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Opto-triac output circuit

Homework Statement
Referring to the opto-triac 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
Opto-triac minimum operating current = 2 mA
Opto-triac 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 opto-triac 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
Opto-triac minimum operating current = 2 mA
Opto-triac LED forward voltage drop = 1.6V

Calculate suitable values for R1 and R2, show all your working.
Homework Equations: Ohm's law

Opto.PNG


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 opto-isolator 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
 

Tom.G

Science Advisor
2,700
1,521
You say you are familiar with series circuits and these circuits are in parallel. So far I agree.

How about if you think of each of the series circuits by itself, one at a time; even though they are in parallel with each other?

Pick one of them and let's see what the calculation is.

Cheers,
Tom
 
Thank you for the reply, Tom.

D1 resistance
R = V/I
RD1 = 1.8/0.01
RD1 = 180 ohms

Calculate total resistance (for one parallel path but treating as a series circuit)

If treat this parallel path as a series circuit:

Rtotal = Vtotal / Itotal

Rtotal = 3.3 / 0.01

Rtotal = 330 Ω

R1 resistance

R1 = 330 - 180 = 150 Ω


I am not sure about the other parallel path because the phrase "minimum operating current" is used rather than "operating current".
 
I may have misinterpreted the LHS circuit.

I thought it was a parallel circuit because of the wire labelled with 0 V.

If current splits, it is parallel, but there is no current flowing into that path? So the LHS is a series circuit.

The current passing through D1 is 0.01 A, because it is a series circuit, the current flowing through the whole LHS is 0.01 A.

Calculate Opto-triac LED resistance

Ropto-triac = 1.6 / 0.01
Ropto-triac = 160 Ω

D1 resistance

R = V/I
RD1 = 1.8/0.01
RD1 = 180 ohms

Calculate Rtotal

Rtotal = Vtotal / Itotal

Rtotal = 3.3 / 0.01

Rtotal = 330 Ω

R1 + 2 resistance

R1 + 2 = 330 - 340 = -10 Ω

That can't be right!
 
Last edited:
I will retry but when calculating the opto-triac resistance, the minimum operating current will be used.

Calculate Opto-triac LED resistance

Ropto-triac = 1.6 / 0.002
Ropto-triac = 80 Ω

D1 resistance

R = V/I
RD1 = 1.8/0.01
RD1 = 180 ohms

Calculate Rtotal

Rtotal = Vtotal / Itotal

Rtotal = 3.3 / 0.01

Rtotal = 330 Ω

R1 + 2 resistance

R1 + 2 = 330 - 260 = 70 Ω
 
If R1 + R2 = 70 Ω, what are the individual values of R1 and R2?

The operating current is 10 mA but only 2 mA is needed through the opto-isolator 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 = 20 Ω

If R1 + R2 = 70 Ω
R1 = (R1 + R2) - R2
R1 = 70 - 20
R1 = 50 Ω
 
Last edited:

NascentOxygen

Mentor
9,203
1,050
What would you say is the function of D1 in this circuit? Explain what it does here.

Ohm's Law says the value of a resistance is equal to the voltage across that resistor ÷ the current through it.

What value is the desired voltage across R1 in this design?
 
The purpose of the LED D1 is to show current is flowing?

If voltage drop across LED D1 is 1.8 V, then the voltage across the resistor is 3.3 - 1.8 = 1.5 V

R = V/I

R1 = 1.5/0.01

R1 = 150 Ω
 
The operating current is 10 mA but only 2 mA is needed through the opto-isolator LED.

Therefore, 10 mA − 2m A = 8 mA will need to be removed by the use of R2:

R2 = 1.6/0.008
R2 = 20 Ω
 

NascentOxygen

Mentor
9,203
1,050
The operating current is 10 mA
10mA is given for the current through D1, but I can't see 10mA specified for anything else.

You are given 2mA as the minimum current for the triac diode, so you could use that as the design parameter, or you might say, well, to allow a margin, I'll design for at least 3mA.
 
If the LHS circuit is a series circuit, 10 mA will flow through the whole circuit.
 

NascentOxygen

Mentor
9,203
1,050
If the LHS circuit is a series circuit, 10 mA will flow through the whole circuit.
But it isn't a series arrangement. The driver supplies current to two diode networks in parallel, so each has a current independent of the other’s.
 
I thought Logic signal (3.3 V) was the only source of voltage in the LHS circuit.
 
If the LHS circuit is a parallel circuit, then the current in the other path needs to be calculated.
 
Is the LHS circuit, a series or parallel circuit?

If it is a series circuit, the same current will flow around it.
If it is a parallel circuit, the paths will NOT have the same current flowing around it.
 
The purpose of R2 is to reduce the current to the Opto-triac LEDs minimum operating current (i.e. 2 mA).

Therefore, it is necessary to calculate the current in that path THEN calculate the resistance required to achieve this reduction.
 
In a parallel circuit, the voltage drops across each of the parallel paths is the same as the voltage gain in voltage source.

If the voltage drop in the Opto-triac is 1.6 V, then the voltage drop in R1 must be 1.7 V (1.6 + 1.7 = 3.3).
 
I am baffled how to calculate the current in that part of the circuit.

I have total voltage but not total current or total resistance.
 
I assumed that the current flows in an anti-clockwise direction.

However, the components indicates the current splits at the first node.
 
If so, then R2 doesn't perform the function outlined above.

The voltage across R2 = 3.3 - 1.6 = 1.7 V
So, R1 is simply:

R1 = 1.7/0.002
R1 = 850 Ω
 
The direction of the current is given in the diagram.

Current passing through a diode can only go in one direction, called the forward direction.

The diode allows current to flow in the direction of the arrow.

So, the current does split at the first node.
 

NascentOxygen

Mentor
9,203
1,050
So, the current does split at the first node.
The current from the logic driver splits and travels to the 0v line via two independent parallel paths. Independent means independent, so the choice of current through R2 here has nothing whatever to do with the value of current through R1.
 
Yes.

The current from the logic driver splits and travels to the 0v line via two independent parallel paths. Independent means independent, so the choice of current through R2 here has nothing whatever to do with the value of current through R1.
Then:

If so, then R2 doesn't perform the function outlined above.

The voltage across R2 = 3.3 - 1.6 = 1.7 V
So, R2 is simply:

R2 = 1.7/0.002
R2 = 850 Ω
 
Last edited:

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