### Application Note 45

```VISHAY SEMICONDUCTORS
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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
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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,
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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