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enter image description here

I know this is a noob question but I just can't find an answer for it on google search results.

Well my question is basically this: How come the amplifying occurs before the transistor [red line] and does not occur after the transistor [blue line]? I was thinking that the current had to go though the transistor to be increased.

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    \$\begingroup\$ The transistor only acts as a switch, it's either (full) on or off. Yes, everybody is talking about a transistor as an amplifier, but that's because with your "low" current signal on the base (B) you can "switch" a higher current (because C and E will conduct). \$\endgroup\$ – Evert Nov 11 '14 at 13:44
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    \$\begingroup\$ @Evert Not true. Transistors act as an amplifier until the input reaches the saturation level. This allows them to work as a switch because once you hit the saturation level they are "full on", but they were initially used as amplifiers in hearing aids and transistor radios. \$\endgroup\$ – Jason Goemaat Nov 12 '14 at 1:10
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    \$\begingroup\$ The risk of thinking of a transistor in saturation as a switch as opposed to a more accurate transistor model is that eventually you come to a configuration where your "switch" isn't switching, and you won't understand why. \$\endgroup\$ – Scott Seidman Nov 12 '14 at 14:25
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    \$\begingroup\$ If you think about the electrons flowing, "before the transistor" is the blue side. This doesn't make it a lot easier to understand, though. \$\endgroup\$ – MSalters Nov 12 '14 at 16:07
  • \$\begingroup\$ Current flows from E to B which "turns on" the transistor allowing current to flow from E to C. The reason you are confused is that you are thinking about classical current flow which was + to - which was wrong (as assumption made hundreds of years ago). \$\endgroup\$ – Peter Quiring Nov 12 '14 at 17:32
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The collector of the amplifier (the pin labeled "C") is actually AFTER the transistor (using the asker's reference frame, which is somewhat misleading.). The Base (B) is the input, and the active transistor creates a situation where the current on the collector is many times that of the base. So, the Base current is the input, and the Collector current is the output.

In this case, completing the circuit at the base with your finger creates a small current through the base. The transistor works things out so as to create a current many times that at the base, lighting the LED.

Why? I'll punt on the mechanism, but this is what transistors do for a living.  from media.tumblr.com/tumblr_luy74c89IH1qf00w4.png

The figure is taken from http://media.tumblr.com/tumblr_luy74c89IH1qf00w4.png, but probably originates from Horowitz and Hill, The Art of Electronics. "Transistor Man" looks at the current at the base, and adjust the current at the collector so as to be a multiple of the base current. Of course, all of this has to do with properties of silicon junctions, but that's a bit beyond the scope of your question.

The current at the Emitter is the sum of the base and collector currents.

This really is an oversimplification, but it gets at the essence of your question.

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    \$\begingroup\$ +1 for the great diagram even if it is at bimbo level LOL \$\endgroup\$ – Andy aka Nov 11 '14 at 15:51
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    \$\begingroup\$ @Andyaka If no other lesson is learned, its that Transistor Man wears a hat. \$\endgroup\$ – Scott Seidman Nov 11 '14 at 16:20
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    \$\begingroup\$ The hat is important of course!! \$\endgroup\$ – Andy aka Nov 11 '14 at 20:06
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    \$\begingroup\$ IMO the first sentence of this answer is misleading - it makes it sound like the asker's reasoning is correct, apart from having "before" and "after" backwards. \$\endgroup\$ – immibis Nov 12 '14 at 7:00
  • \$\begingroup\$ @immibis modified to show I was translating from the original terms \$\endgroup\$ – Scott Seidman Nov 12 '14 at 16:23
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You are looking at a loop of current, so after and before don't have much meaning. Other than the tiny base current, the current in the red and blue paths are the same.

Your concept of before and after in electronics isn't applicable. You need to understand some basics before you can make sense of this, but that would be too much to try to teach in a answer here.

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I was thinking that the current had to go though the transistor to be increased.

This highlights a dangerous misconception. Current is the flow of charge. Charge, like energy, is never created nor destroyed. Thus, you will never find a device where the total current flowing into the device is not equal to the total current flowing out. In more formal terms, this is called Kirchhoff's current law.

This makes sense. Is there any device you can put in a hose such that the water exiting is greater than the water entering? If so, it would be an infinite water machine. Likewise there is no infinite charge machine.

In your circuit, current enters through the base and collector, and exits through the emitter. The emitter current is exactly equal to the base current plus the collector current. Because of the transistor's gain, the base current is much smaller than the collector current by a factor of 100 or more -- this parameter is called \$h_{FE}\$ in the datasheet.

Because the emitter current is the sum of the base and collector current, and thus is also much larger than the base current, it's perfectly valid (and frequently useful) to attach things to the emitter of the transistor to make use of the transistor's gain. See Why would one drive LEDs with a common emitter?

