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I'm a beginner with electronics, anyway I'm experimenting a lot with it. I still don't understand how you can use the same transistor for example to amplify an audio signal, or to act as a switch for turning on and off the current of a motor. Do you use different types of transistor for different purposes? Any help is appreciated, please be tolerant and try to explain it in a simple way, I'm new to electronics

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    \$\begingroup\$ (1) when it is cut off or saturated, (2) when it is in between those states. \$\endgroup\$ – Brian Drummond May 21 at 19:43
  • \$\begingroup\$ The transistor switch is an "overdriven amplifier". \$\endgroup\$ – Circuit fantasist May 22 at 19:38
  • \$\begingroup\$ It is always an amplifier. It amplifies up to the point it clips on the rail and then that is all it can deliver, making it look like a switch. There is no magic. \$\endgroup\$ – old_timer May 22 at 21:21
  • \$\begingroup\$ @old_timer, it is not an amplifier when is saturated or "cut off". Then, if we are talking about a series switch (a transistor in series to the load), it is just a "piece of wire" in the first case and "nothing" in the second case. If we are talking about a common-emitter stage, it is a voltage source with zero voltage (ground) in the first case and a voltage source with constant Vcc in the second case. The transistor is an amplifier only when operating in the active mode (between rails, as you said). Of course, you know that... \$\endgroup\$ – Circuit fantasist May 22 at 21:39
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    \$\begingroup\$ @Circuitfantasist well I think what old_timer is trying to say is that it's always an amplifier, and the transition from amplifier to "switch" is entirely arbitrary. Your comment seemed to imply that this was incorrect, when it actually is correct. It is not an amplifier with zero gain, it is an amplifier with a gain that gets smaller and smaller, and it never stops being an amplifier. \$\endgroup\$ – BeB00 May 23 at 7:10

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schematic

simulate this circuit – Schematic created using CircuitLab

We know that for a BJT the collector current increases with the increase in base current. For example, a 0.01mA increase in base current has caused a 10mA increase in collector current.

Now, let's assume, you have connected the collector and emitter through a copper wire (i.e. short-circuited). Then the current through 'RL' will be: \begin{equation} i_L= \frac{VCC}{R_L} \end{equation}

This is the maximum value of the current that can be passed through the resistance if the collector & emitter works as a short circuit. And the condition when the collector & emitter works like a short circuit is called the saturation condition. And the current at that condition is called saturation current, which is defined as: \begin{equation} i_c(sat)= \frac{VCC}{R_L} ; when V_{CE} = 0 \end{equation}

But in real life collector-emitter voltage will never be zero. So the equation will be: \begin{equation} i_c(sat)= \frac{VCC-V_{CE}}{R_L} \end{equation}

So with the increase in base current, the collector current will increase until it reaches saturation. As soon as the transistor reaches saturation, it is fully on.

Similarly, if you reduce the base current, the collector current will decrease. For a certain base current, the collector current will be almost equal to zero. That point is called the cutoff. At that point, your transistor is fully off.

The range between the cutoff and saturation can be used as an amplifier. Because at that region collector current changes with base current.

And the cutoff and saturation condition acts as a switch.

Output characteristics curves of a typical bipolar transistor

Image source: Output Characteristics Curves of a Typical Bipolar Transistor from Electronics Tutorials

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    \$\begingroup\$ I always found these types of diagrams really show the system response in an intuitive way. \$\endgroup\$ – J... May 22 at 13:50
  • \$\begingroup\$ I wanted to add that. But didn’t add it thinking that it may make the answer complicated. \$\endgroup\$ – Sadat Rafi May 22 at 14:28
  • \$\begingroup\$ And inbetween the amplifier and switch regions: messyness \$\endgroup\$ – Cort Ammon May 23 at 3:17
  • \$\begingroup\$ What have you meant? \$\endgroup\$ – Sadat Rafi May 23 at 6:23
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    \$\begingroup\$ Nice explanation... I have read it with pleasure (maybe because I am a teacher and the way of explaining is important for me). To pique the interest of my students, when reached the point of saturation, I usually ask them, "How can we continue increasing the collector current beyond the saturation? Let's dream..." And we consider the situation in terms of resistances... \$\endgroup\$ – Circuit fantasist May 23 at 7:22
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Acting as a closed or open switch is just an extension of it acting as an amplifier at its limits. Imagine you use your weak little fingers to push some buttons to control a massive flood gate. Anything in between full closed and fully opened is throttling the flow of water somehow, but when fully open or fully close it's just acting as a switch to block or pass for the water.

