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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.

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  • \$\begingroup\$ well, it's a potential difference of at least 0.7 V. When you measure the voltage between these points, it needs to be > 0.7 V. Also, that means if you measure the Base voltage relative to any other point, the voltage needs to be 0.7V higher than that between the emitter and that same point. I mean, it's voltage: a potential thing. \$\endgroup\$ Jan 26, 2021 at 20:13
  • \$\begingroup\$ put simpler/backwards: whatever voltage you put on the base, you get the same-0.7v at the emitter. Conditions like current and temp can shift this slightly, but the emitter will tend to "follow" the voltage of the base, no matter what the voltage of the collector (within limits of course). Often the emitter is connected to ground, so we use resistors to keep the base voltage at 0.7v at a usable current. \$\endgroup\$
    – dandavis
    Jan 26, 2021 at 20:17
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    \$\begingroup\$ A textbook with poorly-worded explanations? Whoda' thunk!!! . Really they should have said: "The input of the NPN transistor is basically a diode, where the diode's "forward voltage" always appears between its base and emitter. The transistor won't respond unless its base-emitter voltage-drop is above a certain threshold (roughly 0.5V to 0.7V for silicon diodes.)" In other words, in order to turn on the transistor, we must turn on its internal base-emitter diode. (Also see my extremely simplified physics explanation: amasci.com/amateur/transis.html ) [Edited by a moderator.] \$\endgroup\$
    – wbeaty
    Jan 26, 2021 at 23:02
  • \$\begingroup\$ what does this mean? ... Is it the voltage remaining after all voltage drops? ... it may be the reason for your misunderstanding \$\endgroup\$
    – jsotola
    Jan 27, 2021 at 6:46
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    \$\begingroup\$ @wbeaty, it might be worth pointing out that the semi-fixed forward voltage doesn't just appear between the transistor terminals out of nowhere, but there needs to be an external source providing the voltage to begin with (even if the transistor does try to keep it roughly at a particular value). Which is what you mean when you say the B-E diode needs to be turned on, but since it's a question on basics, perhaps better be clear about it. \$\endgroup\$
    – ilkkachu
    Jan 27, 2021 at 15:09

4 Answers 4

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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.)

enter image description here

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.

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    \$\begingroup\$ this software is amazing, I have used it to clarify so many confusion I had. Thank you! \$\endgroup\$
    – Vaaal88
    Jan 27, 2021 at 17:07
  • \$\begingroup\$ @ikkachu I have played a bit with the software, which lead to another question I hope you can answer to. In your circuit I can see that with "high enough" resistors (at the base of the transistor) I can switch off the transistor. E.g. with 10M resistor Vbe is <500 and the transistor is off. How to calculate what transistor value to use to obtain a specific value of Vbe? Just pointing me to the right source (for my level) would be really appreciated :) \$\endgroup\$
    – Vaaal88
    Jan 27, 2021 at 19:21
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    \$\begingroup\$ @Vaaal88, depends a bit on what we actually mean with the transistor turning off... If you set the base input voltage at 5 V and the resistor to 10 MΩ in the simulator, you should still see some 45 µA going through the collector. That's way more than the few pA of leakage it shows if you disconnect the base entirely, or connect it to the ground (or a logic port driven low). The transistor still amplifies, it has a β factor of 100, and the collector current is hundred times the base current. The current isn't much, but it is there. \$\endgroup\$
    – ilkkachu
    Jan 27, 2021 at 20:43
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    \$\begingroup\$ @Vaaal88, On the other hand, already at 5 V and 10 kΩ at the base, the simulator shows the transistor is "fwd active", meaning that the β of the transistor limits the collector current. With a resistance lower than that, it shows "saturation", i.e. that the collector resistor is what limits the current. I'm not that familiar with the analogy stuff, but I think this is the point where we should look at load lines. \$\endgroup\$
    – ilkkachu
    Jan 27, 2021 at 20:50
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    \$\begingroup\$ @Vaaal88, then again, just assuming the base-emitter voltage is always 0.7 V, and then calculating the base current and the amplified collector current gets around the right ballpark. (For 5 V and 10 MΩ, that would be \$ I_c = 100 \cdot (5.0\ \mathrm{V} - 0.7\ \mathrm{V}) / 10\ \mathrm{MΩ} = 43\ \mathrm{µA} \$, a bit less than what the simulator shows.) \$\endgroup\$
    – ilkkachu
    Jan 27, 2021 at 20:54
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The sentence is talking about the voltage between two points.

