# Low Frequency Power Amplifiers Tutorial

THEORY AND DESIGN

Figure 1. Basic circuit amplifier in common emitter connection.

In figure 1 we have an amplifier whose input is between the base-emitter and the output between collector-emitter, the emitter is now common to entry and exit and the amplifier is called common emitter amplifier. As in the case of the amplifier in connection CB transistor (NPN) must have the base-emitter diode biased and the base-collector diode reverse biased, this is accomplished with batteries Vbb and Vcc and the resistance Rb, the resistance Rc used to obtain the output voltage, without the Rc we would not have any voltage at the output. Because the base-emitter diode is forward-biased, the resistance RI that exhibits is small, and that is approximately given by the relation:

The resistance RI is the input resistance of the transistor in common emitter connection. Notice that RI is equal to the input impedance of the amplifier in connection common-base because the wiring CE input current is the base current (IB), while the connection CB input current of the transistor is the emitter current (Ie). If at the input of the amplifier in Figure 1 we apply an alternating voltage (Vi), then as shown by the alternating amplifier circuit in Figure 2 (in this circuit we have continuous voltages Vcc = Vbb = 0V) the voltage is equal to the voltage across the resistor RI = Re x B, and is given by the relationship:

The alternating current of the base IB will cause the release of a larger current in the collector (Ic = B x IB) and the current Ic will create a voltage drop across the Rc which is given by the relation:

and the gain voltage is given by the relation:

Because Ic = Ib x B we get:

The amplifier of Figure 2 we have current gain (Ai). Since the output current is Io = Ic = Ib x B and the input current is Ii = Ib, we have:

Ai = (output current) / (input current) or:

Figure 2.

So, the current gain is equal to the parameter b or hFE of the transistor. If b = 50, then the output current will be 50 times greater than the input current. The amplifier provides power gain (AP) given by the relationship:

Figure 3.

The input resistance (Ri) of the CE amplifier is equal to the parallel connection of resistors: RI = (b x re) and RB, which means:

The output impedance of the amplifier (Ro) is approximately equal to Rc, because the output impedance presented by the transistor (r0) is much longer than the Rc. When the alternating voltage at the input of a common-emitter connection amplifier, (CE), increased (Figure 3), the input current and the total output current also increases.

Figure 4.: Basic amplifier circuit with common collector.

Because when the output current (Ic) increasing the output voltage of the amplifier (voltage between collector-emitter, Vce decreases. Note that the battery voltage is equal to the voltage across Rc and the voltage between the collector-emitter (Vce) is:

Because the battery voltage Vcc is stable, when the voltage Vrc increasing voltage Vce decreases and vice versa.

Eg if Vcc = 12 V, and the voltage across Rc = 5V, the voltage VCE = 7V

If now the collector current (lc) increase and the voltage drop across the Rc will increase (Vrc = Rc x lc), lets say VRC = 6V, then the voltage will be reduced and will be Vce = 6V. Therefore, in a CE amplifier when the input voltage increases (or the current), the output voltage is reduced, i.e. We have a phase difference of 180° between the input voltage and output voltage.

COMMON COLLECTOR AMPLIFIER

Figure 4. Basic Common collector circuit.

The connection of the amplifier in Figure 4 is called common-collector (common collector) (CC). Excluding the low internal resistance of the battery Vcc, the collector is directly attached to the ground, so we say that the input of the amplifier is between base-collector and the output between emitter-collector, so the collector is common for the input and the output.

Figure 5. Common collector circuit with alternating current.

If now we apply in the input of the amplifier an alternating voltage, then from Figure 5 is shown that the voltage Vi, is equal to:

the voltage between base-emission boom (Vbe) and the voltage across the resistor RE (VE), namely:

Note that the output voltage of the amplifier is the voltage across Re (Ve). The voltage gain is given by the relationship:

Av = (output voltage / input voltage), so:

Av = VE / Vj

Because Vbe is usually much smaller than the VE, the above equation becomes:

So one common collector amplifier does not reinforce voltages. The current gain (Ai) is given by the relationship:

Ai = (output current / input current)

And because Ie is almost equal to Ic:

The input impedance (Ri) that apears at the common-collector transistor is given by the relation:

The total input impedance of the amplifier (Ri) is equal to the parallel connection of the Ri and Rb, namely:

The output impedance of transistor is given by the relationship:

Where Rs is the internal resistance of the source of the signal that we want to amplify. The total output resistance of the amplifier (Ro) is given by the relationship:

So is equal with the parallel connection of Ro and Re. At the amplifier there isn't any phase difference between input and output voltage.

TRANSISTOR CHARACTERISTICS CURVES

As we have already explained, a transistor can be seted by three basic ways, (CB, CE, CC). For each of these configurations we have characteristic curves for input and output and are usually given by manufacturers.

Figure 6.

The function of amplifiers and other circuits can be explained graphically using the dedicated characteristic curves.

Figure 7.

CURVES OF COMMON EMITTER

Figure 8.

