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This question is connected to the link. Why are base current and emitter current the input current in common emitter and common base configurations respectively?

  1. Is voltage drop across the base-emitter junction in the common base configuration considered the input voltage because this voltage drop mainly triggers electrons/holes to flow from emitter to collector via base?

  2. Why is the voltage drop across the collector-emitter junction in the common emitter junction considered the output voltage but not the voltage drop across the load resistance?

  3. Is the dynamic resistance of the base-emitter junctions considered the input resistance?

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  • \$\begingroup\$ 1) draw the circuit (yeah wel all know what it looks like but we're discussing a circuit so just draw it, a pic form the internet is also fine). 2) Consider that I can either use Vce or V(Rload) as the output voltage. Can we know / determine Vce when we know V(Rload) and the supply voltage ? Is that a simple relation? So does it matter if I use Vce or V(Rload)? \$\endgroup\$ Commented Oct 20, 2021 at 18:42
  • \$\begingroup\$ You were asking why the voltage at the collector is considered as output voltage but not the voltage drop across the load resistance. And my answer was: Both are identical because the load is connected to the collector. \$\endgroup\$
    – LvW
    Commented Oct 20, 2021 at 20:50

3 Answers 3

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Is voltage drop across the base-emitter junction in the common base configuration considered the input voltage because this voltage drop mainly triggers electrons/holes to flow from emitter to collector via base?

Let's break this up into its two parts:

  • Is voltage drop across the base-emitter junction in the common base configuration considered the input voltage

No, not really. A forward voltage is required in order for the common base configuration to work. That's a given. But the input is taken to be a "small signal" that allows the tangent at the local operating point to be used as a linearization (simplication) for analysis purposes. So the input is usually thought of as the small signal itself, not the DC operating point forward voltage that is also required in order to set up the right situation. Though, I suppose, some might decide to just be argumentative about it.

  • because this voltage drop mainly triggers electrons/holes to flow from emitter to collector via base?

That's an issue for microelectronics engineers and physicists (aka "going down a rabbit hole.") It's just a no-nevermind for the first part of the question. Treat it as an unrelated addition designed to direct your mind away from the issues at hand, and to confuse you instead of enlighten you.

As a final note, I like to think of the common base situation as a place where you use the input signal to "pull" current out (or push it in depending on which bipolar you are considering and how you like to think about things) of the emitter, where this current is then propagated across the BJT and generates matching changes in the collector current. The difference is that the emitter "looks like" a voltage source to the input signal (very low impedance) while the collector "looks like" a current source (very high impedance) to the output signal. This concept makes configurations like the cascode very easy to follow.

(But keep in mind that this isn't actually what is taking place. Active mode BJTs always have a collector current that is related to the base-emitter voltage. It's just that this viewpoint is somewhat less helpful here. You can wrestle with it and in doing so you'll understand even more. But for a first-approximation, you don't need to burrow in like that.)

Why is the voltage drop across the collector-emitter junction in the common emitter junction considered the output voltage

The CE voltage drop is usually not considered the output voltage. It is usually the collector voltage taken with reference to ground that is considered the output voltage here. But it could just as well be the collector voltage taken with reference to the collector supply rail on the other side of the collector resistor, too. There's no difference in meaning between the two. So, technically, it's your call here.

The BJT is a 3-pin device, not a 4-pin device. But signals in and signals out are always 2-wire. So the input and the output must share a common reference, leaving 1 remaining wire for the input and 1 remaining wire for the output (which adds up to 3.)

Or, in different lingo, the input and the output are "single-ended" and not "differential."

It usually is easier to just say that there is a common reference between the input and the output signal that is the ground reference. And so the collector output is assumed to be referenced to ground. However, a voltage source is basically the same thing as a dead short to a signal. So there is no difference between thinking of the ground as the common and thinking that the supply rail is the common. They are equivalent for this purpose.

but not the voltage drop across the load resistance?

As I just wrote, you can take it across the collector load resistance. That's just another way of saying the same thing. So, no problem. Do that if you want. It's fine and it's no different.