It says the voltage at the base must be 0.7V higher than the voltage at the emitter. This is comparing the voltage between two points - the base and the emitter. You can measure this voltage by putting one multimeter probe at the base, and the other at the emitter, and seeing if the multimeter reads higher than 0.7V.

You could also measure the same voltage by putting one multimeter probe on the negative terminal of the battery, and the other one on the base, reading the display, then putting the second probe on the emitter, reading the display again, and then subtracting the emitter reading from the base reading. Or you could use the positive terminal. Or any other good "fixed-voltage" point in your circuit (an audio signal would not work because it would change in between the readings).

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It gets rather tedious when describing a circuit if you have to keep saying what two points you are measuring between. It makes life easier to pick a point in the circuit and call it "ground". If the circuit is actually connected to the Earth, then there is a strong convention that the earthed point is ground. Otherwise, it's up to you. The negative terminal of the battery is a popular choice.

By definition, "ground" is at 0V. If you measure the voltage between ground and any other part of the circuit, then you find the voltage "at" that point.

The description of how a transistor works is valid whichever point you pick as ground.

Edit: You could pick the positive terminal of a battery as ground. But that would mean that all the other parts of the circuit would be at a negative voltage. This is valid, if a bit odd. But it's quite common to see dual-rail circuits with a ground, a positive supply and a negative supply.

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  • \$\begingroup\$ So what establishes whether the voltage at the base is at least 0.6v higher than the voltage at the emitter? The set of transistors before the base and after the emitter? Can I get an example of a set of resistence where the transistor is off and one when it's on? \$\endgroup\$
    – Vaaal88
    Jan 26, 2021 at 20:30
  • \$\begingroup\$ Resistance is indeterminate or at least non-linear in current-controlled devices like a BJT. For MOSFET transistors, you can find an RDS curve in the datasheet. Think of it as your job to make sure that base is 0.7, not the BJT's... \$\endgroup\$
    – dandavis
    Jan 26, 2021 at 20:33
  • \$\begingroup\$ @Vaaal88 That's a very hard one to give a simple answer to. Everything that feeds the base of the transistor determines what voltage it's at. But here's just one example petervis.com/GCSE_Design_and_Technology_Electronic_Products/… \$\endgroup\$
    – Simon B
    Jan 26, 2021 at 20:50
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    \$\begingroup\$ @Vaaal88 The simple answer to that question would be "no". It's a lot more complicated than that. \$\endgroup\$
    – Simon B
    Jan 26, 2021 at 21:20
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    \$\begingroup\$ @Vaaal88, Changing the value of a resistor does not necessarily change a voltage or a current near by. It could do, but it depends on the surrounding circuit. For example, if your circuit consists entirely of a resistor connected across a constant voltage power supply, then changing the value of the resistor will change the current in the circuit, but not the voltage. If it's a constant current supply, then changing the value of the resistor will change the voltage, but not the current. And, like Simon B said, there are other cases that are still more complicated. \$\endgroup\$ Jan 27, 2021 at 13:14
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" Voltage is measured across two points, right? So what is the voltage 'at the base' and 'at the emitter'?"

It's measured from any point you choose. For a grounded-emitter circuit (such as is often used when switching), the emitter is at 0 volts (it's grounded, right?), and when the base voltage increases the current through the base-emitter junction will increase exponentially. For silicon, once the voltage reaches about 0.7 volts there will be sufficient current to get useful amplification.

If you want to pick a different reference point, you might wind up with a voltage from emitter to reference of, let's say, 10 volts. Then, when the base to reference voltage reaches about 10.7 volts the transistor will operate properly.

It's not a matter of the absolute voltage at the emitter, it's the difference in voltages between base and emitter (and the current which this implies) which is important.

It might be simpler to think of the situation as taking a voltmeter and putting one probe on the emitter and the other on the base. That voltage is the one to pay attention to.

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