In general, problems in science tend to have just one ‘right’ answer. Electronics isn't always like that. For example, there are an enormous variety of devices & circuits we can use as signal amplifiers. None of them is the ‘best’ in an absolute sense — although hi-fi manufacturers would often like you to believe otherwise! Each amplifier design will have its own combination of good/bad points. Here we'll look at the simple amplifier illustrated in figure 7.4. This arrangement is called a common-source amplifier because the source connection is common to (i.e. used by) both input and output. We can apply a varying input signal voltage, Vin , to the gate of an N-channel J-FET & take an amplified (we hope!) output signal, Vout , from the FET's drain. This circuit isn't, in fact, a very good one but we can use this to see how amplifier's work. Remember that there are lots of other (usually more complicated!) circuits which do a much better job!

One end of a resistor, R, is connected to a fixed (d.c.) positive voltage, Vr . The other is connected to the drain of the FET. The FET's source is connected to ground (zero volts). This means that the resistor & the FET's channel are in series. The current through the resistor must therefore equal the drain-source current, Ids. So the potential difference between the ends of the resistor must be IdsR. Since one end of the resistor is fixed at +Vr volts we can say that its other end (connected to the FET) will be at a voltage

From what we've learned we know that — provided is bigger than about 2 volts — the current, Ids , is determined by the voltage applied to the gate, Vin. This means we can use Vin to alter Ids, changing Vout. To see what happens, consider a specific example.

Let's choose the values, R = 470 Ohms and Vr = +15 Volts. The input signal, Vin, is initially -1 Volts. This means that Ids is initially equal to 18 mA. Using the above equation we can say that this means that Vout = 15 - 0·018 × 470 = 6·54 Volts. The input signal level now changes to -2 Volts. This changes Ids to 6 mA which means that, now, Vout = 15 - 0·006 × 470 = 12·18 Volts.

Notice two things about this. Firstly, changing the input voltage by 1 Volt has caused the output voltage to change by 12·18 - 6·54 = 5·64 Volts. Secondly, the input changed by -1 Volt (from -1 to -2) but the output changed by +5·64 Volts — i.e. the input went ‘down’ and the output went ‘up’. The circuit inverts the signal while making it 5·64 times bigger. The amplifier is said an inverting amplifier with a voltage gain of -5·64.

Having considered this specific result we can now work out the general behaviour of the FET amplifier. What happened was that we changed the input (gate) voltage by an amount,

This changed the FET's drain-source current by an amount,

If you look at books or data sheets on FET's you'll find that one of the transistor parameters listed is the FET's transconductance. This is the ratio of the change in current to the channge in voltage which produced it. Its value varies from transistor to transistor.
For our transistor the transconductance value is
g = (-0·012)/(-1) = 0·012 Ohms-1

Note the units of this ratio. Since we're dividing a current in Amps by a voltage in Volts the result must be in the inverse of Ohms. In many books and data sheets you'll see transconductances quoted in mhos or mmhos. It should be fairly obvious why the term “mho” is used to mean an inverse ohm! The term “mmhos” means milli-mhos. You will also see transconductances quoted in units of Siemens. This is the international scientific unit of one inverse ohm. (As usual, scientists like to name everything after themselves. Engineers tend to prefer mhos since it's a more obvious word & easier to remember!)

The equation we used earlier tells us the amplifier's output voltage when a specific current is passing through the FET. By using the transconductance value (which comes from the slope of the Vgs versus Ids graph) we can define a value for the amplifier's voltage gain, Av

The result confirms the numerical example we considered earlier. The gain value is negative, indicating that the signal voltage's pattern is inverted as it is amplified. Looking at this result we can also see that the FET amplifier's voltage gain depends upon the transistor's transconductance, g, and the value of the source resistance, R, we chose to use. From the curves shown in the above figures we can also see that the g value depends upon the voltages and currents we pick. (The plots of currents versus voltage are curved, not straight lines.) This means we always have to note what voltages and currents a given transconductance was calculated for.

Summary.

You should now know how an N-channel Junction Field Effect Transistor (J-FET) is made and works. How we can use it to control passing though a channel by altering a gate voltage. That the drain-source current along the channel is almost independent of the drain-source voltage provided that voltage is kept above a few volts. You should also know that this type of FET is just one example of a transistor. That there are many other types, each working in a slightly different way, but providing the same ability to control one electrical signal with another.

You should now understand how we can combine a transistor and a resistor to make an amplifier. That the voltage gain of a common-source FET amplifier depends upon the transistor's transconductance and the chosen value of the drain resistor. That transconductance is a ratio of current/voltage and can be specified in units of mhos or Siemens. That this type of amplifier can be used to enlarge input signal patterns, but inverts the signal. It is therefore an example of an inverting amplifier and it's gain has a negative sign.

Content and pages maintained by: Jim Lesurf (jcgl@st-and.ac.uk)
using HTMLEdit2 on a StrongARM powered RISCOS machine.
University of St. Andrews, St Andrews, Fife KY16 9SS, Scotland.