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I have this optocoupler, and I am trying to figure out how it works.

What is the VCC pin supposed to do? How does it supply current to the gate pin of the transistor?

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https://www.digikey.com/en/products/detail/broadcom-limited/HCPL-0531-000E/696030?utm_adgroup=xGeneral&utm_source=google&utm_medium=cpc&utm_campaign=Dynamic%20Search_EN_Product&utm_term=&utm_content=xGeneral&gclid=CjwKCAjw0a-SBhBkEiwApljU0pgvdbqmI2GcPn1GwVyXmIyyVpMqYmgw7JCr73JvQpgOot71IL_LtBoCvF0QAvD_BwE

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3 Answers 3

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The diodes that are attached to the bases of the transistors are photodiodes and are designed to conduct in "reverse" when illuminated. The light from the LEDs on the input side of the optocoupler dislodges/promotes charge carriers in the depletion zone of the photodiode which then move under the influence of the applied voltage (from VCC) to create a current which is amplified by the NPN transistors.

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  • \$\begingroup\$ Why is there no current-limiting resistor between VCC and the base pin? won't that cause a short circuit to GND when the photodiode conducts? \$\endgroup\$ Commented Apr 5, 2022 at 22:33
  • \$\begingroup\$ Because the current through the photodiodes at maximum illumination is still very small (microamps). \$\endgroup\$
    – vir
    Commented Apr 5, 2022 at 22:35
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The main purpose of the separate VCC pin is to apply a reverse bias voltage to the photodiode (this is called photoconductive mode). This decreases the junction capacitance and thus the response time. Optocouplers with a plain phototransistor manage about 10 kbps for digital signals; this optocoupler can do about 1 Mbps.

The current generated by a photodiode is very low; the amplification of the transistor is needed to get a usable signal out of it. As shown in the datasheet, ICCL is at most 400 µA, so it is not necessary to limit the base current with a resistor.

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  • \$\begingroup\$ Would it still work without the VCC connected (and just the pull-up on the VO1 pin)? \$\endgroup\$ Commented Apr 6, 2022 at 12:07
  • \$\begingroup\$ The photocurrent must be able to flow, so VCC must be connected somewhere. And the electrical characteristics are guaranteed (tested) only with a proper bias voltage. If you did not want a separate VCC, you could just as well use a cheaper phototransistor. \$\endgroup\$
    – CL.
    Commented Apr 6, 2022 at 12:19
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The VCC pin needs to be provided with a few volts relative to the GND pin. This may require a floating power supply.

When the LED is activated it will cause light to hit the photo-diode that will then pass current into the base of the transistor. That current is provided by the VCC pin.

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  • \$\begingroup\$ Why is there no current-limiting resistor between VCC and the base pin? Won't that cause a short circuit to GND when the photodiode conducts? \$\endgroup\$ Commented Apr 5, 2022 at 22:38
  • \$\begingroup\$ As air commented - the worst case photocurrent is presumably low enough that it is not an issue. \$\endgroup\$ Commented Apr 5, 2022 at 22:42

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