How do I select the accompanying components for an optocoupler?

Question

I'm using an optocoupler (MOC3021) to sense the On/Off state of an electrial appliance using a microcontroller ATmega16L. How do i go about doing this? My mains supply specs are 230V, 50Hz. How do I design the surrounding circuit and select component values, like the resistors?

EDITED on 13th June 2012 Note: This is the first time I'm solving a circuit like this. Please send any helpful feedback. (including things I did wrong or any improvements)

Referring to the above schematic. The idea is to use this circuit to determine whether the load is on or off. The output pin from the optocoupler connects to an external interrupt of the Microcontroller I'm using which is ATmega16L. The interrupt will Monitor the state of the load. After monitoring I can toggle the state of the load using a relay (relay acts as a Control mechanism) which connects to the same microcontroller.

Now, I tried calculating the resistor values for R1, R2 and Rc. Note, microcontroller's VIL(max) = 0.2xVcc = 660mV and VIH(min) = 0.6xVcc = 1.98V and VIH(max) = Vcc+0.5 = 3.8V.

To calculate Rc is quite easy. When the transistor is not conducting the output is high (at 3.3V). When the transistor conducts the output is pulled low. so from microcontroller's point of view, output high means load is switched OFF and output low means load is switched ON.

Looking at the datasheet for SFH621A-3, using 34% minimum CTR at IF = 1mA. Therefore, at 1mA input, the output is going to be 340uA. So in order for the microcontroller to detect low voltage from the output of the optocoupler can I use resistor value of 1Kohm? So that the output from the optocoupler will have a voltage of 340mV (which is below VIL(max))

More on this later, been a long day.

EDITED on 15th June 2012

Note: Solving for resistors on the power line (R1 and R2). Please check my calculations and any appropriate feedbacks.

Aim: the aim is to keep the LEDs *ON** for maximum period of time in a 10mS half period (20mS full period of 50Hz). Lets say LEDs have to be ON for 90% of the time, that means LEDs require at least 1mA of current for 90% of the time for that half period which means LEDs will be active for 9mS in a 10mS half period. So, 9mS/10mS = 0.9 * 180(half period) = 162 degrees. This shows the current will be 1mA between 9deg and 171deg (and less than 1mA from 0deg to 9deg and 171deg to 180deg). Did not consider ON time to be 95% as working with whole numbers is neat and 5% doesn't make any difference not in this application at least.

Vpeak-peak = 230V x sqrt(2) = 325V. Taking tolerances into account. Minimum tolerance of 6%. 325 x 0.94 (100-6) x sin(9) = 47.8V

So, R1 ≤ (47.8V - 1.65V) / 1mA = 46.1 Kohms Choosing a value one smaller than 46.1 Kohms of 39 Kohms (e12 series). Now that a smaller value resistance is chosen compared to what was calculated, means current through the diodes will be greater than 1mA.

Calculating new current: ((325V x 110%) - 1.25V)/39 Kohms = 9.1mA (too close to max If of diodes). Coming back to this in a moment [Label - 1x]

First calculate the power ratings of the resistor (considering 39 Kohm) ((230 + 10%)^2) / 39K = 1.64 Watts (too high).

Going back to calculation [Label - 1x] Lets choose two 22 Kohm resistors. Together they add upto 44 Kohm which is quite close 46.1 Kohm(calculated above)

checking power rating of the two resistors combined: ((230 + 10%)^2) / (2 x 22) Kohm = 1.45W. choose 22 Kohm resistors each with 1W power rating.

Now, after all this the initial CTR was 34% which means 1mA in will be 340µA out. But now because of 2x22 Kohm resistors the current will be slightly more on the output. That means higher potential across the pull up resistor Rc. Would there be an issue to get a volt drop below 500mV on the output of the optocoupler??

Answer 1

The MOC3021 is an optocoupler with a triac output. It's used to drive a power triac typically to switch mains operated appliances. Triacs can only be used in AC circuits.

You need an optocoupler with a transistor output, preferably one with two LEDs in antiparallel at the input. The SFH620A is such a part.

The two LEDs in antiparallel ensure that the transistor is activated on both half cycles of the mains. Many optocouplers only have 1 LED, that would work, but give you an output pulse of 10ms in a 20ms period for 50Hz. You would need to place a diode in antiparallel to the input also in that case, to protect the LED from overvoltage when reverse polarized.

Important is CTR or Current Transfer Ratio, which indicates how much output current the transistor will sink for a given LED current. CTR is often not very high, but for the SFH620A we can choose a value of 100% minimum, only that's at 10mA in, at 1mA it's only 34% minimum, so that 1mA in means at least 340\$\mu\$A out.

