NPN and PNP transistors and load

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SUMMARY

The discussion centers on the behavior of NPN and PNP transistors in circuit configurations, specifically regarding load placement and current flow. It is established that while circuit simulators may show similar current flow regardless of load placement, real-world applications reveal discrepancies, particularly with PNP transistors. A significant factor is the base voltage requirement, which must be approximately 0.7 volts more positive than the emitter. Additionally, the necessity of a flyback diode across the relay coil is emphasized to prevent transistor damage when deactivating the coil.

PREREQUISITES
  • Understanding of NPN and PNP transistor operation
  • Knowledge of circuit simulation tools
  • Familiarity with relay circuits and inductive loads
  • Basic electronics principles, including voltage and current relationships
NEXT STEPS
  • Research the role of base-emitter voltage in transistor operation
  • Learn about the function and importance of flyback diodes in relay circuits
  • Explore the differences between circuit simulation and real-world circuit behavior
  • Investigate the characteristics of inductive proximity sensors and their integration with transistors
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Electronics enthusiasts, circuit designers, and engineers working with transistor circuits, particularly those involving relays and inductive loads.

Trespaser5
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Hello,
I've been playing with transistors on a circuit simulator and found that it doesn't matter on an NPN transistor whether I put the load on the collector or emitter side, the load still has the same current flowing through it. The same is true for the PNP transistor, however when I physically wired a pnp inductive sensor up to a relay with the load on the collector side there was not enough current flowing through the relay to activate it, can anyone explain why ?
I have wired up PNP and NPN components in certain ways for years but don't really know why.
Kind regards
Kyle
 
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NPN (and PNP) transistors are not symmetric in their layout. While it might be possible to operate them in the wrong direction in some cases, don't expect the same performance all the time.
 
If you have a large resistor in the emitter circuit then there will be a large voltage across it.
So, you need a larger voltage than this on the base. This is because the base voltage has to be about 0.7 volts more positive than the emitter.

An emitter follower has a voltage gain of less than one so it needs a very large input voltage compared to a common emitter circuit.
 
The circuit simulator may be giving you an unrealistic zero source-impedance power source, which can mask some circuit errors. You really need to provide more detail of the circuit - and in particular show how you're driving the base.
 
I've attached some examples of the circuits on the circuit simulator, when I wired the circuit in reality I used a inductive proximity so the base was obviously driven in a different way.
Thanks for your answers
Kyle
 

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  • example pnp npn.jpg
    example pnp npn.jpg
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Those circuits don't really convey anything - they're just using the transistor as a diode, and the only difference between the configurations is the order of the battery and the load - which is irrelevant. For this to make any sense, you'll have to provide details of the base driving.
 
Hi Sylvia,
That is my question really, I have set up these simple on the circuit simulator and observed that arrangement of the load and DC voltage source doesn't matter to the current in the circuit. However when I tried to replicate the PNP circuit shown using a bench power supply unit (the DC voltage source), and inductive proximity sensor(the PNP transistor) and a relay (the load), it would not work. Are you saying that if I just used a PNP transistor and wired them the same as the circuit simulator diagram I have attached they would behave in the same way ?
 
I would expect the relay coil to be energised in both cases, as soon as the power is applied.

Note, BTW, that you need a diode across the relay coil in real circuits, or you'll quite likely destroy the transistor when you try to turn the coil current off.
 

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