Calculating R1 and R2 for Opto-Triac Output Circuit

[Michael Neo]

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.

Relevant 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

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]

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

 

[Michael Neo]

Thank you for the reply, Tom.

D₁ resistance
R = V/I
R_D1 = 1.8/0.01
R_D1 = 180 ohms

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

If treat this parallel path as a series circuit:

R_total = V_total / I_total

R_total = 3.3 / 0.01

R_total = 330 Ω

R₁ resistance

R₁ = 330 - 180 = 150 ΩI am not sure about the other parallel path because the phrase "minimum operating current" is used rather than "operating current".

 

[Michael Neo]

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

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

D₁ resistance

R = V/I
R_D1 = 1.8/0.01
R_D1 = 180 ohms

Calculate R_total

R_total = V_total / I_total

R_total = 3.3 / 0.01

R_total = 330 Ω

R_{1 + 2} resistance

R_{1 + 2} = 330 - 340 = -10 Ω

That can't be right!

 

[Michael Neo]

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
R_D1 = 1.8/0.01
R_D1 = 180 ohms

Calculate R_total

R_total = V_total / I_total

R_total = 3.3 / 0.01

R_total = 330 Ω

R_{1 + 2} resistance

R_{1 + 2} = 330 - 260 = 70 Ω

 

[Michael Neo]

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

 

[NascentOxygen]

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?

 

[Michael Neo]

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

 

[NascentOxygen]

R1 = 150 Ω

That looks right. Now determine R2.

 

[Michael Neo]

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]

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.

 

[Michael Neo]

If the LHS circuit is a series circuit, 10 mA will flow through the whole circuit.

 

[NascentOxygen]

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.

 

[Michael Neo]

I thought Logic signal (3.3 V) was the only source of voltage in the LHS circuit.

 

[NascentOxygen]

I thought Logic signal (3.3 V) was the only source of voltage in the LHS circuit.

It is.

 

[Michael Neo]

If the LHS circuit is a parallel circuit, then the current in the other path needs to be calculated.

 

[Michael Neo]

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.

 

[Michael Neo]

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.

 

[Michael Neo]

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).

 

[Michael Neo]

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.

 

[Michael Neo]

I assumed that the current flows in an anti-clockwise direction.

However, the components indicates the current splits at the first node.

 

[Michael Neo]

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

 

[Michael Neo]

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]

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.

 

[Michael Neo]

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

 

[NascentOxygen]

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

R1 = 1.7/0.002
R1 = 850 Ω

That may be sufficient for this exercise, but in practice you wouldn't design for operation right at the minimum, you'd go for a higher current to allow a margin for reliability. Further, you're not going to specify 850 ohms, you'll choose one of the preferred values.

 

[Michael Neo]

This is a very straightforward question:
(a) if you relax and read the diagram first and understand what is happening (especially the impact of the diodes on the direction of the current);
(b) because of the direction of current flow, the function of R₂ is not the same as the example given in the notes.

What do you mean by "... you'll choose one of the preferred values"?

 

[NascentOxygen]

the function of R2 is not the same as the example given in the notes.

What does the example in your notes say is the function of the series resistor for the LED in the opto-triac?

 

[Michael Neo]

In the example, the purpose of R₂ is to reduce the current to the Opto-triac LEDs minimum operating current (i.e. 2 mA).

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

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

R2 = 1.6/0.008
R2 = 20 Ω

But the direction of current flow in the new question is different because there are two diodes in the LHS circuit.

 

[NascentOxygen]

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

It sounds as though that example involved a fixed current driver, not the fixed voltage driver shown in this exercise. Was R shown connected in series or in parallel with the opto-triac's LED in your notes?

 

[Michael Neo]

Here's the diagram with calculations:

 

[NascentOxygen]

As suspected, it's a different arrangement.

 

[Michael Neo]

R₁ = 150 Ω
R₂ = 850 Ω

This question was easier than the example question.

It is important to read and understand the diagram before throwing things onto the page in panic.

 

[NascentOxygen]

What do you mean by "... you'll choose one of the preferred values"?

If you go into a city electronics store and ask for an 850 ohm resistor, the salesperson will say, "Sorry, but no one stocks that value."

 

[Michael Neo]

Thank you for the excellent advice and support, NascentOxygen.