Does changing the collector resistance of a common base amplifier have any effect on the current?

Question

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, I_c and I_e 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 R_c 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 V_cb.

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

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.

Answer 1

If you concentrate on the V_BB, R_E 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 R_C and the load resistance.

I think your intuition is telling you that current source output would vary as R_C 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 R_C becomes so large that the output current saturates the transistor, the output current will fall, as it's now controlled by R_C, not the transistor.

Answer 2

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)).

Answer 3

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.

^{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.

^{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.

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.

^{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).

^{simulate this circuit}

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

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").

^{simulate this circuit}

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

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).

^{simulate this circuit}

"Shifting" the output voltage down

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

^{simulate this circuit}

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

Final amplifier circuit

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

^{simulate this circuit}

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

^{simulate this circuit}

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

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.