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I have a circuit where a 3.3 V DATA signal is optically isolated using the VO0600T IC. The circuit is as follows: VO0600T circuit

In the schematic, VEE = 5 V (pin 8). Note that there are 2 separate grounds on either side of the optical isolator IC. The datasheet recommends a 0.1 uF capacitor between pins 5 and 8 - it is not indicated in the above circuit diagram, but it is present on the actual circuit that I am testing.

From the VO0600T's datasheet, it is indicated that pin 7 can be used for strobing. In my application the output state of the isolator circuit is constantly monitored, therefore pin 7 is connected to VEE so that the output of the isolator IC is always active. The truth table on page 2 of the VO0600T's datasheet confirms that if pin 7 is HIGH, the output (on pin 6) will be the inverse of the DATA signal. Thus, when DATA is LOW, DATA_OUT should be HIGH, and vice versa.

When I measure the output at pin 6 with a multimeter (when DATA is LOW), it reads a fluctuating value between 0.01 V and 0.04 V. When DATA is HIGH, the multimeter reads 0.00 V (which would be expected due to the inverting output of the IC). Thus, my issue is that DATA_OUT does not go HIGH when DATA is LOW.

I have had brief access to an oscilloscope when this was tested. The following image illustrates what is happening with DATA_OUT when DATA is LOW:

DATA_OUT signal measured

It is clear from the above image that DATA_OUT is oscillating, which explains the fluctuating reading on the multimeter. Unfortunately I did not take a measurement with the oscilloscope when DATA is HIGH (I also do not have direct access to an oscilloscope to make that measurement now).

Can anybody explain why I am experiencing this issue? Am I missing something obvious in my circuit layout? Any assistance will be greatly appreciated.

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1 Answer 1

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Not the cause of your problem, but something that may confuse some users: Vee generally implies a negative supply, but I assume you have a positive supply there - we'd normally call that Vcc.

The output is "open collector" (really, open drain) - the drain of the output FET is only connected to the output pin, not to any internal circuitry, so you need an external pull-up resistor to Vcc to pull the output High. The datasheet recommends a 330 Ohm to 4K Ohm pull-up.

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  • \$\begingroup\$ Thank you for your answer, Peter. Indeed, VEE in the circuit is a positive 5 VDC supply. I have totally missed the fact that the circuit requires a pull-up resistor. I have been misled by the functional diagram in the sense that it looks like there is internal circuitry, making up a logic element of which the output could directly be used without any pull-up resistor. However, the "open drain" specification in the datasheet should have been my clue from the start :) \$\endgroup\$
    – wave.jaco
    Aug 4, 2017 at 4:40

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