Furthermore, your use of "before" and "after" suggests that you think you can start at the + terminal of the battery, then work your way to the - terminal following some linear cause-and-effect reasoning. You can't. It doesn't make sense, anyway. We call them circuits because they are just that:

cir·cuit (sûrkt) n. 1. a. A closed, usually circular line that goes around an object or area.

Current flows through the battery just like everything else. The electric charge moves in a circle. A circle doesn't have a start or an end, so you can't have a "before" or "after".

You don't need anything so complex as a transistor circuit to illustrate this; just an LED and a resistor will work. Try this:

schematic

simulate this circuit – Schematic created using CircuitLab

Is there any functional difference between these circuits? If you really want to get into the cause-and-effect chain, then you need to think at the speed of light, and read How does the current know how much to flow, before having seen the resistor?

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Perhaps it's a little easier if you think of the circuit as being drawn something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Your finger acts as a resistor (a largely-unpredictable variable resistor), connecting the positive terminal of V2 to the base of the transistor. That allows a little current to flow through the base/emitter part of the transistor (both the emitter and the battery's negative terminal are grounded, so they're also connected to each other). The transistor then multiplies that current by some (more or less) constant factor, and lets that proportional current flow between the emitter and collector. Since that circuit also includes the LED, current flows through it, and it emits light.

Your schematic is basically the same, except that they've combined V1 and V2 into a single power supply. You simply need +9V in two places, so it connects both those places to the same power supply. The rest of the circuitry doesn't really care where that power comes from, or that the same power supply is being used in both the "input" side and "output" sides of the circuit.

In other words, the input circuit is basically the base/emitter, and the output is the collector/emitter. That's why this is called a common emitter circuit--the emitter is shared (common) between the input and output. There are also common base and common collector circuits, though common emitter is (no pun intended) much more common.

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Replace the transistor with a simple switch, so you just have a battery, an LED, a switch, and a current-limiting resistor.

Note that when you close the switch, current flows on both "sides" of the switch. "Why does the switch control current "before" it as well as "after" it?" Because it has to be: current in a simple closed loop has to be the same everywhere. Closing the switch doesn't result in the switch "producing" current. It allows the switch to pass current.

Back to the transistor. Do you see? The transistor acts as an amplifier, but it doesn't really "amplify" the current. The current in the C-E path has to be the same on both sides of the transistor, because the transistor has nothing in it that will add to the flow of electrons. The transistor simply allows changes in a small current (in the base-emitter path in this case) to control the flow in the larger current (in the collector-emitter path).

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LED would indeed be brighter if you put it after because on the before transistor there will be hfe*Ib ammount of current, and it will have (hfe+1)Ib after the transistor. You will probably not notice this because hfe 100 or more in most cases and 1% extra current will not cause visibly more light. If you have a transistor with hfe of say 5, then you will notice it, but in that case you will not be able to turn it on with your finger because the "1" comes from the your finger. The hfeIb must come from somewhere and it comes from before the transistor out of the 9V "battery".

Electric current is like the water current, it must come from somewhere, if there is current AFTER the transistor, there must be current BEFORE the transistor!

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A transistor may be used to amplify current, voltage, or both. In the indicated circuit, it is being used to amplify both. When using the emitter as an "output", the output voltage will vary almost 1:1 with the input voltage. In some circuits, that 1:1 voltage behavior is very desirable, since the accuracy of the unity gain won't be significantly affected by component characteristics. For a transistor to scale output voltages up or down, however, it is necessary to use the collector as the "output".

With the indicated circuit, the base voltage will remain at about 0.7 volts, so the voltage across the finger will remain at about 8.3 volts regardless of the voltage across the LED or its current-limiting resistor (so the input voltage changes by almost nothing, while the output voltage changes greatly--thus amplifying voltage). If the LED and transistor were on the collector, then every volt dropped by the LED or its resistor would reduce the voltage that could flow through the finger.

Note that the circuit as shown is slightly "dangerous" since shorting out the finger contacts could drive a nearly-unlimited amount of current through the transistor. Adding a moderate-value resistor (using the same value as for the LED resistor might be convenient) in series with the base and/or supply-side finger contact would make it safer.

Note that when given a touch which is barely sufficient to turn on the transistor, the circuit behavior will be very sensitive to a transistor characteristic called "beta", which can vary considerably from one transistor to the next. Basically, that means that some transistors may amplify the amount of finger current by a factor of 50, while other transistors might amplify it by a factor of 200. For this particular application that may not matter, but some applications require a more predictable level of amplification. Adding a small-value resistor to the base would make the transistor pass a current which was proportional to the amount by which the base voltage exceeds 0.7; connecting a series combination of a diode and resistor would make the base voltage exceed 0.7 by an amount which was proportional to the finger current. The transistor's amplification would then be limited to the ratio of the two transistors [e.g. a 560-ohm resistor from base to ground via the diode and a 56-ohm resistor from the collector to ground would limit current amplification to a factor of roughly 10:1].

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