VISHAY SEMICONDUCTORS www.vishay.com Optocouplers and Solid-State Relays Application Note 45 How to Use Optocoupler Normalized Curves A phototransistor optocoupler provides signal transfer between an isolated input and output via an infrared LED and a silicon NPN phototransistor. When current is forced through the LED diode, infrared light is generated that irradiates the photosensitive base collector junction of the phototransistor. The base collector junction converts the optical energy into a photocurrent which is amplified by the current gain (HFE) of the transistor. The gain of the optocoupler is expressed as a Current Transfer Ratio (CTR), which is the ratio of the phototransistor collector current to the LED forward current. The current gain (HFE) of the transistor is dependent upon the voltage between its collector and emitter. Two separate CTRs are often needed to complete the interface design. The first CTR, the non-saturated or linear operation of the transistor, is the most common specification of a phototransistor optocoupler and has a VCE of 10 V. The second is the saturated or switching CTR of the coupler with a VCE of 0.4 V. Figures 1 and 2 illustrate the normalized CTRCE for the linear and switching operation of the phototransistor. Figure 1 shows the normalized non-saturated CTRCE operation of the coupler as a function of LED current and ambient temperature when the transistor is operated in the linear mode. Normalized CTRCE(SAT) is illustrated in figure 2. The saturated gain is lower with LED drive greater than 10 mA. 2.0 Normalized CTR 1.5 TA = 25 °C TA = 50 °C TA = 70 °C TA = 100 °C 0.8 0.6 0.4 Normalized to: IF = 10 mA, VCE = 10 V TA = 25 °C 0.2 0.0 17486 0.1 1 10 100 IF - LED Current (mA) Fig. 2 - Normalized Saturated CTR The following design example illustrates how normalized curves can be used to calculate the appropriate load resistors. PROBLEM 1 Using an IL1 optocoupler in a common emitter amplifier (figure 3) determine the worst case load resistor under the following operation conditions: VCC IF RL Normalized to: IF = 10 mA, VCE = 10 V TA = 25 °C 1.0 VCE(SAT) = 0.4 V HC04 V OL 17487 Fig. 3 - IL1 to 74HC04 Interface 0.5 TA = 70 °C, IF = 2 mA, VOL = 0.4 V, logic load = 74 HC04 0.0 0.1 17485 1 10 IF - LED Current (mA) Fig. 1 - Normalized CTR vs. IF and TA Rev. 1.5, 18-Oct-11 100 IL1 Characteristics: CTRCE(NON SAT) = 20 % min. at TA = 25 °C, IF = 10 mA, VCE = 10 V Document Number: 83706 1 For technical questions, contact: [email protected] THIS DOCUMENT IS SUBJECT TO CHANGE WITHOUT NOTICE. THE PRODUCTS DESCRIBED HEREIN AND THIS DOCUMENT ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT www.vishay.com/doc?91000 APPLICATION NOTE TA = 25 °C TA = 50 °C TA = 70 °C TA = 100 °C 1.0 Normalized CTR An optocoupler provides insulation safety, electrical noise isolation, and signal transfer between its input and output. The insulation and noise rejection characteristics of the optocoupler are provided by the mechanical package design and insulating materials. Application Note 45 www.vishay.com Vishay Semiconductors How to Use Optocoupler Normalized Curves SOLUTION VCC Step 1. IF Determine CTRCE(SAT) using the normalization factor (NFCE(SAT)) found in figure 2. RL HC04 Normalized CTRCB 1.5 I CB SCTRCB - 25 Normalized to: SCTRCB - 50 IF = 10 mA, VCB = 9.3 V TA = 25 °C SCTRCB - 70 SCTRCB - 100 1.0 VO R BE 17489 0.5 Fig. 5 - Optocoupler/Logic Interface with RBE Resistor 1.5 0.0 1 17488 10 IF - LED Current (mA) Fig. 4 - Normalized Saturated CTR CTRCE(SAT) = CTRCE(NON SAT) NFCE(SAT) CTRCE(SAT) = 20 % x 0.36 CTRCE(SAT) = 7.2 % SCTRCB - 25 SCTRCB - 50 SCTRCB - 70 SCTRCB - 100 100 (1) Normalized CTRCB 0.1 1.0 Normalized