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I have got an optocoupler to switch an output between 0V and 3.3V. That output is also inverted with a NOT gate, but for simplicity, we could focus on just Out_Enable1 signal. As far as I understand, I must read a logic 1 (3.3V aprox) when I set GND in Signal1 input. But I don't get that. I am reading more than 2 volts between pins 4 and 3 in U5 (always), and in R107 I cannot measure a logic 1 or something near 3.3V when I connect Signal1 to the GND for 28VDC power line. I don't know why the optocoupler is not switching its output, or, at least, that's the problem I see.

enter image description here

EDIT: I have tried changing R106 from 5.6kohms to 2.2kohm (more than 10mA for If) but nothing. With a 2200 ohms value, measured voltages are: for OFF status Voltage across pins 4 and 3 = 3V and for ON status (Signal1 = GND) voltage turns into 2.7V.

EDIT2: I have the exact same circuit except for the NAND gate below that one and it is switching properly. The NAND's footprint is right and the pinout and electrical values too. But it is impacting the behaviour of the swithing.

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  • \$\begingroup\$ What voltage do you get across R106 when signal1 is GND? What votlage do you get across pins 1 and 2 when signal1 is GND? And what do you get when signal1 is floating? That will show if the diode in U5 is broken. \$\endgroup\$
    – Puffafish
    Commented Sep 4 at 6:46
  • \$\begingroup\$ For question 1: 26.6 volts. For question 2: 1,14V. For question3: I am connecting 28VDC and GND to a power supply. If signal1 is floating, 0.2V are measured. \$\endgroup\$ Commented Sep 4 at 7:54
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    \$\begingroup\$ The problem seams not to be in the schematic, but in the implementation. Can you provide a board layout or a picture of the assembly? \$\endgroup\$
    – Jens
    Commented Sep 4 at 13:53

3 Answers 3

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From the VOM613A datasheet, page 3, I read that the current transfer ratio of this device is at worst 100%. You should not expect obtain more current through its transistor, than you push through its LED.

LED current \$I_F\$:

$$ I_F = \frac{28V - V_F}{R_{106}} = \frac{28V - 1.6V}{5.6k\Omega} = 4.7mA $$

The expected saturation current \$I_C\$ for the transistor is:

$$ I_C = \frac{3.3V}{R_{107}} = \frac{3.3V}{1.87k\Omega} = 1.7mA $$

You are well within the operating limits of that opto-coupler, and I would expect it to work. Therefore my initial thoughts are:

  1. OUT_ENABLE1 is connected to something else holding it low. Whatever it is must not sink more than \$I_{C(MAX)} - I_{R107} \approx 4.7mA - 1.7mA \approx 3mA\$. To be safe, don't draw more than 1mA from OUT_ENABLE1.

  2. R107 isn't 1.87kΩ, R106 isn't 5.6kΩ or input voltage isn't +28V.

  3. U3 has no power, and its input protection diodes are clamping OUT_ENABLE1 near to ground.

  4. The opto-coupler is damaged.

  5. U3 (inverter) is damaged.

  6. Inadequate supply decoupling causing oscillations somewhere.

Of course there are other obvious pit-falls, like a wiring error, or +3.3V not being +3.3V etc, but I assume (naively) that you already checked those.

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  • \$\begingroup\$ Hi, Simon. Thanks. 1. It is connected to a NAND Gate (HiZ input). 2. All resistors are checked, power supply too. 3. It is check. 4. Maybe, but diode voltage is 1,14 as in datasheet. 5. Maybe. 6. Maybe. \$\endgroup\$ Commented Sep 4 at 8:00
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You are getting some effect from the optocoupler, but not as much as expected. You say you've checked the resistors- I'll assume that means you've actually checked the resistor values not just the markings (if that's not true, then check them out of circuit).

It's possible that the gate is damaged (protection diode shorted) which would yield the symptoms described. It's also possible there's a hair short on the PCB. Or that the opto is not working properly.

I suggest removing the gate and measuring the change at node Out Enable1 with no gate. If that works, then replace the gate with a fresh one from a reliable source (following ESD precautions). If it does not work, then inspect the PCB under a microscope or magnifier, preferably with a strong light behind it.

If nothing can be found, bridge the opto pins 3 and 4 with a resistor of 200Ω or so and see if that causes Out Enable1 to change to about 3V. If it does not, keep looking. If it does, then replace the opto with a fresh unit from a reliable source.

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The optocoupler might not be saturating because it does not have enough "gain", called the Current-Transfer-Ratio (CTR). Your part is rated for a CTR of 100-200%. That is, for every mA of current through the input diode, expect 1 to 2 mA of current through the output transistor.

Let's start with the initial condition of 5 mA into the LED. If the output current limiting resistor limits the maximum output circuit current to less than 5 mA, the output transistor will saturate. If the maximum output circuit current can be greater than 5 mA, the output transistor will not saturate. To determine the peak output circuit current, replace the output transistor with a dead short across the collector and emitter terminals. This simulates a perfect output transistor with a Vcesat of 0.0 V.

When using an optocoupler in a logic circuit, it never is a good thing to have it run right at the edge of its specifications. For example, if the minimum CTR is 100%, design the circuit as if it were only 50%. This will assure firm saturation.

As a quick test to see if this is the issue, increase R107 to 10 K.

Note: If the only reason for the logic gate is to invert the logic polarity of the signal from the opto, you can change the connections to opto pins 3 and 4 and eliminate the gate.

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