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I have an encoder; the circuit is in the image below:

Datasheet

The one that I’m interested in is the PNP one. I want to use a 0-5 V digital voltage, the datasheet says this:

  1. Applied voltage: 30 VDC max.
  2. Source current: 35 mA max
  3. Residual voltage: 0.4 V max. (at source current of 35 mA)

By my calculations using Kirchhoff's law I can't obtain 5 V because of the 35 mA max; the pull-down resistor should be very low, because the emitter resistor is very low. Am I missing something? How can I dimension the resistor correctly?

enter image description here

This is the calculation that I made, what is wrong?

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2 Answers 2

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It says the maximum current is 35 mA and, that would mean the transistor might be on-the-cusp of failing so, choose a more reasonable current like 5 mA and calculate a resistor value that's 1 kΩ. You could probably go as low as 1 mA with a 5 kΩ resistor.

But, there is a minor problem pointed out by @periblepsis; the minimum supply voltage for this device is 12 volts so, in effect, you'll have to use two resistors and form a potential divider like this: -

enter image description here

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    \$\begingroup\$ The transistor is operating as a switch so, it'll push the current defined by the resistor and the power supply. Use a 5 volt supply is my recommendation and it's very easy. OK upper logic level might end up being about 4.8 volts but that's fine for logic. \$\endgroup\$
    – Andy aka
    Commented Apr 5, 2023 at 17:04
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    \$\begingroup\$ @Andyaka The datasheet appears to specify +12 to +24 for the emitter (and 3.3 Ohm) end of the PNP. Elsewhere on the datasheet, for other devices, they do allow +5 V. But not in this case, for the specific E6B2-CWZ5B device. How do you read it? \$\endgroup\$
    – periblepsis
    Commented Apr 5, 2023 at 19:32
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    \$\begingroup\$ @periblepsis yes you are right. Good spot. It needs a potential divider to produce 0/5 volts from a 12 volt supply. \$\endgroup\$
    – Andy aka
    Commented Apr 5, 2023 at 20:04
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    \$\begingroup\$ ok now its clear, so my calc were not wrong, i just had to use a voltage divider to obtain my voltage without burning the transistor \$\endgroup\$
    – Sergio Piccione
    Commented Apr 5, 2023 at 20:14
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    \$\begingroup\$ @SergioPiccione correct and sorry for being late in realizing it couldn't run from 5 volts. \$\endgroup\$
    – Andy aka
    Commented Apr 5, 2023 at 20:35
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The one that I’m interested in is the PNP one. I want to use a 0-5 V digital voltage, the datasheet says this:

Applied voltage: 30 VDC max.
Source current: 35 mA max
Residual voltage: 0.4 V max. (at source current of 35 mA)

I think you may have purchased the wrong encoder (or are contemplating such), if you are stuck with a \$5\:\text{V}\$ source voltage. Here's why:

enter image description here

Note that in the specific case of the only device they make as PNP, that they specify \$12\:\text{V}\to 24\:\text{V}\$, which appears to me to exclude \$5\:\text{V}\$. Strengthened also by the fact that the other three versions they make do support \$5\:\text{V}\$.

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  • \$\begingroup\$ yes i know i buyed the wrong one, too late for that, i was asking if i could fix that, so i have to put a voltage divider, i was thinking to a op amp inverter with negative reaction \$\endgroup\$
    – Sergio Piccione
    Commented Apr 5, 2023 at 20:07
  • \$\begingroup\$ @SergioPiccione Then you will need a 12 V to 24 V supply. No getting around that, without deciding to first ignore the specifications and then perform some testing on the device to see if it can be made to work on 5 V. Do you have access to the appropriate voltage supply? Or only to 5 V? \$\endgroup\$
    – periblepsis
    Commented Apr 5, 2023 at 20:08
  • \$\begingroup\$ nono i have access to the correct voltage supply, just interested in the output \$\endgroup\$
    – Sergio Piccione
    Commented Apr 5, 2023 at 20:09
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    \$\begingroup\$ @SergioPiccione Oh. Then Andy's suggestion about a divider could work. Or a different circuit using another BJT. Personally, I'd use an added BJT circuit. But it can be made to work with dividers. \$\endgroup\$
    – periblepsis
    Commented Apr 5, 2023 at 20:10
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    \$\begingroup\$ @SergioPiccione The resistors mentioned by Andy are fine. You 'could' add a BJT (or two or three, etc.) The extra BJT allows me another degree of flexibility and the possibility to protect your I/O pin from a single part failure (short, for example.) But you've not mentioned anything to suggest a need for that. I think the divider will be fine. Just don't use a screwdriver and accidentally short out the upper resistor in the divider! \$\endgroup\$
    – periblepsis
    Commented Apr 5, 2023 at 21:48

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