Figure 6 shows a cluster of input characteristics for an (NPN) transistor configured as a common emitter. These curves represents the varying of the input current (base current, IB) when we change the input voltage VBE (base-emitter) and we keep stable the the voltage VCE (collector-emitter).

Figure 9.

FEATURED OUTPUT CURVES

Figure 10.

The characteristic output curve of a NPN transistor in common emitter configuration shown in Figure 10 represents the change in output current Ic (collector current) when the the output voltage Vce change, keeping the current IB (base) constant. We can engrave different characteristics output for various steady base currents. Figure 11 shows output curves.

Figure 11.

The sum of voltages Vce (voltage drop across the collector-emitter) and Vrc (voltage drop at the ends of Rc) at the output of the amplifier in common emitter configuration has to be always equal to the voltage Vcc (battery voltage), namely:

The above relation represents a straight line on a graph with axes Vce and Ic, and the straight line is called the continuous load at DC.
To draw the straight loading at the chart Vce / Ic, we need to identify two points. As such points are easy to take as the intersection with axes Ic and Vce. The intersection of the straight line with the axis Ic exists, if at the relation (Vcc = Vce + Vcr), we set Vce = 0 so Ic = Vcc / Rc. The intersection point with the axis Vce can be fount if we set Ic = 0 so Vcc = Vce. The line load will be the straight line joining these two points as shown in Figure 9.

Eg. If Vcc 12V and Rc = 1Kohm then Ic = Vcc / Rc = 12V / 1Kohm = 12mA and Vce = 12V.

In the same way, we can draw straight load lines for amplifiers CC and CB, using the output characteristic curves. Notice that for different value of Rc or Vcc, the line load will be different. The straight load line that is drawed for a certain output circuit (ie for a certain values of Rc and Vcc) gives us all the possible values that the collector current (Ic) can take for a particular voltage value between collector and emitter Vce. Eg At figure 10, the point A of the straight load line, represents a state in circuit where the output current is Ic = 5mA and voltage Vce = 6V.

Point B represents another circuit condition, Ic = 2mA, Vce = 2,4 V. The output circuit can have values for Ic and Vce given only from the points of the line load. The output circuit can not obtain values for Vce and Ic that not correspond to points of a straight load. Eg if the current is lc = 5mA, then the voltage Vce must have a value of 6V, and conversely if the voltage is Vce = 6V, then the current Ic must have a value 5mA. If now we draw the straight load on the output characteristic (Vce - Ic) of a transistor in common-emitter configuration, we can explain graphically the operation of the amplifier, and to choose the appropriate resistors and DC voltages for proper biasing of the transistor. In Figure 11 is plotted the linear load for the amplifier of Figure 12. The maximum current collector elsewhere specified current saturation (saturation) of the transistor and is denoted by Ics and is given approximately by the relation: Ics = Vcc / Rc.

Figure 12.

The voltage between collector emitter when we have the current saturation is called saturation voltage Vces. and is about a few tenths of a volt. When now the base current IB is = 0, then the collector current lc = 0. I.e. to make the collector current (Ic) zero, the base current (Ib) has to be zero. In practice when the base current (IB) is zero, the current of collector (Lc) is not zero but is equal to the small reverse saturation current Iceo. The current IB for which lc = 0 is called cutoff current of the transistor.

Figure 13.

The cut points and saturation as will be seen, has to be avoided because in an amplifier causes distortion in the output signal. An amplifier want to have low distortion of the input signal at the output. In order, therefore the output signal to obtain large amplitudes without exhibiting deformation, should the rest point around which the operation of the amplifier to be somewhere in the middle of the line load, ie Vce = Vcc / 2. Note that point resting mean the collector current (lc) and the corresponding voltage Vce, when we have not apply any signal (voltage) to the input of the amplifier. The resting point should always be on the straight load. Eg resting point Q in Figure 13 corresponds to a resting tension Vce = 12V, a current collector resting 10mA, and the base current is Ia = 50mA, as seen from the figure the output characteristic curve corresponding to Ib = 50mA, intersects the line load at Q rest point.

Figure 14. The output voltage has no distortion. The resting point is in the middle of the load line.

If now the input amplifier we apply an alternating voltage (input signal) the voltage will cause a current input (base) as the voltage and the input current will vary the voltage Vce and current lc will also change from the point resting Q we have chosen. If the rest point (Q) is in the middle of the straight load and the input signal is too large then the voltage fluctuations across collector-emitter (Vce) will not exceed the maximum value of Vce = Vcc, nor minimum value of Vce = Vces. Also the fluctuations around the bias current will not exceed the minimum value of Ic = 0 and the maximum value Ic = vcc / rc. In this case the output voltage Vce will yield a true input signal (the signal will not deform) as shown in Figure 14. If the resting point placed near cut-off, then input signals having a large amplitude will occur at the output and deformed because each instantaneous value of the voltage Vce greater of Vcc, is cutted (sheared), as shown in Figure 15. The same will happen if the resting point is placed near saturation, as shown in Figure 16.

Figure 15. Resting point near cut-off. The output shows distortion. and Figure 16. Resting point near saturation. The voltage shows output distortion