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I am a high school student learning about semiconductors (diodes and transistors). I have been facing a lot of problems to understand the configurations of BJTs and the simplest amplifier circuit. These confusions have risen perhaps due to unclear explanations in my textbook for which I have assumed some of the explanations. Now it seems that the assumed explanations are mostly incorrect.

Questions:

  1. Why is \$I_B\$ considered to be the input current in common emitter/collector configuration? Is it because in the common emitter configuration the collector current is dependent on the base current and is multiple times of that? That seems reasonable for an amplifier circuit to consider collector current as the output current, again emitter current cannot be considered input since in an amplifier circuit if input signal is linked to wire directly connecting emitter then both the base biased voltage and collector-emitter voltage will variate with time?

  2. Why is emitter current the input current in common base configuration but not collector current?

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  • \$\begingroup\$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. \$\endgroup\$
    – Community Bot
    Oct 20 '21 at 15:58
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    \$\begingroup\$ You should split this into separate questions. Also try not to title questions "questions about X" - instead, title your questions with a brief version of the actual question. \$\endgroup\$
    – JYelton
    Oct 20 '21 at 15:59
  • \$\begingroup\$ Do you understand the fundamentals of circuit analysis yet? Are you able to perform things like mesh and nodal analysis? it's kind of pointless to push on BJTs if you don't. Once you do, then you can move onto substituting a circuit model of a BJT for a BJT in circuits which helps clear things up. \$\endgroup\$
    – DKNguyen
    Oct 20 '21 at 15:59
  • \$\begingroup\$ I know about kirchoff' law and ohms law thats it... But my highschool syllabus also includes this section(semiconductors). @DKNguyen \$\endgroup\$
    – MSKB
    Oct 20 '21 at 16:04
  • \$\begingroup\$ @MSKB Hmmmmm...it's rather futile to try and understand transistor amplifiers if you can't apply Kirchoff's laws on a wider scale (which is what mesh and nodal analysis are). \$\endgroup\$
    – DKNguyen
    Oct 20 '21 at 16:05
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In the examples you quote, "Input" and "Output" and based on designer intent.

It's like this wrench:

enter image description here

The "input" is the side you hold, the "output" is the side that's on the bolt, but that's not a property of the wrench, it's decided by the way you use it: if you flip the wrench around, the labels will switch sides too.

Now look at this emitter follower:

enter image description here

On schematic #1, if the intent was to drive a votage on the load resistor, then the output would be the emitter. But if the intent was to make a current source, and drive a current into R5 (schematic #2) then the output would be the collector.

Now, if the emitter resistor R4 was not constant but instead variable, and the base voltage was constant, and the intent was to have the transistor set a constant voltage across this resistor to measure its value by measuring current flowing through it, said current being sent into the collector resistor... then it would be called "common base", but the circuit didn't change much.

#3 If you put an AC current source on the emitter to measure its output impedance, then the output also becomes an input.

#4 If you put an AC voltage source on the collector to measure the Early effect, or the supply rejection ratio, of your emitter follower, then the collector is also an input.

#5, the differential/LTP. What we're interested in is collector currents, so we decide it's the output. We input some signals on the bases, so they get an "input" label. The transistors themselves don't care, and besides, the question of what the emitter is in this case is quite interesting: it is an output because each transistor works as a follower, but it is also an input because each transistor influences the other's current through its emitter. So it combines common collector and common base.

So, basically, all three pins are both input and output, and it's more about impedances, gain, and intent:

Collector: For BTJs in forward active mode (or FETs), Ic depends very strongly on Vbe, but not much on Vce. This means the collector is useless as an input: if you put some signal there the transistor will try its best to ignore it. Low dependence of Ic on Vce means the collector behaves as a high impedance current output. It is only considered an input if you're interested in the effects of some of the transistor imperfections like Cbc, Early effect, etc, on your circuit.

