Electronics Made Easy - a Complete Introduction to Electronics - Martin Denny [6]
xc = 1/2pfC.
The bandwidth of this amplifier is specified at the frequencies where the output voltage drops to 50% of the mid range design value (3db point), ie when xc = Rin, and at the stated top end frequency of the transistor.
The voltage gain will always be less than the current gain of the transistor due to the resistor loading. If the load on the output is equal to R2 then the gain will be reduced by 50% (see fig 6).
The gain of an amplifier, or conversely the attenuation of a circuit is sometimes quoted in terms of decibels.
For voltage gain: Gain db = 10LogG*
For power gain 20logG is used.
For Example: A 50% drop in output voltage is equivalent to a gain of 0.5*.
Gain db = 10Log0.5 = 10 * -0.30103 = -3db
A stage gain of 1* = 10 * 0 = 0db
A stage gain of 10* = 10 * 1 = 10db
A stage gain of 100* = 10 * 2 = 20db
A stage gain of 1000* = 10 * 3 = 30db
Figure 5b shows an example of collector to base bias. The advantage of this configuration is that it offers better gain stability than the fixed biased example. In this case Ic and Ib pass through R4 so:
R4 = (Vs - Vc)/(Ic + Ib)
R3 = (Vc - Vbe)/Ib
Current Gain hfe = Ic/Ib
To select C2
Rin º IcVc and Xc = 1/2pfC (Xc should be at least 0.1*Rin).
Figure 5c shows an example of self bias. This configuration has several advantages over the previous examples, higher input levels can be accommodated as the emitter resistor allows the bias voltage level of the emitter and hence the base to be raised significantly above 0V.
With the capacitor C4 removed the output Vout can be connected across R8 (emitter follower), thus giving a low impedance output, (at the cost of gain usually reduced to approximately unity) and the Vce voltage can be reduced if this is critical.
R6 = (Vs - Vc)/Ic
Note with C4 in circuit ( with xc considerably greater than R8 ) the bias set voltage at the collector (Vc) can be set to the same value as in previous circuits.
R8 = Ve/(Ic + Ib) hfe = Ic/Ib
R7 = (Ve + Vbe)/I, Where i is chosen ³ Ib
R5 = (Vs - Vbe - Ve)/(i + Ib)
To select C3
Rin º IcVc, and xc = 1/2pfC (at least 0.1*Rin)
To select C4
xc = 1/2pfC (at least 0.1*R8 at the lowest frequency)
With capacitor C4 removed Vout will be elevated with respect to (wrt) 0V. The value of Vc can be adjusted to take this into account. The voltage gain will be reduced significantly.
Feedback in Transistor Circuits
When the input level of a single stage transistor amplifier rises, the increase in input current causes a proportional increase in collector current which reduces the output therefore the amplifier is a inverting amplifier. If two stages are used the output becomes non-inverting. It follows that odd stages are inverting and even stages non-inverting.
Feedback can either be positive (regenerative) or negative (degenerative). Positive feedback can be demonstrated by placing a microphone connected to an audio amplifier close to the amplifier loudspeaker. Initially the loudspeaker is silent but the smallest sound is sufficient to start the process for every increase in sound is accompanied by a further increase from the speaker. Positive feedback is used in oscillator circuits. Negative feedback is used in amplifier design to provide gain control and stability.
Figure 7 demonstrates the use of voltage series feedback in an amplifier design. The design is of a two stage non inverting amplifier where negative feedback is fed from the output of stage 2 to the emitter of stage 1. This requires two stages as this circuit would provide positive feedback in a single stage.
Overall Gain = G1 x G2 x b, where b = Re / (Re + Rf)
Re is the emitter resistor
Rf is the feedback resistor
To design a non-inverting amplifier with gain variation 250 to 500, and a low frequency response down to 10Hz.
Transistors BC108, hfe 110 to 800, Vce 5v, Ic max 100mA, Pt 300mW
Assume hfe 150.
Vs = 13.6V, hfe = 150, hfe = Ic/Ib
Stage 1 (also basis for stage 2)
Let Ic = 2mA, then for 6v across R3, R3 = 3KΩ, and 2v across R4, R4 = 1KΩ.
For Ic = 2mA, Ib = 13.3µA
Voltage across R2 = 2v + 0.6v = 2.6v and