Transistor Amplifier Design for Beginners.

A good starting point is the single transistor amplifier stage shown next. Although gain, noise, distortion, supply rejection, output impedance and so on are relatively poor these can be improved by additions and modifications once the basic operation is understood.

The first topic is to see how the DC and AC behaviour are analysed separately. The AC input signal is applied to the transistor base via a capacitor, so if we start with a zero signal we can ignore the input and just look at the DC voltages in the circuit.

What do we mean by a 'DC voltage'? That may already be a source of confusion, DC is short for 'direct current', so what could a 'direct current voltage' mean? This sounds wrong, but it is such a widely used convention that we really just have to accept that it means constant voltages as opposed to varying voltages, which we call 'AC voltages' even though 'alternating current voltages' sounds just as wrong. It would appear to be simpler to just talk about 'voltage', after all there is only one sort of voltage, and for example V3 in the above circuit will be more or less fixed, maybe 9V, with no input signal, and will be 9V plus some amplified signal when we apply an input. It is often easier however to keep DC and AC analysis separate. If we looked at the input V1 and the collector voltage V3 on a dual trace direct coupled oscilloscope then we would see the actual voltages including both DC and AC components as shown next:

Note that the input used has been chosen to have no DC component, it just alternates about zero, but V3 has a DC component of 9V to which is added the amplified AC component. Note that the AC gain of this circuit is about -3, the negative gain just means that the AC signal component is inverted. A more positive input signal makes the transistor conduct more, so more current through the 3k resistor makes the collector voltage fall relative to the supply voltage. (The more astute reader will perhaps realise that the signal levels shown are not actually possible with the circuit as shown, the positive peaks of the output signal are above the supply voltage, but I didn't want to redraw it.)

The next diagram is included to explain why the base and emitter voltages are the values shown in the circuit. The starting point could, for example, be to decide to use an emitter current of 1mA and an emitter voltage of 1V. The current chosen will depend on how much current we need to drive any following stages or load, but not too high otherwise the transistor could get hot. The emitter voltage needs to be high enough to give a predictable current in the presence of thermal drift effects, but not too high because it uses up some of the available supply voltage, and it may be important to maximise available output voltage swing. From Ohm's Law our 1V and 1mA voltage and current tell us the resistor value needed is 1k. The base voltage is higher than the emitter voltage by Vbe, which is a function of collector current, and the next diagram is for a typical small signal transistor, showing that at Ic = 1mA Vbe is close to 0.6V, giving base voltage 1.6V.

The value of Vbe for a given Ic is a function of junction temperature, falling about 2mV for every degree increase, and is also affected by collector voltage, so we need to avoid making our circuits too dependent on an exact value of Vbe.

Next we need to choose the base bias resistors, shown as 16k and 100k. We know there is 1.6V across the lower resistor, so to find its value we need first to choose how much current is to flow through it. Again we don't want it too low or too high. With low current the error added by the transistor base current could upset our calculations The base current is the collector current divided by the current gain, which can vary over a big range for a chosen transistor type, but will be taken to be 250 for the present example. This gives base current in our example of 4uA (that's micro-amps, it should be the Greek letter mu rather than u, but that is easier to type.) Using a high current through the resistors would give a low value resistor and increase the current input required from the signal source, but our choice of 0.1mA current and 1.6V gives a reasonable resistor value 16k.

The upper resistor has voltage 12V - 1.6V = 10.4V, but its current included both the 0.1mA (100uA) through the 16k and also the transistor base current, about 4uA, making a total of 104uA. Again Ohm's Law tells us that for 10.4V to give a current of 104uA we need a resistor value of 100k.I have of course carefully chosen this example to give standard resistor values, a different current gain transistor would lead to a less convenient resistor value, and then we need to check whether a standard value is near enough, or if we need to use a series or parallel pair of resistors to get closer to the theoretical requirement.

Finally we have the 3k collector load resistor, and this has been chosen to drop 3V with its 1mA current, giving a collector voltage 9V. The collector current is actually slightly less than 1mA because the 1mA emitter current is the sum of base current and collector current, so a more accurate figure for Ic is 996uA, but this is close enough to 1mA given that we will anyway be using resistors with 1% tolerance, and there will be other errors including thermal effects, so there is no point trying to be too exact. Unless a very low gain transistor is being used it is usually accurate enough to assume the same values for emitter and collector currents.

AC analysis

When we add the AC input V1 the current through the transistor will be modulated by this signal, and from the graph of Vbe against Ic we could find the corresponding variation in Vbe. Even for Ic varying from 0.5mA to 1.5mA Vbe only varies from about 590mV to 610mV, but the voltage across the 1k emitter resister varies from 500mV to 1.5V, so the Vbe variations are relatively small, and for a first approximation we can regard Vbe as a constant 600mV. What this means is that for AC input V1 the base voltage is 1.6V + V1 and the emitter voltage is 1V + V1. The AC current through the transistor emitter is then V1/1k, and for a high gain transistor the collector current is almost identical, but the higher 3k collector load with the same current will drop 3 times the voltage, so the AC output will be 3V1. Taking into account the inversion mentioned earlier the output is -3V1, so the voltage gain is -3V1/V1 = -3.

