In many applications, a single amplifier cannot furnish all the gain that is required to drive a particular kind of load. For example. a speaker represents a hell load in an audio amplifier system, and several amplifier stages may be required to “boost” a signal originating at a microphone or magnetic tape head to a level ‘ufficient to provide a large amount of power to the speaker. We hear of preamplifiers, power amplifiers, and output amplifiers, all of which constitute stages of amplification in such system. Actually, each of these components may itself consist of a number of individual transistor amplifier stages. Amplifiers that create voltage. current, and/or power. gain through the use of two or more stages are called multistage amplifiers.
When the output of one amplifier stage is connected to the input of another, the amplifier stages are said to be in cascade. Figure 11-1 shows two stages connected in cascade. To illustrate how the overall voltage gain of the combination is computed, let us assume that the input to the first stage is 10 mVrms and that the voltage gain of each stage is AI = A2 == 20, as shown in the figure. The output of the first stage is AIVil == 20 (10 mV rms) = 200 mV rms. This, the input to the second stage.
Figure 11-2 shows an arbitrary number (n) of stages connected in cascade. Note that the output of each stage is the input to the succeeding one (VIII := V,,!, VII! := Vi’, etc.), We will derive an expression for the overall voltage gain V,,)Vil in terms of the individual stage gains A” A2, All’ We assume that each stage gain A” A , A” is the value of the voltage gain between input and output of a stage with all other stages connected (more about that important assumption later). by definition stage gains can be negative, signifying, as usual, that the stage causes a 1800 phase inversion. It follows from equation 11-4 that the cascaded amplifiers will cause the output of the last stage (VUI!) to be out of phase with the input to the first stage (Vii) if there is an odd number of inverting stages, and will cause VUI! to be in phase with Vii if there is an even (or zero) number of inversions.
To find the overall voltage gain of the cascaded system in decibels, we ignore the algebraic signs of each stage gain and compute Equation 11-5 shows that the overau vouagc gain in dB is the sum or the individual stage gains expressed in decibels. Equations similar to (11-4) and (11-5) for over all current gain and overall power gain in terms of individual stage gains are easily derived.
Equation 11-5 shows that the overau vouagc gain in dB is the sum or the individual stage gains expressed in decibels. Equations similar to (11-4) and (11-5) for overall current gain and overall power gain in terms of individual stage gains are easily derived. Our derivation of equation 11-4 did not include the effect of source or load resistance on the overall voltage gain. Source resistance rs causes the usual voltage division to take place at the input to the first stage, and load resistance r[ causes voltage division to occur between rc and the output resistance of the last stage. Under those circumstances, the overall voltage gain between load and signal source becomes
A three-stage amplifier and the ac rms voltages at several points
in the amplifier. Note that V, is the input voltage delivered by a signal source having zero resistance and that V3 is the output voltage with no load connected.
1. Find the voltage gain of each stage and the overall voltage gain U3/VI’
2. Repeat (1) in terms of decibels.
3. Find the overall voltage gain u,Jus when the multistage amplifier is driven by a signal source having resistance 2000 n and the load is 25 n. Stage 1 has input resistance I kn and stage 3 has output resistance 50.
IMG
4. What would be the gain v,.Ivs in dB for the conditions of (3) if the voltage gainof the second stage were reduced by 6 dB?
5. What is the power gain in decibels under the conditions of (3) (measured between the input to the first stage and the load)?
6. What is the overall current gain hli, under the conditions of (3)?
It is important to remember that the gain equations we have derived are based on the in-circuit values of A” A2, ••• , that is, on the stage gains that result when all other stages arc connected. Thus, we have assumed that each value of stage gaIn lakes into account the loading the stage causes on the previous stage and the loading presented ‘to it by the next stage (except we assumed that A I did not include loading by rs and All did not include loading by rd. If we know the open-circuit (unloaded) voltage gain of each stage and its input and output resistances, we can calculate the overall gain by taking into account the loading effects of each stage on another. Theoretically, the load presented to a given stage may depend on (/1/ of the succeeding stages lying to its right, since the input resistance of anyone stage depends on its output load resistance, which in turn is the input resistance to the next stage, and so forth. In practice, we can usually ignore this cumulative loading effect of stages beyond the one immediately connected to a given stage, or assume that the input resistance that represents the load of one stage to a preceding one is given for the condition that all succeeding stages are connected.
To illustrate the ideas we have just discussed, Figure 11-4 shows a three-stage amplifier for which the individual open-circuit voltage gains A”” A, , and A”.1 are assumed to be known, as well as the input and output resistances of each stage. From the voltage division that occurs at each node in the system, it is parent that the following relations hold:
As might be expected, equation 11-7 shows that the overall voltage gain of the multistage amplifier is the product of the open-circuit stage gains multiplied the voltage-division ratios that account for the loading of each stage. Notice that a single voltage-division ratio accounts for the loading between any pair of stages.
In other words, it is not correct to compute loading effects twice: once by regarding an input resistance as the load on a previous stage and again by regarding the output resistance of that previous stage as the source resistance for the next stage
The open-circuit (unloaded) voltage gains of three amplifier stages and their input and output resistances are shown in Table II-I. If the three stages are cascaded and the first is driven by a lO-mV-rms signal source having resistance 12 kO, what is the voltage across a 12-0 load connected at the output of the third stage?