Let's suppose the output goes to a 5V microcontroller and that you would use the 2k\$\Omega\$ pull-up resistor shown in the diagram. If the transistor is off it won't draw current, except for a small leakage current, 100nA maximum, according to the datasheet. So that will cause a voltage drop of 200\$\mu\$V across the resistor, which is more than safe.

If the transistor is on, and it draws 340\$\mu\$A then the voltage drop across the resistor is only 680mV, and that's way too low to get a low level. We'll have to increase either the resistor's value or the current. Since we had a lot of margin on the leakage current we can safely increase the resistor value to 15k\$\Omega\$ for instance. Then 340\$\mu\$A will give a sufficiently low output voltage. (Theoretically 5.1V voltage drop, but there's only 5V available, so it will go to ground.) The voltage drop because of the leakage current is still well within limits at 1.5mV.

If we want to have a CTR of at least 34% at 1mA we have to use the SFH620A-3.

If this would be controlled from a DC source we would almost be done. Just add R1 in series with the LEDs, R2 will probably not be needed. Then R1 \$\leq\$ (\$V_{IN}\$ - \$V_{LED}\$) / 1mA.

But we have to deal with a 230V AC input signal. At the zero crossings there won't be any current, there's little we can do about that. How can we get at least 1mA for most of the cycle without wasting too much power? This is a trade-off. You can have the 1mA for just the maximum voltage, and that will give you only a small pulse, but you'll waste the least power. Or you can go for 1mA for most of the cycle, but then you'll have more current when the voltage is highest. Let's say we want at least a 9ms pulse in a 10ms half period (50Hz). That means the current has to be 1mA at a 9° phase until 171°. 230V AC is 325V peak, but we have to take a -6% tolerance into account, so that's 306V minimum. 306V \$\times\$ sin(9°) = 48V. R1 \$\leq\$ (48V - 1.65V) / 1mA = 46.2k\$\Omega\$. (The 1.65V is the LED's maximum voltage.) The closest E24 value is 43k\$\Omega\$. Then we have more than 1mA at a 9° phase, but what at the voltage's maximum. For that we have to work with the positive tolerance, max. 10%. Then peak voltage is 230V \$\times\$ \$\sqrt{2}\$ \$\times\$ 110% = 358V. Maximum current is then (358V - 1.25V) / 43k\$\Omega\$ = 8.3mA. (The 1.25V is the LED's nominal voltage). That's well below the optocoupler's limit.

We won't be able to do this with just 1 resistor. It probably can't stand the high voltage, and may have power dissipation problems too, we'll come to that in a minute. Peak voltage across the resistor is 357V. The MFR-25 resistor is rated at max 250V, so we'll need at least 2 of them in series. How about power? 230V+10% in 43k\$\Omega\$ is 1.49W. The MFR-25 is only rated at 1/4W, so two of them won't do. Now you can choose to have more of them in series, but that would have to be at least 6, or choose a higher rated resistor. The MFR1WS (same datasheet) is rated at 1W, so 2 in series will do. Remember that we'll have to divide the resistor value by 2: 21.5k\$\Omega\$, which is not an E24 value. We can choose the closest E24 value an d check our calculations, or choose an E96. Let's do the latter.

That's all, folks. :-)

edit
I suggested in comment that there's a lot more which has to be accounted for, this answer could well be 3 times as long. There's for instance the input leakage current of the AVR's I/O pin, which can be ten times as high as the transistor's. (Don't worry, I checked it, and we're safe.)

Why didn't I choose an optocoupler with Darlington output? They have a much higher CTR.
The main reason is the Darlington's saturation voltage, which is much higher than for a common BJT. For this optocoupler for instance it can be as high as 1V. For the ATmega16L you're using the maximum input voltage for a low level is 0.2 \$\times\$ \$V_{DD}\$, or 0.66V at a 3.3V supply. The 1V is too high. That's the main reason.
Another reason could be that it may not really help. We do have enough output current, it's just that the 1mA input current is so high that we need power resistors for them. Darlingtons don't necessarily solve this if they're also only specified at 1mA. At a 600% CTR we'd get 6mA collector current, but we don't need that. Can't we do anything about the 1mA in? Probably. For the optocoupler I mentioned the Electrical Characteristics only talk about 1mA. There's a graph in the datasheet, fig.5: CTR versus forward current, which shows a CTR of more than 300% at 0.1mA. You have to be careful with these graphs. While tables often give you minimum and/or maximum values, graphs usually give you typical values. You may have 300%, but it may be lower. How much lower? It doesn't say. If you build just one product you can try it, but you can't do that for every optocoupler if you want to run a 10k/year production.
It might work. Say you use 100\$\mu\$A in, and at a relatively safe CTR value of 100% you would have 100\$\mu\$A out. You would have to do the calculations again, but your major advantage will be that the input resistors will only dissipate 150mW, instead of 1.5W. It way be worth it.

