1
\$\begingroup\$

I was looking at an old schematic and I came across this strange circuit for low voltage detection section of a larger circuit (not shown). The operation of circuit is straightforward, Q1 is normally conducting and the voltage divider R2/R3 set a threshold voltage, above which, Q2 conducts. When Q2 is normally conducting, the Q3 is off due to the voltage divider R1/R5. Upon depletion of power source,R2/R3 can no longer provide the base voltage for Q2 and it turns off. The Q3 turns on and discharges C2 to provide a pulse. Does anyone know why Q1 is used in this configuration without the emitter connected to anywhere?

enter image description here

\$\endgroup\$
  • \$\begingroup\$ It might have been used to save having to use an extra part type. When getting PCBs assembled, each part you use adds overhead, so if you can use a kind of transistor you're already using a bunch of, instead of a kind of diode you're not, that saves money. \$\endgroup\$ – immibis Feb 21 '18 at 22:26
-1
\$\begingroup\$

Yes it is just a diode.

There are so many clever but inefficient design choices here, I wouldn't spend much time on it.

The person obviously doesn't know how to make a power on reset even with a miswired pot and 3 transistors, one of which is experimental.

Pulse period is about 1x~2x RC or 32 ms

\$\endgroup\$
  • \$\begingroup\$ The pot was my mistake, it's corrected now. \$\endgroup\$ – Baphomet Feb 21 '18 at 21:42
  • \$\begingroup\$ Diode are only a penny in volume , there many ways to create a POR. This is very obsolete. \$\endgroup\$ – Sunnyskyguy EE75 Feb 22 '18 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.