Designing with Transistors - Chapter 5

5.0.  OVERVIEW OF CHAPTER 5

The main theme of this section is methods of achieving high input impedance in amplifiers. The Darlington amplifier is a simple but costly way of producing the desired result whilst the much more subtle "bootstrapping" technique gives excellent results with the addition of only a capacitor and resistor to the basic emitter follower. After this we take a quick look at negative feedback techniques.

We have seen an example of negative feedback in the emitter follower, but it is usual to apply negative feedback over a stage or a series of stages in order to increase stability, reduce distortion and noise, and increase impedance.

5.1.   DESIGNING FOR VERY HIGH INPUT RESISTANCE

Some signal sources (crystal microphones, photo-diodes, etc.) only deliver very tiny currents because of their very high source resistance. An attempt to amplify such signals will only be successful if the input resistance of the amplifier is also very high.
Although emitter followers have this property, the possibility of achieving R_IN of the megohm order is still difficult without the use of a FET.

A simple answer is the Darlington pair, the basic circuit of which is shown in Fig. 5.1. remembering a previous equation for a single stage emitter follower rin = hfe(re + R_E) and assuming re is negligible in relation to R_E, we may simplify this to rin = hfe(R_E) However, this circuit uses two stages of amplification so that the formula becomes rin=(hfe1 * hfe2)R_E where hfe1 and hfe2 are the respective gains for the two transistors.

Assuming they are both the same, the equation becomes h²fe * R_E For example if R_E is 5k and h_fe is 100, rin becomes about 50MΩ which is so high in relation to the shunting of R1 and R2 that it can be ignored in the calculation of R_IN

It is easy to see therefore that the R_IN of the stage is almost entirely determined by Rl, R2 values. The R_E resistor is to complete the divider chain for the base of TR2 and also to pump a little more collector current into TR1. Without it, the base current of TR2 would be the collector current of TR1 which would reduce its h_fe too much.

A possible set of components is shown in Fig. 5.2. This circuit gives, an R_IN of nearly 0.7MΩ (R1 and R2 in parallel).

5.2.   INPUT BOOTSTRAPPING

There is a novel method of increasing the input resistance of an emitter follower stage known as "bootstrapping". The purpose is to increase R_IN by effectively removing the bias resistors from the signal path. Two extra components are needed as Fig. 5.3

BOOTSTRAPPING ACTION

Resistor R3 is made small enough not to interfere with the d.c. bias. Note that the signal voltage V_IN is at one end of R3 and the output voltage (via C1) is at the other end of R3 and the output voltage (via C1) i at the other end.

Now Vout is in phase with with Vin and has almost the same amplitude (voltage gain = 1). thus it follows that Both ends of R3 are at the same potential and so no signal can pass through. This means that the signal can carry out its proper function of changing the base current without wasting half of its energy flowing down R1, R2. Bootstrapping thus effectively removes R1 and R2 from the stage input resistance formula. Thus R_IN = rin, and rin can be made very high.

CHOICE OF R3

The choice of R3 is not critical providing the base current does not produce any appreciable voltage drop (much less than the 0.6V is the usual criterion). Capacitor C1 should have a reactance (Xc) less than R3 at the lowest expected frequency. This ensures that the high input resistance is maintained at signal frequencies.

5.3   COUPLING LOSSES

Suppose we have one amplifier stage which on its own, gives a gain (A) of 100 and we feed the output into another stage having a gain A = 50. A miserable disappointment is in store for those who are hoping the overall gain will be 5000. The output resistance of stage 1 and input resistance of stage 2 form a voltage divider which causes a coupling loss between the stages.

OVERALL GAIN

To calculate the overall gain between signal emf and final output it is customary to treat coupling losses as "gains". For example, if we lose half half the signal at some point due to coupling loss, we can say that the 'gain' of the coupling is 0.5.

In this way we can work methodically through a multistage amplifier as if it consisted of isolated blocks, and by multiplying all the individual gains we can find the total gain.

EXAMPLE

To illustrate the procedure, consider the circuit shown. in. Fig 5.4. In this example there are five "gains" in all. G1, G3, and G5 are fractional gains (losses) and G2 and G4 are actual stage gains.

Let G1 = 0.2, G2 = 100, G3 = 0.1, G4 = 100 and G5 = 0.5.
Then the total gain = 0.2 x 100 x 0.1 x 100 x 0.5 = 100.
It is advisable to use this method of analysis in all multistage amplifiers.

5.4   MINIMIZING COUPLING LOSSES

To illustrate the seriousness of the problem consider the following system:
A pick-up has an output e.m.f. of 5mV, a source impedance of 5kΩ and is feeding an amplifier having a gain A of 100. The R_IN of the amplifier is l0kΩ and Rout is 2kΩ. The final load on the amplifier is 5kΩ.
Now the coupling loss between the pick-up and amplifier is:

G1 = 10k/(10k+5k) = 0.67
The gain of the amplifier gives:
G2 = 100
The coupling loss between the amplifier output and final load is
G3 = 5k/(5k+25) = 0.71k
The total gain is G1 x G2 x G3 = 0.67 x 100 x 0.71 = ±48.
Thus a gain of supposedly 100 has been reduced to a mere 48.

RULES ON COUPLING

Some rules and tips on coupling designs are clearly worth memorizing.

• Keep output resistances low and input resistance high.
• The R_IN tends to rise if Ic is lowered
• The Rout tends to fall if Ic is increased.

Thus we have a conflict of requirements to satisfy the requirement of Rin and Rout.
Mathematical analysis indicates that Total losses will be minimal if the input and output coupling losses are arranged to be equal.

Total losses will be minimal if input and output losses are arranged to be equal.

There is of course an easy way out-just put an emitter follower between every stage. This is a little extremist however, and could be rather costly if adopted as a general principle.

5.5   NEGATIVE FEEDBACK

DEFINITION

Negative feedback is the feeding of part or the whole, of the voltage or current from some later stage of an amplifier back to an earlier stage. The phasing of the voltage must tend to reduce the gain.

CLASSIFICATION

Series voltage feedback - the voltage feedback is in series with the input. This increases R_IN.
Parallel voltage feedback - voltage fed back is in parallel with the input. This reduces R_IN.

EQUATIONS

If Ao is gain before feedback, Ac is gain with feedback, and β = fraction of output which is fed back then
A'c = Ao/(1+β * Ao)
If βA is large, the equation reduces to
A'c = 1/β

This indicates a remarkable property - the gain of the amplifier can be made independent of the components inside it, including the transistors themselves! In addition to this obvious advantage, negative feedback will reduce distortion and internally generated noise, and change input and output resistances in the same ratio as the gain is reduced. In view of this, it is good design practice to produce a gain higher than required and reduce it by negative feedback to the desired value.

The circuit in Fig. 5.5 is an excellent example of the use of d.c. and signal negative feedback. By altering component values, gain can be increased to 100.

FEEDBACK LOOPS

There are two negative feedback loops and their subtlety should be carefully studied. Loop 1 (from the collector of TR2 to the emitter of TR1) is producing signal negative feedback and the feedback being:

β = R2/(R2+R3)
The total closed loop gain A' is given by:
A' = 1/β = (R2+R3)/R2 = 3.6

The open loop gain is of course much higher because it is basically two grounded emitter amplifiers in series.
Note that the feedback is series type because it is applied back to the emitter of TR1
Loop 2 (from the junction of R6 and R7 to the base of TR1) is not acting on the signal because C1 is keeping TR2 emitter at ground. Therefore this loop is stabilizing d.c. conditions only.