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I am trying to use transistor as switch and getting confused with the workings.

For a NPN transistor, Collector needs to be connected to +ve and emmiter to -ve.

If we look at the schematics below, in both the diargrams, I am connecting the collector to + and emitter to the - .

schematic

simulate this circuit – Schematic created using CircuitLab

schematic

simulate this circuit

Now, to provide a signal to the base,when I connect base to +ve terminal of bread board, it works for scenario , but not for scenario 2 (diagram on right).

Simsons
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    you've not considered the base. whether or not the transistor turns on depends on the current that is flowing into the Base. With the emitter sitting at 0V, it's pretty simple to determine what current might be flowing into the base. In your second example, the emitter is sitting up in the air. The voltage on the emitter must be lower than the voltage on the base for current to flow and turn the transistor on. – Kartman Jan 17 '22 at 10:21
  • Tip: the base connection is at the 9 o'clock position on the transistor symbol, not an extension of the junction as you have drawn it. There is a CircuitLab button on the editor toolbar that makes it very easy to create legible and editable schematics. Hit **Save and Insert** to add it into your post. No CircuitLab account is needed. – Transistor Jan 17 '22 at 10:24
  • @Transistor, Thanks, updating with circuit lab. – Simsons Jan 17 '22 at 10:33
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    Please note that both of your drawings, i.e. the left and right ways of using the transistor, both are valid and work as expected when the base is properly driven. How you use the circuits and how you expect them to work may not be correct. Commonly you use the left one, which is called common emitter circuit where the transistor works as a switch, and commonly you don't use the right one unless you need an emitter follower circuit. – Justme Jan 17 '22 at 10:45
  • @Kartman, updated the question with bit more detaill. In the second diagram, the emitter is connected to -ve terminal of battery and I am connecting the base to +ve terminal to get voltage and current flowing. Am I missing anything? – Simsons Jan 17 '22 at 10:51
  • The first circuit can never work because the base-emitter region (being a diode) will short out the 9 volt supply and either cause the LED never to illuminate or burn the transistor. Clarity is definitely needed. You need a base resistor urgently in circuit 1. – Andy aka Jan 17 '22 at 11:13
  • @Andyaka, added the resistors now. Was doing them on bread board but forgot to add while drawing. – Simsons Jan 17 '22 at 11:31
  • What specific LED are you using - part number please. – Andy aka Jan 17 '22 at 11:40
  • In the second diagram, what's the emitter voltage? – Momo Jan 17 '22 at 11:44
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    @Simsons I tried to answer a very similar question [here](https://electronics.stackexchange.com/a/602308/38098). – jonk Jan 17 '22 at 18:13

1 Answers1

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) The NPN circuit redrawn. (b) The PNP circuit redrawn.

  • In (a) the low-side NPN is driven hard and is fully saturated. As a result there is a very low voltage drop across it leaving most of the voltage available to the load.
  • In (b) the high-side NPN is working in voltage follower mode which means the emitter tracks the base voltage with about 0.7 V drop due to the base-emitter junction. There is only 8.23 V available to the load.

For a 9 V, low-current application it works OK. If you were running lower voltage then the loss in available voltage may become a problem. If you're running a high current load then you will find it difficult to drive the PNP base hard enough without additional loss in base voltage. In addition, don't forget that you'll generally be driving the base from another transistor so there will be further voltage loss in that.

Note also that the power dissipated in the transistors will differ in proportion to the voltage across them. The low-side NPN is saturated so will dissipate little power. The high-side is not and will run hotter.

Low-side NPN switching is reliable, efficient, solves level shifting from the driving voltage to the driven voltage and is very flexible.

Transistor
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