0
\$\begingroup\$

I'm building a dimmer, and got some questions about the resistor in the zero cross circuit.

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?

\$\endgroup\$
  • \$\begingroup\$ 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. \$\endgroup\$ – Marcus Müller May 30 '16 at 19:08
  • \$\begingroup\$ Equations reordered for readability and MathJAXed. \$\endgroup\$ – Transistor May 30 '16 at 19:26
  • \$\begingroup\$ @BrunoAraujo don't worry; it was really just a question :) \$\endgroup\$ – Marcus Müller May 30 '16 at 19:31
  • \$\begingroup\$ @MarcusMüller sorry about that.. \$\endgroup\$ – BrunoAraujo May 30 '16 at 19:31
  • \$\begingroup\$ 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. \$\endgroup\$ – Michael Karas May 30 '16 at 20:02
0
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\begingroup\$ @SpehroPefhany Could you please take a look at my calculations? With this I can say "it is OK" or I'm missing something? \$\endgroup\$ – BrunoAraujo May 31 '16 at 13:23
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – efox29 May 31 '16 at 13:42
  • \$\begingroup\$ Looks okay. I would use 1.5W or 2W resistors for ~1W nominal dissipation. Say line is high by 5-10%. \$\endgroup\$ – Spehro Pefhany May 31 '16 at 13:48
2
\$\begingroup\$

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.

\$\endgroup\$
  • 3
    \$\begingroup\$ CTR == Current Transfer Ratio == Ratio of output collector current versus the input diode forward current. <== For the benefit if @BrunoAraujo. \$\endgroup\$ – Michael Karas May 30 '16 at 19:32
  • \$\begingroup\$ @SpehroPefhany How did you get 300uA for output current? It was from 3V / 10k? \$\endgroup\$ – BrunoAraujo May 30 '16 at 21:29
  • \$\begingroup\$ @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. \$\endgroup\$ – Spehro Pefhany May 30 '16 at 22:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.