Answer 2

In my other answer I explained why I didn't use a Darlington optocoupler there: the main reason is the Darlington's saturation voltage, which is much higher than for a common BJT, it can be as high as 1 V. For the ATmega16L you're using the maximum input voltage for a low level is 0.2 × VDD, or 0.66 V at a 3.3 V supply. The 1 V is too high.

But it's not a thing which can't be fixed, it takes just a couple of extra components. At the same time we'll do something about the 1 mA input current as well.

To start with the input current, we had to use 1 mA because the datasheet didn't mention anything lower, and then you can try things, but you're on your own, there are no guarantees. The datasheet for the FOD816, however, has an interesting graph.

That's the one. This one does give CTR for input currents as low as 100 µA, and it's even high: 350 % (remember this is a Darlington). But you have to be careful with these graphs. While tables will often give you minimum or maximum values this kind of graphs will give you typical values, unless indicated otherwise. So what's minimum? We don't know, but 100 % is safe. Let's go for even more safety and assume a CTR of 50 %. So for 100 µA in we'd get 50 µA out. Let's see if that's enough.

This is the modified output stage. U1's transistor is the photo-Darlington, sourcing the 50 µA when on. Let's choose 10 µA for R4, so its value will be 0.6 V/10 µA = 60 kΩ. I'll come back to the function of R4 later.

Then we have 40 µA remaining for T1's base current. If we pick a BC857A for that we have an \$H_{FE}\$ of 125 minimum, so our collector current is minimum 5 mA. An R5 of minimum 660 Ω is enough to make the output low. Since we're driving a high-impedance microcontroller input we don't need the 5 mA, and might as well choose 15 kΩ, so that limits the current to 220 µA. The 1 µA input leakage of the controller's port will only cause a 15 mV drop, so that's OK.

What's the function of R3? It's normally used to limit the base current, but doesn't the Darlington do this already? Yes, it does, but the value can go pretty high. In my other answer I calculated that the peak LED current could go as high as 8.3 mA, that will become 830 µA in our low power version. We calculated with a safe CTR value of 50 %, but typically it will be 350 %, and maximum maybe even higher. 830 µA \$\times\$ 350 = 290 mA, which is too much for that poor BC857A. So we'll limit that to 100 µA by choosing a 15 kΩ value for R3.

R4 still needs some explanation. Suppose we omit it. Then all the Darlington's current goes to T1. When off the FOD816's leakage current (called "dark current" in the datasheet) can be as high as 1 µA. T1 will amplify that to 250 µA maximum worst case, which is enough to drop 3.3 V across R5. So the output may be permanently low.
We chose a value of 60 kΩ for R4. Then as long as the voltage drop across it is less than 0.6 V all the Darlington's current will go through R4, and none through T1, because the minimum base-emitter voltage isn't reached. That was at 10 µA. So the 1 µA dark current will only cause a 60 mV drop, and no base current.

We have values for all our components, the only thing remaining is to increase the input resistors to 220 kΩ each. You can use 1/4 W resistors for that.

Answer 3

To figure the parameters of the circuit, start with what you need at the output and work backwards. 10 kΩ is a good value for the pullup on the output. Unless you have unusual requirements, like battery operation where low power is important, 10 kΩ is a good tradeoff between low enough to pull the line solidly high against leakage and reasonable noise, but not so low as to require too much current.

When the output transistor in the opto turns on, it will put at most 3.3 V accross Rc. 3.3 V / 10 kΩ = 330 µA, which is the minimum current the transistor must be able to sink. You want some extra so that the line is held solidly low when it is supposed to be low. I'd say at minimum it should be able to sink 500 µA, but I'd use 1 mA unless you have a particular reason to cut it close.

Now that we know that the output has to sink 1 mA, we look in the datasheet of the opto to see how we need to drive it to get that 1 mA out. You are using the "-3" variant of this part, which according to the first page of the datasheet has a minimum guaranteed current transfer ratio of 100%. That means that the transistor can sink at least as much current as you put thru one of the LEDs. However, note the little "±10 mA" above the CTR specs. What this is really saying is that if you put 10 mA thru the LEDs, then the transistor will be able to sink at least 10 mA. It doesn't actually promise anything at any other input current.

