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Why am I getting 5 volt on my transistor output? Why is the output not 12 volt?

How can I make this work? Is it possible to make it work? I mean for the output to be 12 volt while the base is using a lower voltage.

Here's the illustration:

enter image description here

Please explain the cause.

JRE
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Thea Thehe
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    Put the load on the collector, not the emitter. You can't use an NPN transistor for high-side switching without a more complicated driver circuit. – Hearth Jun 15 '22 at 05:13
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    Does this answer your question? [Emitter follower](https://electronics.stackexchange.com/questions/74563/emitter-follower) – ocrdu Jun 15 '22 at 07:36
  • You have built an emitter follower, which won’t allow you up output any higher voltage than your base voltage. Look for an PNP or P-MOSFET solution if you need a high side switch. – winny Jun 15 '22 at 09:47

2 Answers2

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Your circuit is known as an emitter follower.

In an emitter follower, the output voltage is always about 0.7 volts lower than the base voltage.

To make an NPN transistor conduct current from collector to emitter, you must have a voltage difference of about 0.7 volts between the base and the emitter. That's the forward voltage of the base to emitter junction.

You have 5V on the base, so the emitter can never be more than about 4.3V.

You could connect the base to 12V, which would give you about 11.3V on the emitter. Since your 5V is probably from the output of a microcontroller or something, you probably can't easily put 12V on the base.

The usual way to use an NPN transistor is to put it in the low side. That is, between the load and ground.

schematic

simulate this circuit – Schematic created using CircuitLab

When there's 5V on the "5V Digital input," the transistor will conduct and you load connected to "Output" will receive power. This is known as a low side switch.

Note that there's a resistor between the input and the base of the transistor. You need that resistor to protect the transistor and the microcontroller. The base of a transistor acts like a diode. When it conducts it "turns on" and acts like a short circuit. Without a resistor to limit the current, so much current could flow through the base that is will be destroyed. It could also damage the circuit driving the base.

JRE
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Why am I getting 5 volt on my transistor output?

To start to activate an NPN transistor, the voltage at the base needs to be a little more positive than the voltage at the emitter. This is due to the PN junction between base and emitter.

So, if you have 5 volts on the base, then you cannot have much more than about 4.5 volts at the emitter or the PN junction just won't be conducting enough.

How can I make this work? Is it possible to make it work? I mean for the output to be 12 volt while the base is using a lower voltage.

No, this circuit will never work as you want so, no need to keep flogging the dead horse.

Andy aka
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