If the collector is connected to the power supply, then the collector current comes from the power supply but it is not used as signal in the circuit, so the pin is not labeled as "output", because that's not the intent. Unless the intent is to inject current into the power supply, of course, then it's an output ;)

Base has high-ish impedance and low current capability, so it's not generally usable usable as an output. It is however very convenient as an input, so that's what it is used for. Its impedance depends on the circuit.

In common collector (follower) mode, emitter voltage follows base voltage, so the base's input impedance is quite high (ie, hFe * emitter load). Then it behaves as a high impedance voltage input that draws a bit of parasitic current that's annoying but you got to live with it.

In common emitter mode, since Vbe is quite constant, base voltage stays pegged to emitter voltage. So input impedance is low: you can feed current into the base and it won't move much. So it behaves as a low impedance current input.

The emitter is the most versatile since it's low impedance IO.

It behaves both as input and output. As an output, it will dump as much current into the load as necessary such that emitter voltage follows base, minus one Vbe. So the emitter acts as a low impedance voltage output. However, emitter current also appears on the collector (minus base current) which means the emitter acts as a low impedance curent input.

Anyway.

Why is emitter current the input current in common base configuration but not collector current?

Because you use common base when you want to do that, it's simple as that.

Common base is a current conveyor. Say the previous stage outputs a current, but wants to output that current into a constant voltage (or known voltage) node. A common use case is to use a resistor to turn voltage into current: you use the emitter of a transistor to define the voltage on one side of that resistor.

Another use case is the cascode:

enter image description here

Here Q5 acts as a current source controlled by Vin, it will output AC current on its collector equal to Vin/R4. If its collector is directly connecte to a load that has large voltage swings, then perhaps its Early effect or collector to base capacitance will cause problems. So, cascode Q4 takes Q5's collector current as input on its emitter, and just outputs it on its collector. All it does is keep Q5 at a constant Vce, which solves problems caused by Early effect and Cbc.

Why is IB considered to be the input current in common emitter/collector configuration?

To nitpick, "input current" can mean two things:

  • The input signal is actually a current, and the transistor is used for its current gain, in that case the output current would be Ic or beta times Ib at low frequency.

  • The input signal is voltage, it is input on the base, and base current is an annoyance and is labeled "input current" just because it happens on the pin that was previously labeled as "input", but it is not the input signal that we want to use, just a source of error to keep in mind.

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  • \$\begingroup\$ How does Ie fluctuate as output in common collector config. as Ic does in common emitter/base config? \$\endgroup\$
    – MSKB
    Oct 20 '21 at 18:18
  • \$\begingroup\$ In common collector (emitter follower) the emitter sets a voltage on the load, for example a resistor, and the load decides what the current will be. In common emitter, emitter is connected to GND or a power supply. So it is like having load resistor equal to re, which is low, about 26mA/Ie ohms. So if the circuit sets a voltage on the base, the transistor will have huge current gain that needs feedback to control. If the circuit sets a base current, then Ic will just be hFe*Ib. \$\endgroup\$
    – bobflux
    Oct 20 '21 at 19:31
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These are all routine questions frequently asked but perhaps hard to search. Don't doubt your textbook until you fully understand.

The transistor has many non-linear characteristics that with certain configurations can be linearised, somewhat with R ratios or kept nonlinear and used as a switch. As a switch, the current gain reduces up to 90% or 10:1 to 50:1 at best unlike a relay, which can switch with a current gain of >>5000 taken to the limits. e.g. 25A relay with 5 mA coil.

  1. In CE config with an emitter resistor becomes base current controlled collector current with current gain hFE ((h parameter- Forward Emitter) as Vbe stays relatively constant and voltage gain = Rc/Re. But Vbe still increases slightly with Ib and always follows the Shockley exponential equation of Ic vs Vbe even if we do not use it.
  • So by adding the Re which is much larger than the dynamic resistance of Rbe = \$r_{\pi}\$ as it is called in the Pi simplified model, the AC voltage gain is now linearized by Rc/Re ratio and the DC current is controlled by Ic/Ib according to some nominal value which is hard to control in manufacturing so it has a wide tolerance.