Looking thru the datasheet more, you find additional information at the top of page 3. Here they actually show the CTR for 1 mA input. Note that now it's only guaranteed to be 34%. That means to get the 1 mA output sink capability, you have to drive the LEDs with 1 mA / 34% = 2.9 mA, so let's aim for 3 mA absolute minimum.

You say the voltage that has to be sensed is 230 V AC. Since that is a sine, it will have peaks of 325 V. The output signal of the opto is going into a micro, so there is no need for it to be a steady signal when power is on. In fact, it's a good idea for the micro to be able to ride out momentary interruptions and glitches. I'd probably keep a counter that is decremented every ms when the signal is off and reset to something like 50 when on. That means you have to see no signal for 50 ms to declare that power is off. All that is needed is a little blip at the peak of the line cycle and this system will work fine. Note that line cycle peaks occur every 10 ms with 50 Hz power.

So let's see where we're at. We want to have at least 3 mA flow thru the LEDs when the power voltage is 325 V. The LEDs will drop up to 1.65 V (top of bottom table on page 2), and this should still work at the lowest reasonable power line voltage. Let's aim for being able to detect 200 VAC minimum, which is 283 V peak, and 281 V after the LED drop. 281 V / 3 mA = 94 kΩ. In theory, that's all that is needed in series with the LEDs to trigger the output at least a little bit once per power peak.

In practise it's a good idea to add some margin. You want the output to be asserted for a reasonable finite fraction of each half-cycle, not just guaranteed to be on for a little blip. Given all that, I'd roughly halve the resistor to 47 kΩ. That will solidly turn on the output for all reasonable conditions with significant margin.

You might think that's all you need to do, but wait, there's more. Think what will happen at high line voltage, like 240 V. The peaks are 340 V, which would cause 7.2 mA thru the LEDs. You have to check the maximum allowed LED current, which is 60 mA, so that's OK. However, consider the power dissipation in the resistor. If we say the worst case line voltage is 240 V, then the power going into the resistor (ignoring the LED voltage drop) is (240 V)² / 47 kΩ = 1.23 W. That should be at least a "2 W" resistor then, and it will get noticeably warm.

Another issue is that the voltage rating of the resistor needs to be considered. It needs to be able to withstand the 340 V peaks, so overall you need a 47 kΩ resistor rated for 2 W and 400 V. Those can be found, but it might be simpler all around to use several resistors in series. That spreads out the peak voltage and the power dissipation among the series resistors. Four 12 kΩ resistor would do this and will only dissipate 300 mW and see 85 V each. That will be easier to find and cheaper than a single resistor unless this is a volume product where you can buy things in large quantity. So the answer to the question as asked is to put four ordinary 12 kΩ 1/2 Watt resistors in series with the LEDs.

Note that these don't need to be split up on each side of the opto as you show R1 and R2. There only needs to be a single resistance in series with the LEDs somewhere. Since this resistance happens to be made up of four individual resistors in this case, you can split them up any way you like to make things work out best mechanically on the high voltage side of the circuit. Preferably they would be end to end to maximize the creapage path for the high voltage and to spread out the heat.

However, I don't really like this optocoupler for this application since it has such low current transfer ratio, which forces us to provide a lot of LED current, which causes a lot of power to be consumed in the resistor. For this sort of application where high current transfer ratio is useful and speed doesn't matter much, I like the cheap and available FOD817. The D versions of this part have a guaranteed CTR of 3x at 5 mA. They don't say exactly what you get a 1 mA, but it's a pretty safe bet the output can sink at least 1 mA with 1 mA in.

The FOD817 has a single LED, but that is easy to deal with (The FOD814 has back to back LEDs, but is less available and doesn't come in some of the higher gain variants). Using the 50 ms scheme described above, it's no problem if you get a pulse once per line cycle, which is every 20 ms. Put a diode in series with the LED in addition to the resistors, and a high-value resistor accross the LED to make sure it doesn't see high reverse voltage due to a little bit of diode leakage. 100 kΩ is fine, and is high enough for its current to be irrelevant to our other calculations. Another advantage of this is that not only do you get lower power dissipation due to requiring less LED current, but you get another factor of two reduction in power due to the LED only being driven in one direction. A total series resistance of 100 kΩ or so will still give you plenty of current headroom at the power line peaks, but will dissipate less than 300 mW total due to the lower current and being on only half the time.

So here is my final answer:

Answer 4

If you're looking for a very high CTR for this type of application, look at the Liteon LTV-8xxx series. 600% min. at 1mA IF.