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 3
 Homework Statement

Referring to the optotriac output circuit shown below.
...
Given the following parameters:
Logic voltage = 3.3 V
D1 forward voltage drop = 1.8 V
D1 operating current = 10 mA
Optotriac minimum operating current = 2 mA
Optotriac LED forward voltage drop = 1.6V
Calculate suitable values for R1 and R2, show all your working.
 Homework Equations
 Ohm's law
Homework Statement: Referring to the optotriac output circuit shown below.
...
Given the following parameters:
Logic voltage = 3.3 V
D1 forward voltage drop = 1.8 V
D1 operating current = 10 mA
Optotriac minimum operating current = 2 mA
Optotriac LED forward voltage drop = 1.6V
Calculate suitable values for R1 and R2, show all your working.
Homework Equations: Ohm's law
I am baffled. The example in the notes relates to a series circuit.
Calculating R1 and R2
If 3.3 V were connected, then both of these devices would burn out unless resistors limited the current to a safe value (10 mA?).
Therefore, these resistors need to drop (3.3V?V) whilst carrying 10mA.
The circuit is parallel  So I am using the formula used for resistors in parallel and applying it to the voltage drops in parallel.
1/?V = 1/1.8 + 1/1.6
V = 72/85 V
Therefore, the resistors need to drop 3.3  (72/85) = (417/170) V
R1 + R2 = (417/170) / 0.01
R1 + R2 = (4170/17) Ω
The switch current is 10 mA but only 2 mA is needed through the optoisolator LED.
Therefore, 10 mA − 2m A = 8 mA will need to be removed by the use of R2 with a value:
R2 = 1.6/0.008
R2 = 200 Ω
So, R1 = (4170/170)  200 = (770/17) Ω
...
Given the following parameters:
Logic voltage = 3.3 V
D1 forward voltage drop = 1.8 V
D1 operating current = 10 mA
Optotriac minimum operating current = 2 mA
Optotriac LED forward voltage drop = 1.6V
Calculate suitable values for R1 and R2, show all your working.
Homework Equations: Ohm's law
I am baffled. The example in the notes relates to a series circuit.
Calculating R1 and R2
If 3.3 V were connected, then both of these devices would burn out unless resistors limited the current to a safe value (10 mA?).
Therefore, these resistors need to drop (3.3V?V) whilst carrying 10mA.
The circuit is parallel  So I am using the formula used for resistors in parallel and applying it to the voltage drops in parallel.
1/?V = 1/1.8 + 1/1.6
V = 72/85 V
Therefore, the resistors need to drop 3.3  (72/85) = (417/170) V
R1 + R2 = (417/170) / 0.01
R1 + R2 = (4170/17) Ω
The switch current is 10 mA but only 2 mA is needed through the optoisolator LED.
Therefore, 10 mA − 2m A = 8 mA will need to be removed by the use of R2 with a value:
R2 = 1.6/0.008
R2 = 200 Ω
So, R1 = (4170/170)  200 = (770/17) Ω