  • But when the CE is used as a switch , we choose Re=0 then the gain ratio is now with the very small dynamic resistance Rbe , Av=Rc/Rbe = voltage gain. But when collector base voltage drops to 0 and becomes forward biased say Vce <0.7 the effective current gain still called hFE drops to 10% or more of it's maximum linear current gain when \$V_{ce}=V_{ce(sat)} "value" @ I_C= "spec" for ~ Ic/Ib=10 \$

  1. For CB the Vbe is held "relatively" constant with some Rb Vb and caps to short out the ac Base voltage then the collector current is modulated by Vbe from a low impedance emitter driven from a voltage source low impedance by injecting current at a relatively constant Vbe. Thus the voltage gain is again Rc/Re but but keeping Vbe constant Ie = Ic (+Ib=ignore) and thus the voltage gain occurs simply from R ratios drive by a voltage source at the emitter for high speed but no current gain.

CB details CB Simulation At the risk of too much info, here is a CB amp with a 50 Ohm gen. Note that Vbe still modulates Ic but Vb is constant and Vbe is relative small change. So effectively we say it is linearized with Av=Rc / (Rbe+ Re) but Rbe is controlled by Ic is the small nonlinear dynamic resistance. Thus gain here resulted at 85% of Rc/Re by choosing Ic >1mA Vc near Vcc/2 and Rth base Thevenin Req/Re about 75% of hFE. Your mileage will vary with these choices.

Assume that the Cap. impedance Xc(f) = 0.

enter image description here

  1. Vbe is a function of the band-gap and minority and majority carriers from PN junction material . P for positive, N for Negative. The doping concentration level is quite different than the Vbc junction yet both have PN diode characteristics for V vs I so the CE behavior is not the same when reversed and thus Vbe has a very low reverse breakdown voltage threshold -5 ~ -6V unlike Vbc.

  2. We use Vce for output rather than Vcc-Vc across the resistor in order to share a common emitter rail voltage e.g. 0V When you change references from 0V to V+=Vcc you are inverting the voltage and subtracting from the supply. Since CE config tends to be a current source/sink, that behavior is defined to be independent of Vcc so using Vcc as a reference doesn't help the designer or student understand.

  3. The input resistance changes for each pin BCE but for CE it is Rin @ base = hFE * (Rbe + Re) and for common collector, CC the emitter sees the base effects as Rout @ emitter = Rbe + Rb/hFE where Rb is the Thevenin equivalent of say R1//R2 and Rbe is often neglected as it is small and insignificant except in the CB config.

Some search results may be helpful.

A Table of input output characteristics:
https://www.etechnog.com/2019/12/transistor-cb-ce-cc-configuration.html

Jonk's introspective perspective.

Confusion regarding signal amplifier

Notice that by swapping ground on CB input Cap. with Re signal gen. I can go back to a CE amplifier. again with a voltage gain of 42. (4.22Vpp/ 100mVpp)

enter image description here

Keep up the great questions. This is the root cause of every great engineer.

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  • \$\begingroup\$ I didn't understand "vbe remains relatively constant" part. \$\endgroup\$
    – MSKB
    Oct 20 '21 at 17:33
  • \$\begingroup\$ My textbook mentions that current gain is the ratio of change of collector current and base current keeping Vce constant. How is Vce kept constant? Is it as mentioned by you that the change is so small compared to the corresponding changes of collector and base current that it is ignored? \$\endgroup\$
    – MSKB
    Oct 20 '21 at 17:35
  • \$\begingroup\$ Vce is only kept constant for standardizing the test result with Rc = 0 Ohms because adding Rc converts current gain into voltage gain but also changes Vce and dynamic resistance , so this is just a test method. \$\endgroup\$ Oct 20 '21 at 18:32
  • \$\begingroup\$ How does Vce remain constant if we are changing Ib and IC? \$\endgroup\$
    – MSKB
    Oct 20 '21 at 18:37
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MSKB - the answer to your question is relatively simple and can be rather short:

The transistor is a 3-terminal device (C,B,E). When properly connected to supply voltages the current between C and E can be varied with a control voltage Vbe between the base and the emitter. The control function is an exponential function as given by W. Shockley Ic=f(Vbe).