When acting as an open (non-conducting) switch it is acting as an amplifier amplifying a signal of zero. When acting as a closed switch (conducting) it is acting as an amplifier trying to amplify the largest signal it can. It is amplifying so hard that it can't amplify any further. In the same way you can throw open the flood gates but that doesn't mean you can pass an infinite amount of water through the flood gates. The flow rate is is capped by the size of the flood gate. If more water wants to pass through the flood gate than in any single instant than the size of the flood gate will allow, it simply can't (you don't want this because it means the switch is the bottleneck which a good switch should not be). If the flow rate is less than the size of the flood gates, then flow is unrestricted by the flood gate and the flood gate is invisible to the flow (this is what you want).

Where's the amplification part? Don't forget that you couldn't possibly control all that water directly with just your little button pushing fingers.

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  • \$\begingroup\$ Your example is great. I've never head this kind of example. Now it will be very easy for me to explain that to school students. \$\endgroup\$ – Sadat Rafi May 21 at 21:45
  • \$\begingroup\$ As usual for you, an intriguing explanation... A clever question that you can ask after the transistor is saturated can be, "How can we get the transistor out of this state?" \$\endgroup\$ – Circuit fantasist May 23 at 7:40
  • \$\begingroup\$ @Circuitfantasist I suspect if you asked that it might carry the connotation that saturation is a "bad" state, which is definitely not what I would want people to think. \$\endgroup\$ – user253751 2 days ago
  • \$\begingroup\$ @user253751, it seems so ... I would not have thought about it ... I ask such questions with increased difficulty to those of my students who want a higher grade :) \$\endgroup\$ – Circuit fantasist 2 days ago
  • \$\begingroup\$ It's a perfectly good question, I just don't think it's good to give students the idea that saturation is bad. You could frame the question in a silly way: Your boss comes past and overhears you talking about saturation. He chimes in that there's a new regulation about saturated fat and this product isn't allowed to have any. You try to explain it's about transistors but he won't listen. What things could we change to prevent this transistor from being saturated? (This way, it doesn't give the impression that saturation is bad, just that the boss is silly) \$\endgroup\$ – user253751 2 days ago
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The transistor can be switched "on" or it can be switched "off." But it also has an infinite number of positions in between "on" and "off." It is those intermediate positions that allow it to act as an amplifier.

If you have a +15V power supply and a -15V power supply, you can use two transistors to apply any voltage between +15 and -15 to a speaker. The signal that controls all this is a much lower voltage (line input, say).

Hopefully that will make some sense.

Also, this is a conceptual outline of how an amplifier works. There are many, many details I have totally glossed over. Real amplifiers require a lot more transistors (or integrated circuits which contain transistors).

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    \$\begingroup\$ You haven't explained anything. \$\endgroup\$ – Jack Danniels May 21 at 21:49
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    \$\begingroup\$ @JackDanniels write an answer! \$\endgroup\$ – mkeith May 21 at 21:56
  • \$\begingroup\$ I just gave an honest opinion. Don't mind. \$\endgroup\$ – Jack Danniels May 21 at 21:59
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    \$\begingroup\$ I upvoted it because I am reading it and it's elaborated. \$\endgroup\$ – Jack Danniels May 21 at 22:04
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    \$\begingroup\$ @JackDanniels Thank you. Upvoting the answers you prefer is a good way to make a positive contribution to the site. Providing good answers and asking good questions are also good ways to contribute. Keep up the good work. \$\endgroup\$ – mkeith May 21 at 22:08
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A transistor acts as an amplifier based on it's characteristic of a small increase in it base current resulting in a larger increase in its collector current. The ratio of collector current to base current is known as the current gain of the transistor.

With the base current at zero, the collector current would be zero and the transistor said to be 'cut-off'. When, with an increase in base current, the transistor is fully conducting and there is no further increase in collector current, the transistor is said to be 'saturated'.

The transistor acts as a switch when driven from 'cut-off' to 'saturation' and vice-versa with an instantaneous change of base current.