Is the dynamic resistance of the base-emitter junctions considered the input resistance?

I'm not sure you are ready for this, as I'm now uncertain you understand the concept of linearizing complex curves in order to simplify analysis. But the dynamic resistance is, once again, just the tangent line to an operating point along an exponential larger-scale curve. It's a simplification. But it only works for one operating point. Different operating points will have different tangent lines and therefore different values.

But yes, it is considered as part of the input resistance when computing it for the purposes of small signal analysis. But it is not considered when trying to work out the DC operating point because it has no relevance there. (Doesn't exist as a concept for DC operating point calculations.)

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    \$\begingroup\$ Oh, jonk... I miss you! SE.EE is not the same without you! \$\endgroup\$ Commented Apr 2, 2023 at 21:13
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I dedicate this story to @jonk - the amazing professional, enthusiast and Great Man with the soul of a child, with the hope that he will come back to us!


Understanding through concepts

No matter how hard we try to explain something in detail, for it to be truly understood, we need to show the basic ideas... the concepts on which it is built.

In electrical circuits, such powerful concepts are resistor voltage-to-current converter, resistor current-to-voltage converter and their derivatives - variable resistor (rheostat) and voltage divider (potentiometer). They have been well known since the 19th century; we use them for their purpose now... and they will be used in the future because they are eternal and immortal.

By these concepts we can indirectly explain the more abstract transistor circuits. Indeed, electrical resistances are linear and transistor resistances are non-linear, but this does not particularly matter for the intuitive understanding of the most basic principles on which transistor amplifier stages are built. Furthermore, non-linear resistances can be represented as changing (static) resistances and thus a connection between electrical and transistor circuits can be made.

That is why, in order to truly answer the OP's fundamental question, I have examined, step by step, the evolution of the elementary transistor stage in parallel with its electrical (conceptual) implementation. I have done it by the help of CircuitLab simulations but I have also created a similar story with real experiments.

Building transistor amplifier

STEP 1: Controlling current by a variable resistor

The amplification is a kind of regulation. The simplest form of such an operation is the current control. We can do it, according to Ohm's law, by changing either the voltage or the resistance or both. In amplifiers we use the second way.

Resistor current "amplifier". So, to regulate the current in the simplest Ohm circuit, we insert a variable resistor Rin and start changing its resistance. Thus, the resistance is the input quantity of our "amplifier" and the current Iout is the output quantity. Let's set an initial (quiescent) current value of 5 mA. We can do it in two ways - by calculation from Ohm's law or by adjustment (changing Rin in parameters window while looking at the ammeter).

schematic

simulate this circuit – Schematic created using CircuitLab

It is interesting to see the transfer curve Iout = f(Rin) of our "amplifier". As you can see, it is non-linear because the input quantity (the resistance) is in the denominator but if we change Rin within small limits, the dependence will be close to linear.

R-I resistor amplifier

It would also be better to start with the largest resistance value (lowest current value), but the CircuitLab DC sweep simulation does not allow this.

Transistor current amplifier. Transistors can be functionally thought of as a type of "resistors" which, like ordinary resistors, dissipate heat and power, and limit the current. So let's replace the "manually-controlled resistor" Rin with such an "electrically-controlled resistor" - the transistor Q. Now the voltage Vin applied to the base-emitter junction is the input quantity of our "true amplifier" and the current Iout (Ic) is the output quantity.

schematic

simulate this circuit

A little more dexterity is required to set the quiescent current of 5 mA since at some point (about 0.6 V), the base-emitter junction begins behaving as a voltage-stabilizing non-linear resistor - it decreases its resistance when we increase the voltage and v.v. As they say, it has "low differential resistance"; I would say it behaves as a "dynamic resistor". Practically, we can adjust Vin as above - opening the Vin parameters window and changing Vin while looking at the ammeter.

The transfer curve Iout = f(Vin) of our amplifier is non-linear again, now because of the base-emitter junction properties. But we can apply the same linearization trick as above - if changing Vin within small limits ("wiggling" around the quiescent input voltage aka "bias voltage") the dependence will be close to linear (as they would say, "Vin is a small-signal quantity").