  • For each of the three possible configurations (CE, CB, CC), one of the nodes of the controlling pair (B,E) is held at a constant potential - and the other node receives the input signal voltage that can vary the voltage Vbe - and, hence, the current Ic resp. Ie through the device.

  • As far as the input current is concerned (your question) - this is the current driven by the voltage which is conected to the corresponding input node (B or E). This current has a DC part (Ib resp. Ie, fixing the DC operational point) and an ac part (ib resp. ie) determined by the signal voltage.

  • The input signal current (ib resp. ie) plays NOT a major role for amplification purposes. Together with the input voltage it determines the signal input resistance. This resistance should be not too small because we normally wish that the internal source resistance of the signal source should have a minor influence on the amplification factor only.

  • Example: The main (if not the only) purpose of the common-collector configuration (emitter follower) is to provide a rather high input impedance (due to the emitter feedback resistor). The voltage gain is only unity.

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  • \$\begingroup\$ Considering the 300% dynamic range of hFE with Ic and Ib, your 3rd item on Ib not playing a major rule has some caveats. But we know a better design is less dependent on hFE and thus Ib but it must be chosen within certain ranges for optimal performance. \$\endgroup\$ Oct 20 '21 at 18:38
  • \$\begingroup\$ The tolerance of hfe concerns the input resistance only - not the amplification (provided that there is no source resistance which lowers the gain). The amplification factor is determined by the transconductance only. The tolerance of B=Ic/Ib (DC values) can have small influence on the DC operational point, but a clever design with DC feedback can keep this influence within small limits. \$\endgroup\$
    – LvW
    Oct 20 '21 at 18:50
  • \$\begingroup\$ That's what I meant about hFE dependency. It's too bad that gm is not plotted in datasheets. \$\endgroup\$ Oct 20 '21 at 18:55
  • \$\begingroup\$ But gm can be directly derived from the DC quiescent current Ic. I am sure you know about this: gm=Ic/Vt. So - what do you expect from a gm plot in the data sheet? \$\endgroup\$
    – LvW
    Oct 20 '21 at 18:59
  • \$\begingroup\$ I would expect gm curves to show variations wrt. Ib, Vbe, Vce due to IS, NS,ISC, NF etc \$\endgroup\$ Oct 20 '21 at 19:07
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The main theme pervading your questions is "why consider something an input or an output in a transistor circuit?". The principal answer is "the output is the one where current and/or voltage appear amplified in relation to the input, because otherwise why bother with a transistor?".

Now of course transistors are not linear devices, so there may be a number of other reasons for using them that cannot easily be classed as input/output (for example, a transdiode, namely just tying together basis and collector of the transistor, obviously has no separate input/output, and circuits like a constant current source stretch the input/output definition). But for the standard amplifier configurations, one can with some justification determine a pin working as input and an output pin that carries in some form an amplified version of the input.

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    \$\begingroup\$ Transistor as a switch does not need amplification as far as I know. \$\endgroup\$
    – MSKB
    Oct 20 '21 at 16:21
  • \$\begingroup\$ And for common collector configuration why do we consider emitter current as output though is not the amplified form of the base current which is the input? \$\endgroup\$
    – MSKB
    Oct 20 '21 at 16:22
  • \$\begingroup\$ @MSKB In common collector, the collector is common (the hint is in the name), so it's not considered an output. \$\endgroup\$
    – Neil_UK
    Oct 20 '21 at 16:25
  • \$\begingroup\$ @MSKB: a transistor used as a switch certainly does provide amplification - a small base current controls a large collector current. \$\endgroup\$ Oct 20 '21 at 16:54

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