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I have come across this equation for understanding the voltage gain of a common base amplifier: $$ A_v = \frac{I_c R_c }{I_e R_e} $$ Because base current is small, Ic and Ie are roughly equal. Therefore, the gain is essentially just a ratio of the resistances on either end of the transistor.

What I'm struggling to understand is how changing Rc in the diagram below would affect the overall circuit. My understanding is that current between the emitter-collector results from electrons leaking into base-collector depletion zone which has an electric field which ultimately pushes the electron to the collector (for NPN BJTs). From this point, the electron will naturally flow to Vcb.

So, as I understand it: the collector current is dependent only on the emitter current and this depletion zone crossover, but independent of Rc. I am skeptical of this however. that significantly increasing Rc would not have an effect on the amount of current flowing, but I am having a hard time working through this.

Common Base Amplifier using an NPN Transistor from Electronics Tutorials

Image source: Electronics Tutorials - Common Base Amplifier using an NPN Transistor

I am half-hoping that my attempt at explaining this may illuminate a point of confusion I'm going through to someone more knowledgable.

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  • \$\begingroup\$ Note that's \$r_e\$ in the relation, not capital, as the input voltage is defined to be \$V_{IN}\$. In general, a resistive (or other impedance) source is more practical for this circuit (and transresistance the more meaningful expression of gain). The diagram could also be better labeled, as Vcb is usually understood as the measurement at the transistor itself, not the supply to it (which is traditionally the repeated letter, i..e. Vcc for collector supply, Vee for emitter supply). \$\endgroup\$ Commented Jun 8, 2023 at 4:40

3 Answers 3

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If you concentrate on the VBB, RE part of the circuit, then the transistor behaves as a current source into the collector load.

For an ideal current source, its output resistance is infinite, so the voltage gain is controlled entirely by parallel combination of RC and the load resistance.

I think your intuition is telling you that current source output would vary as RC varies, and you would be right, it doesn't stay exactly constant.

It stays fairly constant. There is small feedback from the collector voltage to the collector current, which can be described as a finite output impedance to the current source. It's due mainly to the Early effect. The variation is down in the one or a few percent region for most transistors, so negligible in all but the most precision work.

In the limiting case, when RC becomes so large that the output current saturates the transistor, the output current will fall, as it's now controlled by RC, not the transistor.

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  • \$\begingroup\$ Nice. This is what I'd wanted to say, but hadn't taken the time. Better than the other answers, so far! +1 \$\endgroup\$ Commented Jun 8, 2023 at 7:48
  • \$\begingroup\$ Thank you! Is there a name for the point at which the limiting case mentioned in the last paragraph occurs? And I believe you mean Vbe (not Vbb) in the first paragraph, is that right? \$\endgroup\$
    – shafe
    Commented Jun 15, 2023 at 3:03
  • \$\begingroup\$ @shafe I mean the voltage that you have confusingly labelled VBE, but that every other engineer in the universe would call VBB. If you want to avoid confusing people, use the appropriate terms. VBE is the voltage between base and emitter terminals. VBB is the battery voltage, the sum of VBE and Vre. The name for that point is 'the limit of output compliance', though 'saturation' would also be well understood. \$\endgroup\$
    – Neil_UK
    Commented Jun 15, 2023 at 5:25
  • \$\begingroup\$ @Neil_UK indeed! Thanks for pointing that out. I think I understand - this is my attempt to try and say it back to you: Vbb and Vre will dictate the electron flow from emitter to base. The current through collector (assuming an appropriate amount of output impedance) is mostly just a function of the base current and the ratio Beta (which is a property unique to the specific transistor). The voltage at Vc will thus be: (Ic * Rc) + Vcc (because from there it is connected to ground) \$\endgroup\$
    – shafe
    Commented Jun 15, 2023 at 23:24
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    \$\begingroup\$ @shafe with the caveat that I am over-simplifying, the best way to think about this is - VBB dictates the voltage across RE, as VBE is essentially constant. The fixed voltage across RE dictates the current through RE. Almost all of this current flows to the collector. The difference between a transistor beta of 100 and 300 is the difference between 99.0% and 99.7% of the emitter current flowing to the collector - so really beta has nothing to do it with it, as long as it is 'big'. And please, don't think about electrons, not in circuit theory anyway, semiconductor physics yes, circuits, no. \$\endgroup\$
    – Neil_UK
    Commented Jun 16, 2023 at 7:58
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You are correct. Unless the transistor is in saturation, IC is nearly equal to IE. Therefore (nearly) the same current flows in RC and RE.

So for a change in the base voltage deltaVB, a change in emitter current of deltaIE=deltaVB/RE occurs. This change also occurs in RC leading to a change in voltage across that R of deltaVC = deltaIE * RC = deltaVB * RC/RE. So the gain is RC/RE -- as in your equation above.

Note this is voltage gain. The current gain is a little less than 1 (actually beta/(1+beta)).