to: IF = 10 mA, VCB = 9.3 V TA = 25 °C 0.5 Step 2. Select the minimum load resistor using the following equation: RL(min.) = CC V VOL 0.0 0.1 1 17490 CTRCE(SAT) IF 100 % - IL (2) 10 100 IF - LED Current (mA) Fig. 6 - Normalized CTRCB vs. LED current - RL(min.) = 5 V 0.4 V (0.072) 2 mA 100 % - CTCB = 50 µA APPLICATION NOTE RL(MIN) = 48.94 kΩ, select 51 kΩ ± 5 % The switching speed of the optocoupler can be greatly improved through the use of a resistor between the base and emitter of the output transistor. This is shown in figure 5. This resistor assists in discharging the charge stored in the base to emitter and collector to base junction capacitances. When such a speed-up technique is used the selection of the collector load resistor and the base emitter resistor requires the determination of the photocurrent and the hFE of the optocoupler. The photocurrent generated by the LED is described by the CTRCB of the coupler. This relationship is shown in equations 3 and 4. Equation 5 shows that CTRCE is the product of the CTRCB and the hFE. The hFE of the transistor is easily determined by evaluating equation 4, once the CTRCE(SAT) and CTRCB are known. The normalized CTRCB is shown in figure 6. Equations 5, 6, and 7 describe the solution for determining the RBE that will permit reliable operation. Rev. 1.5, 18-Oct-11 ICB IF ICB = IF (3) 100 % CTRCB (4) 100 % CTRCE(SAT) = CTRCB HFE(SAT) hFE(SAT) = RBE = RBE = (6) CTRCB VBE (7) ICB - IBE VBE HFE(SAT) RL ICB HFE(SAT) RL - [VCC- VCE(SAT)] VBE RBE = CTRCE(SAT) CTRCE NFCE(SAT) CTRCB NFCB IF CTRCE NFCE(SAT) RL 100 % (5) (8) RL (9) - [VCC- VCE(SAT)] Document Number: 83706 2 For technical questions, contact: [email protected] THIS DOCUMENT IS SUBJECT TO CHANGE WITHOUT NOTICE. THE PRODUCTS DESCRIBED HEREIN AND THIS DOCUMENT ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT www.vishay.com/doc?91000 Application Note 45 www.vishay.com Vishay Semiconductors How to Use Optocoupler Normalized Curves PROBLEM 2 Using an IL2 optocoupler in the circuit shown in figure 6, determine the value of the collector load and base emitter resistor, given the following operational conditions: TA = 70 °C, IF = 5 mA, VOL = 0.4 V, Logic load = 74HC04 IL2 Characteristics: CTRCE = 100 % at Tamb = 25 °C, VCE = 10 V, IF = 10 mA CTRCB = 0.24 % at Tamb = 25 °C, VCB = 9.3 V, IF = 10 mA SOLUTION Step 1. Determine CTRCE(SAT), and CTRCB. From figure 2 the CTRCE(SAT) = 55 %, [NFCE(SAT) = 0.55] From figure 6 the CTRCB = 0.132 %, [NFCB = 0.55] Step 2. Determine RL. From equation 2 RL = 1.7 kΩ Select RL = 3.3 kΩ TABLE 1 IF = 5 mA, VCC = 5 V RL = 3.3 kΩ RBE = ∞ Ω RL = 3.3 kΩ RBE = 220 kΩ tdelay 1 μs 2 μs trise 4 μs 5 μs tstorage 17 μs 10 μs tfall 5 μs 12 μs tphl 3.5 μs 7 μs tplh 22 μs 12 μs Pulse distortion 50 μs pulse 37 % 10 % Not only does this circuit offer less pulse distortion, but it also improves high-temperature switching and common mode transient rejection while lowering static DC power dissipation. Step 3. Determine RBE, using equation 9. 0.65 V RBE = (100 %)(0.55) (3.3 kΩ) (0.24 %)(0.55) (5 mA)(100 %)(0.55)(3.3 kΩ) [5 V- 0.4 V] 100 % (10) RBE = 199 kΩ, select 220 kΩ APPLICATION NOTE Using a 3.3 kΩ collector and a 220 kΩ base emitter resistor greatly minimizes the turn-off propagation delay time and pulse distortion. The following table illustrates the effect the RBE has on the circuit performance. Rev. 1.5, 18-Oct-11 Document Number: 83706 3 For technical questions, contact: [email protected] THIS DOCUMENT IS SUBJECT TO CHANGE WITHOUT NOTICE. THE PRODUCTS DESCRIBED HEREIN AND THIS DOCUMENT ARE SUBJECT TO SPECIFIC DISCLAIMERS, SET FORTH AT www.vishay.com/doc?91000

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