Timeline for In a circuit being switched by an NPN transistor, do power supply and input need the same ground?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2012 at 5:42 | comment | added | stevenvh | @Mason - I guess we have a slight misunderstanding here. Omitting R2 means no connection between base and ground, not replace it by a wire. Connecting the base to ground won't work, because the base voltage will be zero, and you need 0.7 V. I'm sure it will work with the 68k, but also without. Just don't connect base to ground. | |
| Aug 20, 2012 at 1:05 | comment | added | Mason | For some reason, when I just connected a lead from the base to ground, it didn't work, but when I added the 68kΩ resistor in its place, it worked perfectly. | |
| Aug 20, 2012 at 0:47 | comment | added | Mason | In this case, then, I can just connect the base directly to ground? | |
| Aug 19, 2012 at 11:44 | history | edited | stevenvh | CC BY-SA 3.0 |
added 980 characters in body
|
| Aug 19, 2012 at 11:34 | comment | added | stevenvh | @m.Alin - the base voltage will be 0.7 V maximum. In the 0.5 V case the base won't draw any current, but if your 10:1 divider would set that to 0.9 V the B-E junction limits that to 0.7 V. So R2 doesn't set the voltage, and will only draw a limited current, because much larger than R1, so it doesn't really serve a function. Which is why I said you'll rarely need it. | |
| Aug 19, 2012 at 11:22 | comment | added | m.Alin | 'R1 and R2 form a resistor divider, and if we choose R1 = R2 then the 1 V input would become 0.5 V base voltage' But you usually choose R2 >> R1 (x10 times), so the input would still be near 1 V.. | |
| Aug 19, 2012 at 6:58 | history | answered | stevenvh | CC BY-SA 3.0 |