# Amplifier Circuits

## Emitter-Follower (a.k.a. Common-Collector) Circuits

The simplest amplifier is a unity-gain amplifier, which is sometimes called a “follower”. You might this is totally useless, but it totally isn’t.

The primary application of follower circuits is to isolate circuits from one another. In various contexts, we’ve seen cases of circuit loading, where attaching a low-impedance load causes so much current to flow that there starts to be a noticeable voltage drop internal to the circuit, owing to it’s non-zero output impedance. Well, a follower uses a transistor (in this case, an NPN BJT) to source current directly from a power source, and insulate the driving voltage from this current. The result is that the output impedance of the circuit upstream of the follower is reduced by a factor of the inherent gain of the transistor ($\beta \approx 60$), effectively isolating circuits from one another.

An NPN follower (left) and a PNP follower (right)

In practice, both NPN and PNP-based followers rely on the fact that, when conducting, transistors set up a fixed voltage drop between the base and the emitter. For this discussion, we’ll examine the NPN-based follower shown above. Provided that the signal input Vin meets the criteria for conduction (i.e. it is above $V_{BE}\approx 0.6V$) the transistor will begin conducting current. This current sets up a voltage as it flows through RE. As more current flows, this voltage rises until Vout is 0.6V below Vin. At this point, if the voltage were to rise any higher, the transistor would stop conducting. If the voltage were to fall, more current flowing into the base would cause more current to flow from collector to emitter to correct the problem. Hence, Vout is held in a stable equilibrium at

$V_{out}=V_{in} - V_{BE}. \,\!$

Typically, in order to ensure that Vin is centered between V + and ground, we use a bias circuit (with a coupling capacitor and a resistor voltage divider). Similarly, we can remove the DC offset on Vout with another coupling capacitor. The result is that Vout is a copy of Vin with unity gain.

## Amplification

An inverting BJT amplifier circuit

A basic BJT amplifier (illustrated in the figure above) depends on setting resistor and capacitor values such that the basic rules of operation for transistors (which were covered under the “Transistors” subject heading) are met. This is fairly straightforward to do, but nonetheless requires some attention. The basic principle is to use the transistor to enforce that the current that flows out the emitter through RE has to be pulled through the collector, and hence, through RC. Since the same current can produce a different voltage, depending on the value of the resistor it flows through, the ratio of RC to RE sets the gain of the amplifier. We’ll go through the process in detail.

### Setting a Bias Current and Gain

Rather than immediately pick resistor values, we should first decide a center voltage for Vout and how much current (IE = IC) we would like flowing through the transistor. This current is called the ”bias current”. If we suppose Vcc is 5V, we might pick Vout=2.5V. Let’s choose RC to be 1k, so that IC = 1.5mA. Now to cause that current to flow, we have to set up the appropriate voltage over RE. Since the gain of this amplifier is going to be g = RC / RE, we will also want to choose an appropriate RE. Let’s set g = 5, so RE is 200Ω.

### Setting a Bias Voltage at the Base

Now, in order to get 1.5mA flowing through RE, the voltage at the emitter must be 0.3V. For a typical VBE of 0.6V, this means that the voltage at the base must be 0.9V. So the job of the biasing circuit in front of the transistor is to center our voltage signal at 0.9V. For Vcc = 5V, this means $\frac{R_2}{R_1+R_2}\approx \frac15$. To avoid pulling excessive current, we’ll choose R1 = 4k and R2 = 1k. We’ll add a decoupling cap at the input of the bias circuit (let’s just say C1 = 1μF), and we’re all set!

At this point, we should make sure that these biasing resistors will have an adequate output impedance to drive the resistors on the other side of the transistor. As mentioned with regard to followers, impedances attached to the base are divided by the $\beta \approx 60$ of the transistor. Hence, our ˜1k output impedance of the bias circuit looks like a ˜20Ω impedance for anything on the other side of the transistor. Since the input impedance over there is ˜100Ω, we should be ok.

We should also check our RC values. Our coupling cap, connected to our biasing resistors, is a high-pass filter with $RC\approx 1k\cdot 1\mu F = 10^{-3}s$. This gives us f3dB = 130Hz. If this is too high for the signals we want to play with, then C1 should be increased.

### Optional: Increasing Gain for AC Signals

We’ve just built a circuit that omits C2, as shown in the diagram above. We have an amplifier with a modest gain of g = 5. We’ve biased our circuit at 0.9V at the base, which sets the emitter voltage at 0.3V. Unfortunately, this means that our maximum voltage swing at the base can only be 0.3V (otherwise VBE drops below 0.6V and our transistor stops conducting). We could increase RE to give us more headroom for a given bias current, but this hurts our gain. And what if we want more gain? Decreasing RE gives us more gain, but it’s not much use if we have to input a smaller signal at the base, as we would need to do because our emitter voltage is perilously close to ground.

The clever solution is to realize that we could chose a different gain for different frequencies. A capacitor to ground at the emitter (in parallel with RE) creates an impedance that decreases with frequency, allowing a higher frequency signal to see an impedance at the emitter that is much lower than RE. This allows us to separately choose our bias current (which sets the center voltage of Vout and determines the range of outputs allowed), our bias voltage (which sets the base voltage, and determines the range of inputs allowed), and our gain. Cool.