# Electronic – Basic confusion about how transistors work

transistors

I am trying to understand transistors but it doesn't matter what source I find, there is always something that I cant fully grasp.

In Make: Electronics second edition, page 88, it says that
"[a transistor] won't respond unless the voltage at its base is higher than the voltage at its emitter (usually by around 0.7V)"

I have trouble understanding this sentence. Voltage is measured across two points, right? So what is the voltage 'at the base' and 'at the emitter'?
Is it the voltage remaining after all voltage drops?

But them how to change the voltage 'at the emitter'?

I know it's a basic question.

Voltage is the electric potential difference between two points. A single point doesn't have an absolute voltage, only relative voltages compared to other points in the circuit. Of course, often there is an implicit reference point, the "ground" (either something actually grounded, or just the negative or middle point of the power supply marked as "the ground").

So, the voltage "at the base" is either meaningless, or implicitly means the voltage between the base and the ground, and similarly for the voltage at the emitter.

But note that in the quote there is a comparison between two points: the voltage between the points "at the base" and "at the emitter" can be measured directly, without needing a third reference point. This is the voltage that matters, regardless of what the voltages at the base and emitter are compared to other points in the circuit. E.g. in an emitter follower, neither might be grounded, so the voltages against the ground are not relevant.

What the voltage between the base and the emitter actually is, depends totally on the surrounding circuit. If both are tied to ground (or whichever one point), the voltage between them is zero, and the transistor is "off". If the base is driven at least 0.7 V (or so) higher than the emitter, the transistor is "on", and the resulting current will be such that the base-emitter voltage stays around 0.7 V (but really it depends on the current). Unless you connect a voltage source directly across the transistor, of course, in which case the voltage and current would be higher, for a short moment.

It may help to play with it. See the linked circuit in the Falstad simulator. You can vary the voltage that goes to the base (through the resistor) with the slider. (The arrows are current meters, not diodes.)

As for the voltage "remaining after all voltage drops", well, in a way. In the above circuit, the voltage between the transistor base and emitter is whatever the voltage source to the base gives, minus what's lost on the base resistor. That's probably not very helpful in practice, though.