V-I transistor amplifier

STEP 2: Controlling voltage by a voltage divider

The amplifier we invented above produces output current... but we need a voltage output. Ohm's law gave us the idea above to use a resistor as a voltage-to-current converter (I = V/R). Now it can give us another idea - to use a resistor as the inverse current-to-voltage converter (V = I.R).

Resistor voltage "amplifier". So insert a constant resistor Rc in the place of the ammeter (or, a clever trick, increase the ammeter resistance in its parameters window). 1 k is a very convenient value because it eliminates the need for calculations (another clever trick).

schematic

simulate this circuit

Now you can set the quiscent voltage Vout = 5 V by changing Rin in its parameters window while looking at the voltmeter. The result is interesting - Rin = Rc, i.e. the two resistors form a voltage divider with a ratio of 0.5. So our "resistor amplifier" is actually a voltage divider with controlled ratio.

The transfer curve Vout = f(Rin) of the "resistor voltage amplifier" is the same as of the "resistor current amplifier" above because of the clever 1 k trick - [V] = [mA].[kohm].

R-V resistor amplifier

Transistor voltage amplifier... Now let's improve the transistor current amplifier above by converting the output current into voltage.

... by floating output... As in the resistor amplifier, we connect a 1 k "collector" resistor Rc and take the voltage drop across it as an output voltage Vout. As above, a voltage divider is formed where the top resistor is the same but the bottom resistor is a transistor. If you compare the two arrangements - this and the resistor "amplifier" above, you will see that they are functionally identical. The transistor behaves as a 1 k "resistor" with the same voltage as the voltage across Rin above and the same current through it (still there is some difference and to find out what it is, change Rc in both configurations).

schematic

simulate this circuit

When plotting the transfer characteristic Vout = f(Vin), we encounter a difficulty because the DC sweep simulation cannot directly plot a "floating" voltage such as Vout. Therefore, we ask it to plot the result of the expression Vout = Vcc - Vc.

V-V transistor amplifier floating

... by grounded output. Transistor amplifiers and electronic circuits in general are cascaded devices where the input of the next device is connected to the output of the previous one. As a rule, their inputs and outputs are "single ended" (grounded). However, the output Vout of our amplifier is referenced to Vcc. What do we do?

They asked themselves this question many years ago and solved it elegantly. It is much easier for us now... We can even get to this idea via the CircuitLab trick above (Vout = Vcc - Vc). Let's formulate it:

Instead of the referenced to Vcc voltage VRc, we use its referenced to ground complement Vcc - VRc.

schematic

simulate this circuit

Accordingly, the graph of the transfer curve is complementary to the graph of the stage with referenced to Vcc output voltage. It simply represents the Q collector voltage Vc.

V-V transistor amplifier grounded

This famous configuration has two features:

  • Vout is not the same in voltage value but its change value is the same. An AC amplifier, for example, amplifies only the change in voltage.

  • The direction of change is reversed - when VRc increases in value, Vout decreases. That is why this amplifier stage is called "inverting". Note that Vout has the same polarity as Vin but its direction of change is opposite.

STEP 3: Applying emitter voltage

The transistor amplifier we "invented" above has a grounded emitter; so its input (transistor base) is single-ended (referenced to ground). However, the transistor has the unique ability to have its emitter used as a second input. Thus, something like an "amplifier with a differential input" (the base-emitter junction) is obtained. I have known this for a long time, but only now have I realized why it is possible. I will say it briefly here and we will see it in the experiments below.

The reason the emitter can be used as an input is that the transistor output (its collector-emitter part) behaves as a current source that creates the voltage drop VRc across the collector resistor and, accordingly, its complement Vout referenced to ground. That is why, this current and accordingly, the output voltage, does not depend on the emitter voltage.