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    \$\begingroup\$ Sorry, that's overly simplifying. A transistor is not "basically" an amplifier - it rather may act as one, provided a number of requirements are satisfied. The way you put it seems like they're interchangeable terms. \$\endgroup\$ – edmz May 22 at 19:24
  • \$\begingroup\$ Hi edmz, Many thanks for pointing it out. Will amend my answer accordingly \$\endgroup\$ – vu2nan May 23 at 5:44
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A transistor acts as a switch when you ensure that inputs are always either low enough that the output will be sensed as a zero, or else high enough that it will be sensed as a one, but prohibit any "in between" states.

Conversely, it acts as an amplifier when you ensure that the inputs are always in that "in between range" where the output is not normally going to be at (or even extremely close to) one rail or the other.

For example, if we look at the data sheet for the venerable 7404 hex inverter, we see that the maximum input level for a logic 0 is 0.8 volts, and the minimum input level for a logic 1 is 2.0 volts.

So, somewhere in the range between 0.8 and 2.0 volts, an inverter will act like an inverting amplifier. It's not designed for linearity or low distortion, so it'll probably be a pretty crappy amplifier, but an amplifier nonetheless. Oh, and 0.8 and 2.0 are the rated minimum/maximum it's specified to meet. It may only act like an amplifier over an even smaller range than that.

Oh, along with being a crappy amplifier, a typical digital chip would have a pretty serious heat problem if you tried to use it in the linear range much. When used as intended, the transistors in the gates dissipate relatively little power, because they've turned pretty much completely on or completely off. They're only supposed to end up in the intermediate range (where they'd dissipate a lot more) for a short time as the input slews from low to high or vice versa. With the input "in the middle" for very long, there's a good chance of toasting the chip.

TL;DR

An inverter is really an inverting amplifier, but it's not designed to be used in the range where it would be at all linear. But if you did drive it in the right range, it would be a (really poor) amplifier.

Reference

https://www.futurlec.com/74/IC7404.shtml

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Basically...

If your input signal varies from 0-1 volt, and you configure your transistor amplifier to multiply it by 5, then you get an output that varies from 0 to 5 volts.

If your input signal varies from 0-1 volt, and you configure your transistor amplifier to multiply it by 1000... but you only have a 5 volt supply, then you get an output that switches between 0 and 5 volts.

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this transistor 2N3604 circuit u have given it can be used as an amplifier because it's am capacity is 200ma so it can't be used as a switching.and also it's gain value is 300.so it can be used as an amplifier by applying signals to its base.

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shivlal gajjar is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
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I still don't understand how you can use the same transistor for example to amplify an audio signal, or to act as a switch for turning on and off the current of a motor

The very straight answer to your question («how») is because the physics of transistor can be engineered in a such a way to achieve both the things you mention - the transistor need not change, because the I-V curves (namely a higher level view of the physics showing up) have different regions of operation, like 3 or 4 of them. Then, by figuring out what the device may be able to achieve in such regions, but always taking into the account the physics beneath them, you can essentially get the very same device to act differently and then engineering out the best out of them. IOW the physics of the transistor embodies, if you wish, different characters of the same device, provided you're able to extrapolate them.

Please notice this not a straightforward thing: If you take a resistor, that could not happen: no matter how you reason about it, you will never get that behavior because the characteristic curve of a resistor will always be a straight line -- that does not allow it to amplify or switch something. Same for a cap or a diode.

Of the two, I reckon it's easier to view a transistor as a switch from its characteristics -- essentially, it lies down to whether the current sweeps from 0 to something non-zero as the 'input' voltage varies. To act as an amplifier is a bit trickier however: as I was saying above, it turns out that the region suitable to obtain a device that has the properties of an ideal amplifier (and do notice: an amplifier is not determined from its gain) is one. We call it the "active region". So in this case you also have to make sure the transistor does not leave the active region, where again the physics is suitable to obtain those nice properties.

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Yes, you could use different types of transistors for amplifying an analog signal and for switching a motor. For analog signals, you would like a large amplification factor, e.g. bipolar transistor. For switching a motor, you would like a low on-resistance, e.g. power fet.

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My answer is in a form of question:

A box with two terminals is supplied through a resistor by a voltage source. A voltmeter, ammeter and wattmeter are connected (accordingly, in parallel, series and both) to the box terminals.

The voltmeter shows some voltage... the ammeter shows some current... but the wattmeter shows zero power.

How is it possible? What is inside the box?

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