Feb 24, 2011 | By:

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drfae_iconDear Dr. FAE. You’re always going on and on (and on) about how smart you are…perhaps you’d like a challenge where I give you the answer-and you have to figure out the question. Here’s the answer: a 1150 ohm resistor. You don’t feel so smart now, do you, Dr. Smartypants?

  • – Throwing Down in Tampa

Through the application of only a small portion of my intellect, I can tell your question relates to selecting the right pull-up resistor to use for a cost-effective, low-performance optocoupler circuit…to be specific, the half-pitch, mini-flat packaged HMHA281 device. This is a simple circuit, but there are still a few things to consider when designing a reliable and robust detection circuit.

dr-fae_opto_2_11

 

First of all, we need to select a current-limiting resistor for the emitter circuit. With a 12Vin, a forward voltage drop of 1.3V across the LED and a reasonable current of 8mA, let’s figure out what value of resistor should be used…

dr-fae_opto2_2_111

The closest standard value is 1.33K, so we’ll use that.

We’re using a standard CMOS logic gate (with hysteresis) as a detector on the secondary side. The question is, with the optocoupler turned on, what voltage do we need to provide to make sure the gate input detects a logic zero? So, we look at the NC7SZ14 datasheet. In order to be recognized as a logic zero over a wide operating temperature range, the input voltage must be less than 1.30V.

How do we guarantee a voltage no greater than 1.30V when the detector transistor is on? We’re running this device in a switch mode, so we need to know the minimum Current Transfer Ratio (CTR) with a saturated detector transistor. This is not something that can be read directly off the datasheet but we can derive it from the VCE(SAT) transfer characteristic in the HMHA281 datasheet. With an IF of 8mA, we get an IC of 2.4mA; that gives us a CTR of 2.4/8 or 30%. This is a typical minimum at room temperature, so we need to derate it for temperature. By studying Figure 4 of the HMHA281 datasheet, we see that we need about an 80% derating at 55°C, so that gives us a minimum CTR of about 24%.

Being clever and highly experienced, we know that a gently-treated optocoupler’s CTR will degrade by about 1% per year. If we expect this design to work for ten years, then we should derate the CTR an additional 10%…to 21.6%.

By gently-treated, we mean the optocoupler is not exposed to high ambient temperatures, large emitter/detector currents or extreme thermal cycling. With more aggressive environments and usage, we can expect degradation closer to 2% per year.

With 8mA emitted, then the minimum current in the output phototransistor is:

dr-fae_opto3_2_11

So, what pull-up resistor value do we need to make sure the voltage at the inverter input is less than 1.30V when 1.73mA flows through it?

dr-fae_opto4_2_11

We’ll choose 1150 ohms as the closest standard value. Pardon me while I do a little victory waltz. 1150 ohms. Boom, I have dispensed with your challenge.

I don’t anticipate any problems, but we should make sure the ‘off’ current will be detected as a logic ‘one’. According to Figure 8 of the HMHA281 datasheet, we can expect 10uA of dark current at 55°C. This will drop 11.6mV across the pull-up resistor. The resulting voltage at the gate input is much more than 2.2V required to assure a logic ‘one’ is detected. So, all is well.

This was a lot of work to create a low-speed logic interface. If you want us to the hard work, you can use a device with the logic detection built in, like the FODM8061.

There – that is the question to your answer – and yes I do feel smart!

Let me throw a challenge back at you, TDiT…and to my readers around the world. Here’s the answer: Valvular Conduit. What is the question?

The first right answer in the comment section below will win a trendy and fashionable prize package, including my attractive and collectable signed photograph and a 2N7002 transistor.

Sincerely, Dr. FAE aka “Smartypants”

This post is also available in: Chinese (Simplified)