Driving the transistor through the base (common-emitter stage)

For starters, let's "lift" the Q's emitter with some constant voltage Vref. Since we chose a quiescent voltage Vc = 5 V, we need to leave enough "room" for the voltage Vce to change (so that the transistor does not saturate). OK, let's set 1 V and raise the input voltage Vin = 0.701 V of the grounded-emitter amplifier above by that much. Thus the difference Vbe between the two voltages remains the same, the same collector current Ic = 5 mA flows through the resistor Rc and accordingly, the same output voltage Vout = 5 V appears.

schematic

simulate this circuit

As we expected, the graph of the transfer curve is almost the same as the graph of the stage with grounded emitter. The only difference is that the transistor saturates earlier - when its collector voltage becomes 1 V (because its emitter is lifted by 1 V). Our conclusion is that there is no difference between the two configurations; only the output voltage range is reduced in the second one.

V-V transistor amplifier lifted

Driving the transistor through the emitter (common-base stage)

But agree that with the same success we can drive the transistor from the emitter. Why not? We just start varying the emitter voltage Vin as an input while keeping the base voltage Vref constant; their difference is the same as above.

schematic

simulate this circuit

Looking at the graph, we can see that:

  • Regarding the gain, the two configurations are the same because Vbe is the same.

  • The output voltage changes in the same direction as the input voltage, i.e. the circuit is non-inverting; the reason is that in the input loop Vref reverses the sign of Vbe.

  • The output voltage cannot go lower than the input voltage because the transistor saturates.

  • The output (collector) current passes through the input source trying to change its voltage in the opposite direction. So a negative feedback will appear if the source has some internal resistance.

V-V transistor amplifier input emitter

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    \$\begingroup\$ Quote: "...the transistor has the unique ability to have its emitter used as a second input. Thus, something like an "amplifier with a differential input" (the base-emitter junction) is obtained." Yes - since the collector current depends on the voltage difference between B and E, this transistor really can work as a differential amplifier - however, with a big disadvantage: The input resistance at the E node is much lower than at the B node. Problem solution: Connect an emitter follower with its high input resistance to the E node of the transistor. Result: The classical diff. amplifier. \$\endgroup\$
    – LvW
    Commented Apr 5, 2023 at 10:14
  • \$\begingroup\$ @LvW, Exactly! This is a good continuation of this story. I realized why putting a voltage source between the emitter and ground in the common-emitter stage does not change the output signal. We change the output circuit - to the existing three elements in series (Vcc, Rc and CE) we add another fourth element (Vref) in series but still the output current and output voltage do not change. The transistor does this "magic" - by reducing its "resistance Rce", it decreases the voltage drop Vce across itself by as much as the "inserted" voltage Vref. Thus both Iout and Vout do not change. \$\endgroup\$ Commented Apr 5, 2023 at 10:53
  • \$\begingroup\$ Just to clarify: By comparing the two circuits - without source and with source in the emitter, I assume that the voltage Vbe does not change because simultaneously with the insertion of Vref, we also change Vb by this much. In the schematics above, initially Vb =0.7 V and Vref = 0 V (grounded emitter). Then, I insert Vref = 1 V but simultaneously increase Vb = 1.7 V. Thus Vbe is always 0.7 V. \$\endgroup\$ Commented Apr 5, 2023 at 11:02
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Answer to 3.): The voltage between base and emitter (Dc with superimposed ac) controls the collector current. Therefore, it is considered as "input voltage". However, I would not use the term "voltage drop" - because it is an externally applied voltage.

Answer to 4.): When the load RL is connected to the collector, the voltage at the collector (vs. ground) is the voltage across the parallel combination (r_out||RL). In this expresson, r_out is the output resistance at the collector node.

Answer to 5.): It is the input resistance in C-E configuration only

  • without consideration of the external base resistors
  • and when there is no ac resistance in the emitter path (which would create signal feedback)
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  • \$\begingroup\$ Didn't understand the 4th answer😓 \$\endgroup\$
    – MSKB
    Commented Oct 20, 2021 at 19:07
  • \$\begingroup\$ Sorry - my reply was at the wrong place. Here it is again: You were asking why the voltage at the collector is considered as output voltage but not the voltage drop across the load resistance, right? And my answer was: Both are identical because the load is connected to the collector. \$\endgroup\$
    – LvW
    Commented Oct 21, 2021 at 8:28

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