The approximation of assuming Vbe to be constant will, in this example, give an error of about 2.6% in the voltage gain, but with a smaller emitter load resistor than the 1k the error is greater. A more exact calculation is fairly easy, and to see how this is done look at the next example where the 1k is reduced to zero. The DC bias is set by a variable resistor, and this would need a very fine adjustment to set the current through the transistor close to 1mA, so this is not an ideal circuit to use in practice. The problem is that a change in base voltage Vb is in effect a change in Vbe because the emitter is at 0V, and a small change in Vbe can produce a big change in collector current. With the previous 1k emitter resistor most of the change in Vb appears as a change in the voltage across the 1k, and this may be small as a proportion of the 1V across the resistor, so the current change is also small.

Now we can no longer assume Vbe is constant, it will include the AC input voltage added to the DC level of 600mV. A positive change in Vbe of 1mV will increase the collector current, but a 1mV negative change will not reduce the current by exactly the same amount, this is a result of the nonlinear relationship between collector current and Vbe. The percentage difference between the positive and negative effects will however be low for small input signals, so it is common to ignore the nonlinearity for 'small signal analysis'. 1mV is just about small enough to justify treating the circuit as linear for our present purpose. (The harmonic distortion for a 1mV peak sinewave input will be about 1% for this circuit, which may be too much for some applications. The percentage distortion is numerically equal to the signal level in mV to a good approximation, so for example 0.1mV peak input gives 0.1% distortion.)

The equation relating collector current to base-emitter voltage is:

Ic = Is exp ( q Vbe / kT) Where q is the charge of an electron, k is Boltzmann's Constant (1.38 x 10^{-23}joule deg^{-1}), T is absolute temperature. Is is a current which varies for different transistor types, it is small for low power transistors, and usually bigger for high power types. The equation is often simplified to:

Ic = Is exp ( Vbe / Vt ) where Vt = kT/q is a voltage which is a function of temperature, and is about 26mV at T = 293 deg.K ( that is 20 deg.C, which here in the UK is considered to be typical 'room temperature'.) 'Is' is rarely if ever specified, once we have a figure for Vbe at some specified collector current and temperature we could work out Is, but it is usually not very useful to know the value.What often interests us is a quantity known as the 'mutual conductance', gm. The 'g' confusingly refers to 'conductance' (C is already used for capacitance) , and the 'm' more obviously refers to 'mutual'. This is also sometimes called 'transconductance', but then it is even less obvious why we call it gm.

Anyway, gm is the ratio of the change in collector current to the change in base-emitter voltage causing that change. In other words, gm = dIc / dVbe, and is found by differentiating the earlier equation for Ic, the result being:

gm = Ic / Vt , and if Ic = 1mA then gm= 1/26 (amps per volt). This is the conductance of a 26 Ohm resistor, and that leads to a useful simplification. Suppose we had an ideal transistor for which Vbe was really constant at 600mV, but it included a 26 Ohm resistor in series with the emitter, this would then have the same gm as our real transistor operating at 1mA. It would however be perfectly linear, so it is only useful as a small signal approximation. To see how we can use this equivalent circuit redraw the circuit with the 26R included:

We should deduct the voltage across the 26R from the 600mV fixed value of Vbe, but we really only need to use this equivalent circuit for AC analysis, and we can usually ignore any effect on DC levels.

When I first saw this approach explained, in an article in Wireless World many years ago, it was a revelation, suddenly I saw how easy it could be to analyse circuits. The voltage gain for the above example is just the 3k collector load divided by the 26R 'emitter resistor', so we have a voltage gain of -115. We can also work out a more exact figure for our previous circuit with a 1k external resistor. We had asumed a constant Vbe, but including the 26R in series with the 1k accounts for the variations in Vbe. The gain now becomes - 3000 / (1000 + 26) = -2.924, slightly lower than our first approximation of -3.We know that the AC signal across the total emitter resistance is identical to the input signal, but with current gain 250 the input base current is 250 times smaller, so we can deduce that with the same voltage but 250 times lower current the impedance is 250 times higher than the emitter resistance. For the circuit with no external emitter resistor the base input impedance is 250 x 26R = 6500R = 6k5. The base biasing resistors reduce the impedance further, so the total circuit input impedance is about 3k2. With the 1k external emitter resistor the base input impedance is 250 x 1026R = 256.5k, but again the biasing resistors reduce the circuit input impedance.

