# Zero Crossing Circuit Input Resistor

I'm building a dimmer, and got some questions about the resistor in the zero cross circuit. My circuit is like the image above, and I'm wondering if in the case of 100V input the current is to small. I'm using smd resistors, with 1/8W and 300V max overload voltage. For the optocoupler I'm using Vishay H11AA1.

At 240 V, $I = \frac {V}{R} = \frac {240}{240k} = 1~mA$.

Calculating the power on each resistor:

$P_{120K\Omega} = I^2R = (1mA)^2 120k = 0,12~W$ (OK... the resistor has power of 0,125W.)

Otherwise, when 100V: $I = \frac {V}{R} = \frac {100}{240k} = 0,41~mA$.

How can I calculate if 0,41 mA is OK for the zero detection? These resistors affects the 10k pullup resistor?

• what does "240 = 240k * i" mean? there's no units on that? what's 240? your formulas are a little vague, to say the least. May 30, 2016 at 19:08
• Equations reordered for readability and MathJAXed. May 30, 2016 at 19:26
• @BrunoAraujo don't worry; it was really just a question :) May 30, 2016 at 19:31
• @MarcusMüller sorry about that.. May 30, 2016 at 19:31
• The P<sub>240</sub> equation is wrong. It is showing 10mA when it should show 1mA. It shows 0.12W when it should show 0.24W. Missing is explanation of how that 0.24W may be spread across two resistors. May 30, 2016 at 20:02

Only you know the intent of the circuit, but I would say that for a dimmer you want to detect close to the actual zero crossing, say within 10V or so.

So with input voltage of 10V you need output current of about 300uA. Since CTR is minimum 0.20, that means an input current of 1.5mA. So a resistance of 10/0.0015 = 6.7K (total). Power dissipation would be about 7W at 220V. That's very high. Even at 100VAC it's still 1.5W total for the two resistors.

So I suggest you find a better optocoupler that responds reasonably quickly (not a darlington type) and has a CTR of 100% or more.

• CTR == Current Transfer Ratio == Ratio of output collector current versus the input diode forward current. <== For the benefit if @BrunoAraujo. May 30, 2016 at 19:32
• @SpehroPefhany How did you get 300uA for output current? It was from 3V / 10k? May 30, 2016 at 21:29
• @BrunoAraujo Yes, that's right. You can increase the 10K a bit to make it a bit more sensitive but it gets slower and slower as you do do. May 30, 2016 at 22:10

Considering your answer, I searched another optocoupler and redid my calculations as below:

I choose the optocoupler VOS627A-3, which according to datasheet has a CTR equal of higher 100%.

Output Current for Zero Crossing detection:

$I = \frac {V}{R} = \frac {3.3~V}{10~k} = 0,33~mA$

Considering CTR = 100%, we need $0,33~mA$ of input current.

For a detection with 10V input, we get:

$R = \frac {V}{I} = \frac {10~V}{0,33~mA} = 30~k\Omega$

Now, to determine the power dissipation of the resistors:

$P = \frac {V^2}{R} = \frac {(240~V)^2}{30~k\Omega} = 1,92~W$

I'm considering 2 resistors of $15~k\Omega$ with 1 W each, disposed like original question.

• @SpehroPefhany Could you please take a look at my calculations? With this I can say "it is OK" or I'm missing something? May 31, 2016 at 13:23
• This shouldn't be posted as an answer if its looking for validation. OP, I suggest you edit your question to include these calculations rather than post this is an answer. May 31, 2016 at 13:42
• Looks okay. I would use 1.5W or 2W resistors for ~1W nominal dissipation. Say line is high by 5-10%. May 31, 2016 at 13:48