Timeline for Resistor for open-collector transistor
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2023 at 20:35 | comment | added | Andy aka | @SergioPiccione correct and sorry for being late in realizing it couldn't run from 5 volts. | |
| Apr 5, 2023 at 20:14 | comment | added | Sergio Piccione | 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 | |
| Apr 5, 2023 at 20:12 | history | edited | Andy aka | CC BY-SA 4.0 |
added 298 characters in body
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| Apr 5, 2023 at 20:10 | vote | accept | Sergio Piccione | ||
| Apr 5, 2023 at 20:04 | comment | added | Andy aka | @periblepsis yes you are right. Good spot. It needs a potential divider to produce 0/5 volts from a 12 volt supply. | |
| Apr 5, 2023 at 19:32 | comment | added | periblepsis | @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? | |
| Apr 5, 2023 at 17:04 | comment | added | Andy aka | 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. | |
| Apr 5, 2023 at 17:00 | comment | added | Sergio Piccione | nono i was saying i ve updated my question sorry, im not sure why my calc are wrong, but when i force the current as you said also your solution make sense, | |
| Apr 5, 2023 at 16:49 | comment | added | Andy aka | @SergioPiccione how do you want me to update my answer? | |
| Apr 5, 2023 at 16:36 | comment | added | Sergio Piccione | update the answer, thank you btw, it perfectly make sense but my math was off | |
| Apr 5, 2023 at 16:15 | history | answered | Andy aka | CC BY-SA 4.0 |