Designing with Transistors - Chapter 3

3.0   OVERVIEW

In the previous chapter we saw how the characteristics of the transistor must control our circuit design. Now that the parameters of the transistor are more fully understood we can get down to the real business of choosing the components so that the transistor behaves just the way we want it to.
An example of bad design will be given so that the impatient beginner does not fall into the more obvious traps. As well as choosing components, there is also the supply rail voltage to be defined. This is often of major importance to successful circuits.

3.1.   RELEVANT EQUATIONS TO VOLTAGE AMPLIFYING STAGE

There are four equations which must be considered as absolutely essential to the design of a simple amplifying stage. These are shown in Fig.3.1 in a way that should immediately suggest their use.

These equations should be copied onto paper and chanted over and over in the mind before going to sleep.

Note how the variables interact with one another The choice of collector current will affect re which in turn will affect the voltage gain A and the input resistance of the stage R_IN

3.2.  CHOICE OF COLLECTOR CURRENT AND RAIL VOLTAGE

With any particular transistor there are obvious limits at both ends of the collector current scale.
The limit at the lower end is due to the reduced h_FE and also to the encroachment of the base current onto the leakage area. For small silicon transistors similar to the BC108, the lower limit may be taken as around ten microamps.

UPPER LIMIT

At the other end of the scale, there is the absolute limit imposed by the manufacturers known as I_c(max), which. again quoting the BC108, is 200mA. There is a hidden danger in this figure however, unless the maximum power dissipation (Pmax) is also examined.

The BC108 claims 300mW for P(max) therefore if the designer does attempt to use 200mA collector current, he had better make sure that he never allows the collector voltage to be more than one volt, or the poor thing will protest warmly.
Except for pulse circuitry, where large currents can be tolerated for very brief times, the practical maximum for amplifier stages will be lower than Ic(max). Ten milliamps is the recommended maximum for the BC108; beyond this, the h_FE starts to fall again.

CHOOSING THE COLLECTOR CURRENT

Having established the range as 10uA to 10mA, there is still the particular value to be decided upon when designing an amplifying stage.

Other things being equal, a choice of low Ic will:

• Enable the stage input resistance to be higher because R_IN is the parallel combination of the base bias resistors R1, R2 and the transistor input resistance r_IN.
  Rl, R2 will be higher because the base current will be small. Also r_IN will be higher because r_IN = h_fe (re + R_E and both re and R_E can be high if Ic is low.
• Cause the output resistance Rout of the stage to be high because Rc will tend to be high. This is not good, as a low output resistance is desirable to drive the output load.

SUPPLY RAIL VOLTAGE

The higher the supply voltage, the easier the circuit is to design. There is a greater freedom of choice more breathing space, since voltage can always be dropped but not so easily increased. The limit is, of course, V_CE(max), which, in the case of the BC108, is around 20 volts. However we are often forced into a choice by virtue of the available supply ie a 9V or 12V battery.

3.3.  DESIGN PITFALLS

The beginner to the art of design is soon face to face with a frustrating problem: how to meet the conflicting demands of satisfying both the signal and the d.c. bias requirements.

To illustrate this let us assume a simple voltage amplifier is to be designed having a gain of 100. This is a harmless specification, with no mention of input resistances, supply rail voltages, etc. and the beginner might be excused for underestimating the job.

His reasoning might proceed as follows:

• Look up the equation for voltage gain A =Rc/(re + R_E)
• Note that R_E appears a nuisance so why not leave it out altogether?
  (This will save a resistor and make the equation easier)
• Look up the equation for re:
  re = 25/Ic(mA) and note that choosing a nice figure of 1mA will make re = 25Ω.
  Rc can then be found by Rc = 1000 * 25 = 2k5Ω.
• The circuit may now be drawn with as much data as possible included (see Fig. 3.2). The 3V rail is handy because the small family torch has a 3V battery.
• Look up h_FE of a BC108 which will be 300 (typically) which means that the base current will be Ic/300 = 1mA/300 = 0.003mA (approx).
  Since the divider chain should carry a much larger current than the base, let this be 0.03mA.
  Allowing the regulation 0.6V across R2, R2 = 0.6V/0.03mA = 20kΩ.
• Rl must drop the remaining 2.4V therefore R1 =2.4V/0.03mA = 80kΩ.

The paper work is now finished and the designer can now rig up the circuit and try it out. What a bitter disappointment he is in for! Voltage checks will soon reveal that:

• The 0.6V may be disturbingly out, and may drift downwards slowly.
• The transistor is almost or even completely saturated. The output terminal has no room to move since it is grovelling at almost ground voltage.

In short, it is a shocking example of bad design.

3.4.   THE IMPORTANCE OF R_E

A familiar component in most small signal amplifiers is the resistor R_E connecled between emitter and ground. The primary reason for its presence is to provide a simple self-adjusting mechanism for forward bias on the base.
It may be remembered that the transistor will only amplify correctly if the base voltage is held at mean value of about 0.6V higher than its emitter (at normal temperature).
Not only is it difficult to tap off such a precise small voltage from a divider chain, there is also the added complication of the change in this voltage when the temperature changes (remember that VBE falls by about 2mV per °C rise in temperature. The inclusion of R_E will compensate for these effects (see Fig. 3.3).

