delta Z\left( {{t_2}} \right) = 0.605{\rm{ }} \times 0.185 = 0.112^\circ C/W$$

$$Z\left( {{t_2}} \right) = \left( {0.605 \pm 0.112} \right){\rm{ }}^\circ C/W$$

3 Results

The photodetector voltage has been used instead of LED’s radiant power in this new thermal impedance calculation method. There is no optical measurement here. There is no need to use a spectrometer. Thermal impedance can be measured for microseconds. Measurement circuits and methods are simple. The error calculation demonstrates the accuracy of thermal impedance measurement. Calculation has been given for two different LEDs and different values of LED’s currents and voltages to see which difference can be occur.

The slope at the top of the photodetector voltage becomes evident when the LED’s current exceeds 4 amps. After 4 amps LEDs behave like approximately 1 ohm resistance. This internal resistance is sum of bulk and contact resistances and LEDs have been behaved approximately like resistor. The LEDs tested here are general-purpose LEDs. These LEDs have a maximum 5% optical efficiency. Power LEDs have more optical efficiency and because of that the measurement error becomes high. Therefore, this measurement method is valid for pulsed currents of short duration (microseconds) and large amplitude. It has been shown that the junction temperature and thermal impedance change linearly over time at pulsed high currents. LED can be destroyed at more LED’s current amplitudes and pulse durations. The unit of thermal impedance is not standard as used in the world standard JEDEC. The JEDEC standard of thermal impedance is °K/W. In this study k is used and therefore the thermal impedance unit is °C/W.

The voltage amplitude and current amplitude of the LED are constant during the pulse. The maximum pulse duration at which the LED remains intact can be calculated using the junction temperature-time equation.