# PNP switching circuit - How it will be correct?

Following circuit is used as a switch in one my circuits. The collector is connected to the Vcc and emitter is supplying current. Theoretically it is wrong, right? The supply should be connected in emitter for PNP for driving. Do you think it is OK?

PS: It is worked for driving a seven segment!!??

• Many bipolar transistors will still work, although with poorer performance, if the collector and emitter are swapped. – Dave Tweed Jul 29 '13 at 10:59
• It may 'work' but its hardly efficient. Its not really acting a switch anyway. If it were an NPN it would be an emitter follower. As a switch the emitter of an NPN would be connected to the '0V' rail with the load connected between the +5 V and collector. If it was a PNP switch the emitter would be connected to the +5V rail and the base circuit (2 resistors) pulled to the 0V line to turn it on. – JIm Dearden Jul 29 '13 at 11:50
• I have experienced exactly what Dave Tweed mentioned. As as student one lab exercise was to identify the leads of a bipolar transistor. The only way to tell the collector and emitter apart was to compare the gain. You get higher gain when the real emitter is at a lower voltage than the collector for an NPN or at a higher voltage for a PNP. – Joe Hass Jul 29 '13 at 15:55

BJTs are capable of delivering their maximal current in one particular configuration (Forward-Active). For PNP transistors, this configuration requires: $V_{EB} \approx 0.7V$, and $V_B>V_C$.