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Short answer

As shown in the OP's schematic, the input voltage Vin is directly applied to the transistor base-emitter junction, i.e. this is a classic common-base configuration in which the collector current is controlled by the input base-emitter voltage.

Functionally, it is a current source (sink) that passes its current through the resistor Rc. Therefore, the current is determined only by the current source (the transistor) and does not depend on the load (Rc). Only the floating voltage across Rc and, accordingly, its complement Vout referenced to ground depend on Rc.

The emitter current has nothing to do in the configuration shown. A part of it flows through the input voltage source and not entirely through the resistor Re.

Indeed, this is a simplified schematic, but important components are missing to make it workable.

Expanded answer

I had no intention of expanding my answer, but after reviewing the link cited by OP, I realized that I was not going to get away with "inventing" this famous circuit using CircuitLab simulations... simply because clarity needs to be brought to this mess. I will accompany my schematics with short explanations which, if there is interest, I will expand later.

Setting the initial output current

For some reason, amplifier elements (transistors) are transconductance devices - their input is voltage and their output is current. In order for the output quantity (current and voltage later) to change in both directions, initially it (or, as they say, operating point) must be set in the middle.

For this purpose, we apply a reference voltage Vref to an emitter follower Q and connect a constant resistor Re in its emitter. By adjusting its collector (emitter) current Ic = Ve/Re through Re, the transistor makes its emitter voltage equal to the reference voltage so Ic = Vref/Re.

schematic

simulate this circuit – Schematic created using CircuitLab

Setting the initial output voltage

But our amplifiers produce output voltage; so we have to convert the collector current into a voltage. For this purpose, we connect another (collector) resistor Rc. The current Ic flowing through it creates a voltage drop VRc. Thus, the two voltage drops are connected through the common current as with some "electrical transmission" and VRc/VRe = Rc/Re. If we consider VRe = Vref as an input voltage and VRc as an output, Rc/Re = 5 is the gain of this "common-emitter amplifier with emitter generation". Note something very important - the input voltage is not directly applied to Re but indirectly by the mechanism of the negative feedback.

schematic

simulate this circuit

Testing the current source

We will now experimentally answer the OP's question whether changes in Rc resistance cause changes in the collector current Ic. Note that, in contrast to the cited schematcs, there is no input voltage source Vin connected to the emitter.

After running the DC sweep simulation in the schematic above, we see that while the transistor is in active mode, practically Ic does not depend on Rc.

STEP 4

Applying a DC input voltage to the emitter

Our circuit, as they say, is properly biased and we can now apply an input voltage to the base-emitter junction of the transistor (more precisely speaking, we have to add the input voltage to the bias voltage). We can do it in two ways - either from the base side (common-emitter stage) or, our case, from the emitter side (common-base stage).

To understand the input biasing mechanism, we apply the following trick. First we measure the voltage drop across Re and adjust the voltage of a DC input source Vin equal to it. Then we connect the source in parallel to Re and see that there is no current flows and nothing changes in the circuit.

schematic

simulate this circuit

Inserting an AC input voltage source

Now we connect the AC input voltage source in series to the "shifting" DC source. Thus the input voltage variations around ground will be "lifted" to the emitter (around the quiescent 1 V).

schematic

simulate this circuit

After running the time-domain simulation, you will see that the gain (Vout/Vin) is more than 100.

STEP 4

Replacing the source by a capacitor

It is not convenient to use a "floating" DC voltage source; so let's replace it with a capacitor Ce (also, for simplicity, let's initially replace the zero-voltage source with a "piece of wire"). When the power is turned on, it quickly charges through the AC input source (which must be "galvanic").

schematic

simulate this circuit

The voltage across the capacitor tends to and eventually reaches the quiescent emitter voltage of 1 V.

STEP 5

Then the capacitor alternately discharges and charges slightly with the frequency of the input signal, with the average value of its voltage remaining relatively constant (see the simulation).

schematic

simulate this circuit

STEP 6

"Shifting" the output voltage down

Similarly, with a 5 V "battery" we can "shift" the output voltage to ground so that it alternates...

schematic

simulate this circuit

... (descends below and rises above the ground).

STEP 7

Final amplifier circuit

Again, we replace the DC voltage source with the more convenient capacitor Cc...

schematic

simulate this circuit

... which is initially charged through the ("galvanic") load RL to the initial quiescent voltage of 5 V.

STEP 9

schematic

simulate this circuit

So the voltage variations "up" around 5 V are moved down around 0 V (ground).

STEP 7

Conclusions

  • The AC common-base amplifier stage has small DC gain (5 in the example) and high AC gain (> 100).

  • With respect to the load Rc, it behaves as a constant current source.

See my previous and last answer to related questions.

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    \$\begingroup\$ “Only the floating voltage across Rc and, accordingly, its complement Vout referenced to ground depend on Rc.” Thank you! It was very difficult for me to wrap my head around this \$\endgroup\$
    – shafe
    Commented Jun 15, 2023 at 23:28

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