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I am making PCB where I need to turn off PNP BJT with my MCU. But my MCU is only 3.3V and it can't reach emitter voltage. So, it is good practice to turn it off by setting output pin to high impedance or should I waste my precious space on PCB for another NPN to drive PNP?

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    \$\begingroup\$ What is the voltage you are switching? If 5V, is the MCU pin 5V tolerant? Which MCU and which pin? \$\endgroup\$ – Justme May 15 at 22:30
  • \$\begingroup\$ Could you provide a part number for the PNP and load current, as this changes dc biases? \$\endgroup\$ – Ernesto May 15 at 23:14
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    \$\begingroup\$ It is BC807 and load current is 1mA maximum. \$\endgroup\$ – andz May 16 at 0:00
  • \$\begingroup\$ Optocoupler used in those cases \$\endgroup\$ – user263983 May 16 at 0:06
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Even if you set the pin to high impedance, voltage on it must not exceed the maximum ratings in the datasheet, which is probably close to 3.3V. There could be an exception if the pin is labeled "5V tolerant".

So if the base of your PNP is above that voltage, you'll need an extra transistor to control it.

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  • \$\begingroup\$ Emitter voltage it is between 8.4 and 6 V. Pins are not 5V tolerant anyway, but I was guessing maybe high impedance will protect the pin (MCU is esp32), don't know why 😃. \$\endgroup\$ – andz May 15 at 22:51
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Generally speaking you should keep output voltage within Vdd or within some limit that is usually less than 6V. Otherwise the protection diode (when present, typically on non-5V tolerant pins) will conduct which will keep the PNP transistor from turning off, and could cause damage.

I suggest using a dual pre-biased transistor to save PCB space, for example one similar to this one or this :

enter image description here

That way only a single component is required rather than several.

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