Suppose R1, R2 values were a little out due to tolerance errors resulting in a voltage oF say 1.5V instead of 1.6V.
This would produce only 0.5V across base and emitter if the collector current remained the same, but of course it will not. The collector current must fall and the 1V drop across R_E must then fall to say 0.9V. This means that the voltage difference between base and emitter would be back again to 0.6V.
The system is thus almost completely self-regulating if the temperature rises, more current would tend to flow which would increase the 1V across R_E to say 1.1V and that would reduce the forward bias in the base back to 0.5V which holds the current back again.
With regard to the design equation, how big should R_E be?
Compromise as usual!  The higher it is the better the stabilization action, but the more volts are 'wasted' across R_E.
A good rule of thumb is: Make R_E at least five, but preferably ten times greater than re

3.5.   THE EFFECT OF R_E ON VOLTAGE GAIN AND DISTORTION

Apart from stabilizing the d.c. bias conditions, R_E causes the voltage gain to fall.
This can be understood by re-examining the equation for voltage gain A = RC/(re + R_E)
A more convincing way of understanding the effect is to consider the effect of a signal applied between base and ground (see Fig. 3.4).

Consider first the action with the switch closed, i.e. R_E not in circuit.
If a positive-going signal of say 2mV is applied, the whole of the signal is across the base to emitter junction.
With the switch open, the 2mV signal between base and ground would cause an increase of collector current (as before) but the volts drop across R_E would cause the emitter to rise as well as the signal. Thus the effective value of the signal between base and emitter would be less than in the case above.
For example, if the base signal rises by 2mV and the emitter voltage also rises to say l.5mV, then the effective signal between base and emitter is only 0.5mV
From the point of view of voltage gain, it appears that R_E has in this case, reduced the effectiveness of the signal by a factor of four which means the gain is reduced by a factor of four.

This gain reduction caused by a signal voltage, appearing across R_E is an example of negative current feedback.

ADVANTAGES OF NEGATIVE FEEDBACK

Although the gain is reduced, there are certain advantages in this kind of feedback:

• The input resistance r_IN is increased, since r_IN = h_fe(re + R_E).
• The distortion, inevitable to some degree in all transistors, is much less. In fact the distortion is reduced by the same factor as the gain is reduced.
  In the above example distortion would be reduced by four times by the inclusion of R_E.
• The reduced gain now tends to be less dependent on the actual transistor causing the circuit performance to be determined without worrying about tolerance variations in the h-parameters.
  If a high gain is more important than these advantages R_E can be "by-passed" with a capacitor which will provide a signal short to ground.

The capacitor's reactance Xc, should be smaller than R_E at the lowest frequency, a rough and ready equation being C = l/(f * R_E) where C is in microfards, f in kHz and R_E in kilohms.

3.6   SETTING THE DC OUTPUT VOLTAGE

The output terminal of the stage should wherever practical, rest somewhere about in the middle of the available voltage swing. This will ensure that signal variations will have freedom to move in both directions without hitting the supply rail at the top or the saturation area at the bottom.

The above remarks still apply even if the signal variations are known to be small. The percentage distortion introduced by the slight non-linearity in transistor curves is always proportional to the size of the swing.

For example, a signal swing of one volt superimposed on the output will introduce less distortion if the available limit of swing is 10V than it were 5.

3.7   SETTING THE BIAS VOLTAGE

The setting of the output voltage should normally be tackled first. If Rin is important however, some preliminary fiddling is required. Afterwards the required voltage drops across the divider chain , the desired base current, the divider chain current and finally the values for R1 and R2 (in that order) should be determined. Suppose for example we have decided on the output circuit shown in fig 3.8.

BASE CURRENT: Before we can calculate the bias resistors the base current must be estimated. Don't be too fussy rather be pessimistic!
Assume you will be unlucky enough to pick the worst sample of a BC108 ie having a hfe only of about 100. This would demand a base current of 1mA/100 = 10uA.


In case this reasoning appears rough and ready, remember that the only thing of importance about the current down the divider chain is that it should be much larger than the base current.
If we use the "ten times" rule and make the current in R2 = 100uA when the worst transistor is used then the current stability will always be good enough and often better than necessary.
R2 = 1.6V/0.1mA = 16k and
R1 = 7.4V/0.11mA = 67k
The nearest standard values may be used because R_E (1k) will ensure that the VBE 0.6V is maintained within correct limits.

STAGE INPUT RESISTANCE
It may be interesting at this point to calculate the input resistance R_IN.
R1 and R2 have a combined parallel resistance of of 13k. the transistor itself presents an input resistance of hfe(re + R_E) = 100(25+1000) = 100k.
As this is much higher than 13k it will have little effect so the input resistance stays at about 13k.