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Optocoupler schematic

Long time lurker, first time asker. I've been working through a biosensor schematic from a BME class to build an EKG (I'm a physician, aware of risks of these devices etc) (typical voltages being detected in the range of <100mV). After passing through a series of amplifiers and filters, it connects to an Arduino via a optocoupler for electrical safety.

In the original schematic I'm working with, there are 3 generic red LEDs in series with a 220 Ohm resistor with the input side of the optocoupler (CNY17-3) producing about 10mA of bias current through the optocoupler. With this set up, the output waves look crisp with good amplitude. I tried to ditch the LEDs and just up the resistor value to maintain 10mA (in this case, supply voltage of 9V battery w/ resistors with values between 600-1k ohms), but the output waveforms I'm getting have much lower amplitudes compared to with LEDs.

CNY has a CTR of 100% at 10mA, forward voltage of 1.2 at that current as well (max is 60mA).https://docs.broadcom.com/docs/AV02-0772EN datasheet

Any thoughts on if this is a special effect of LEDs (if so, any way to use a different component to replicate this) or whether I'm choosing inappropriate resistor values?

Thanks!

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  • \$\begingroup\$ I would venture to guess that the LEDs are there as a cheap voltage regulator. to supply a more or less fixed drop between the signal and the resistor. In this day and age it is a profoundly cheezy way to implement such a circuit; do you know how old it is? \$\endgroup\$
    – TimWescott
    Apr 2, 2019 at 21:23

5 Answers 5

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The arrangement limits the current through the isolator LED by adding diode drops, thus lowering the voltage that the 220 ohm resistor sees. Without them, you'd likely saturate your phototransistor, or damage the LED. I see no reason why adding a bigger resistor wouldn't do the same thing, though the fixed diode drops to get rid of some voltage probably take out some of the guesswork.

Personally, I tend to use those types of isolators as digital devices. For analog isolation, I'd point you to something more appropriate, like an HCNR200. Yes, you need some support circuitry, but you've already got a full biopotential amp on the left side, so you probably have the resources you'll need in place.

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I think Scott Seidman is close, but not quite on.

Vf for your opto is about 1.5 volts. if the input is at a nominal zero volts, with no external LEDs the opto current will be about 30 mA ((8 - 1.5 - 3x2) / 0.2), and I suspect that your source driver is unable to handle that gracefully.

With the LEDs in place, nominal current is about 2.5 mA, and that sounds like a much more reasonable output level for whatever is driving your input.

Try monitoring your input voltage swings for the two cases.

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I think Scott Seidman and WhatRoughBeast both are suggesting the approach that you tried(replacing leds with larger resistor should work but you are saying it does not work.

So one thing to consider is: If you removed LEDs and increased resistor value then circuit is active with minimal signal on input side. Whereas with 3 red LEDs you need that minimum 3*2V=6V voltage difference for circuit to pass any signal.

So this could be to avoid noise at lower levels and to setup minimum input voltage level that can pass the signal.

I suggest you test with DC levels to see why your signal gets attenuated between two approaches if forward current is exactly 10mA in both cases.

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I forgot to update here, but I was able to contact the creator of the original schematic and ask him, his excellent answer is quoted below

"The LEDs have a fixed voltage drop across them (about 2V each). This means the LED current is determined by I_LED = (Vout2 + 9 - 6)/R, where Vout2 is the output of op amp #2 (the gain + LPF stage). What we care about is the CHANGE in I_LED, which is given by delta_I_LED = (delta_Vout2) / R. Smaller R produces a larger delta_I_LED, which is important for giving a big signal. However, you cannot arbitrarily choose R, since its value determine the baseline LED current (about 14 mA). If you do NOT include the three external LEDs, then you will need a larger R to produce 14 mA of baseline current. A larger R results in a smaller change in current, and therefore a reduction in the size of the recorded ECG signal.

If you don't like the three blinking LEDs, replacing them with a single 6.2V zener should give roughly the same performance. You may need to slightly tweak the resistor values (e.g. use a single 200 ohm instead of a 220 ohm resistor)."

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This circuit is not fully defined. The following points need further consideration:

  • You must describe the driving signal and source impedance. Sin it says LPF out I take it to be an analog signal.You're going to have 4 diode drops (roughly 3x2.2V for LEDS + 1.5V for CNY 17 diode) of dead band in your input range. Your input DC level must overcome this level
  • Your circuit (using CNY 17) works well for isolating digital signals but not so for analog signals. Output voltage Aurduino A0 sees will be proportional to the optically coupled base signal of the CNY 17 transistor and hence to the current transfer coefficient of the CNY 17. However this not a precisely defined parameter and will change with environmental & device conditions besides having nonlinear characteristics. You can observe this from CNY17 datasheet. It may work to some extend but at any precision level needed here. HCNR200 from BROADCOM, as suggested above, is a far more suitable choice but since this is a precision circuit I would suggest using ACPL-C87A-000E also by BROADCOM.
  • If you have an LPF filter to drive the input that may effect the filter characteristics. You must consider using a buffer amp.
  • I suggest you use a different circuit with ACPL-C87A-000E. Since this is an isolated precision voltage sensor, you will also have to insert a buffer amp after the LPF. If your input is predominantly a current source then you may simply insert a fairly high value (>100 kOhm < 1MOhm) high stability resistor to stay clear of overloading the LPF. This will not have any effect on ACPL-C87A-000E since it has 1 GOhm nominal input impedance. On the sensor output side you may also need some voltage gain adjustment by using a single rail to rail precision OPAMP in non inverting mode. For more on ACPL-C87A-000E device please refer to: https://www.broadcom.com/products/optocouplers/industrial-plastic/isolation-amplifiers-modulators/isolation-amplifiers/acpl-c87a
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