Remember the 26R is only correct at 1mA collector current, it is inversely proportional to Ic, so at 100uA it is 260R, and at 10mA it is 2R6. The value calculated becomes increasingly inaccurate at higher currents. Real transistors are more complex than our simple model, but we can add a few more components for greater accuracy. There is a small internal resistance in series with the base, called the 'base spreading resistance' (rbb') which can vary over a wide range, for a 2SC2547 it is around 2R, for a MPSA18 about 800R. It varies slightly with collector current, so the specification should mention the current. Less often specified is a small 'fixed' resistance (ree') in series with the emitter, often a fraction of an ohm, and this makes our earlier equation for gm less accurate at higher collector currents. Adding these components we arrive at the next diagram.

We are not finished yet, there is at least one more resistor to add, and this determines the output impedance of the transistor. The equation for Ic given earlier assumes all the current passes through the load, but in reality the transistor is not a perfect current source, it has a finite internal impedance. So is there something in a typical data sheet to tell us the value? If we are lucky the h parameters are given, hfe, hie, hre and hoe. If we are very lucky there is a graph showing their values as a function of collector current, then we have some useful, but by no means complete, information. An example can be seen in the Toshiba data for the 2SC4117.

The BC560B (pnp) however has no such data given by either ON-Semi or Fairchild, and I had to go back to an ancient Motorola data book to find the 'typical' values given at Ic = 2mA and Vce = 5V:

hre = 3.5 x 10^{-4}

hfe = 330

hoe = 10 uS

hie = 6k (wrong? Should be 4k3 at 2mA if hfe = 330)The next diagram gives some idea what these mean, hie is the input impedance, hfe is the current gain, hoe is the output conductance ( but only in a common-emitter circuit ), and for the BC560B is typically 10uS. The impedance is the inverse of the conductance, which is 100k, so we can add a 100k resistor from collector to the earthed emitter.

I have not explained hre, which has some small effects, including making the output impedance a function of source impedance. Fortunately this is a problem we rarely if ever need to worry about, if we just take output impedance Zo = 1/hoe, which for the BC560B gives Zo = 100k, then that is good enough for most purposes. The values of hoe and hfe can vary over a wide range for a given type of transistor, which is why we try to avoid designing circuits for which the exact characteristics of the transistors need to be known. For our simple amplifier circuit with a 3k collector load if we assume a 100k output impedance is in parallel this reduces the effective load from 3k to 2.91k, so the effect is to reduce the amplifier gain a little, in this case by about 3%.

If we operate a transistor at constant Ic and apply an AC signal to the collector, then we find there is a small AC signal added to Vbe. In other words Vbe is modulated by the collector voltage. I measured this for a few typical small-signal transistors on my Common Mode Distortion page, where for the BC560B I found 320uV modulation of Vbe for 1V at 2kHz added to Vce. The modulation predicted by hre = 3.5 x 10

^{-4}is 350uV, so my result is in close agreement with the 'typical' value. This may however be just a coincidence, what I measured was emitter voltage with earthed base, but hre should be measured as base voltage with earthed emitter. I am fairly certain the two effects have different causes and would be entirely different values if I had compared them at other collector currents.The common emitter stage is rarely used in practice in the exact form shown in the first diagram on this page. If it is used we often want a different voltage gain to that determined by the DC bias components, which in our example gave a fixed gain of -3. Suppose we need more gain, to achieve a much higher gain we can bypass the 1k emitter resistor with a capacitor C2.

At high frequencies C2 has a very low impedance, so the gain is just the ratio of 3k load to internal emitter resistance 1/gm, which at Ic = 1mA is 26R. The ratio is 3000/26 = 115. The circuit inverts the signal, so we have negative gain, -115.

At low frequencies C2 impedance increases and will reduce the gain. I need to write something about capacitance and phase angles, but for now I will just state that the gain reduces to -115 multiplied by 0.707 when the impedance of C2 is equal in magnitude to the internal resistance 1/gm. This is the frequency at which gain falls by 3dB. The impedance of C is given by Zc = 1 / 2pi Cf where f is frequency. For 1/gm = 26R and for example a -3dB frequency of 20Hz we need a capacitor value of 306uF, and in practice would use the nearest available value, 330uF. This would have to be an electrolytic capacitor, and these have typically a tolerance of +/- 20% or more, so this is not a good way to determine the frequency response. Better is to use a much larger capacitance, maybe 1000uF or more, and use C1 to determine the low frequency response. This capacitor will be something like 1uF, and a closer tolerance type such as polyester or polypropylene can be used.

For intermediate gains we could add a resistor in series with C2, for example adding 26R would give half the gain, about -57, and we could also reduce C2 to half its original value. Eventually if we increase the series resistor further the 1k emitter bias resistor starts to have a significant effect, and prevents the gain being reduced further than -3. If we want less than -3 gain we could also reduce the 3k collector load resistor.