__An Audio Amp:__This particular project involved injecting the audio from a TV receiver into a stereo system. (These days even the cheapest TV has that feature, including MTS stereo inputs for digital accessories). Anyways, the audio-output portion of the TV-audio receiver was abandoned because of its poor frequency response and high distortion. Instead, we wanted to come right off the detector into a quality audio amplifier and speaker. So, after picking off the audio at a convenient point in the set (in this case, from a potentiometer), we wanted to feed it to the auxiliary input of the stereo amplifier.

The amplifier we used required an input of 1 volt RMS, but a quick check with an AC VTVM indicated that out picked-off audio signal was only 0.1-volt RMS. Obviously, an amplifier with a gain of 10 was needed.

Scanning the literature on transistor amplifiers reveled that a common-emitter amplifier with a voltage-divider bias circuit would solve our problem nicely. Such a circuit is shown in Fig. 1. Some of that circuit's characteristics include: moderate input impedance, moderate voltage gain, inverted output, and input/output impedance and gain that depend only slight on transistor beta.

There are, of course, several rules that must be followed in using a common-emitter amplifier, including:

- With a positive supply use an NPN transistor.
- With a negative supply use an PNP transistor.
- The supply voltage must not exceed the transistor's Vce rating.
- The power-dissipation rating of the transistor must not be exceeded.

- The beta of the transistor should be 100 or higher.

In our example the following facts are known:

- Our amplifier had a single-ended 12-volt power supply.
- We need a voltage gain of 100.
- The input impedance of the amplifier should be about 15K, the same as the potentiometer from which the audio was taken.
- The impedance of the stereo amplifier's auxiliary input is about 50K.

__Doing the Math:__For maximum undistorted output swing, we will make the quiescent collector voltage 1/2 the supply voltage. See Fig. 2. The drop across R

_{c}must therefore be 6 volts.

The value of R

_{c}, the collector load resistance, is chosen considering output impedance, gain, and collector current. If possible, the output impedance should be lower than the impedance of the circuit we are feeding by a factor of 10 or more. Doing so will avoid circuit loading. So let's make R

_{c}equal to 4700 ohms, which is about 50K/10. Collector current I

_{c}, is equal to 0.5V

_{cc}/R

_{c}, or 6/4700 = 1.28 mA. That current is certainly low enough that we will not exceed any collector-current ratings, so let's go on.

To achieve maximum stability, the emitter resistor should be in the range of 40 to 1000 ohms. Voltage gain (A

_{v}) = R

_{c}/R

_{e}, so R

_{e}= R

_{c}/A

_{v}. In our case R

_{e}equals 4700/10, or 470 ohms. That falls within the range of acceptable values.

The current through the emitter resistor consists of the collector current plus the base current. The base current here is significantly smaller than the collector current, so it can be ignored for the next calculation.

The voltage drop across the emitter resistor = I

_{c}X R

_{e}, or 1.28 mA x 470 ohms = 0.602 volts. The base voltage must exceed the emitter voltage by 0.6 volts for a silicon transistor and by 0.2 volts for a germanium transistor. We'll use a silicon transistor (most if not all germanium types are obsolete) in our circuit, so the base voltage must be 0.6 + 0.602 = 1.202 volts.

The input impedance of the circuit equals R2 in parallel with the emitter resistor times beta; input impedance will vary with the transistor's beta. FOr our example, assume we are using a transistor with a beta of 100. We want the input impedance to be about 15000 ohms. Solving for R2, we find:

ZWe can use a 22K resistor. In general, if input impedance is not critical, for maximum stability R2 can be 10 to 20 times R_{in}= (R2 X R_{e}X beta)/[R2 + (R_{e}X beta)] R2 = (Z_{in}X R_{e}X beta)[(R_{e}X beta) - Z_{in}] R2 = (15000 x 470 x 100)/[470 x 100) - 15000] R2 = 22,030 ohms.

_{e}.

The drop across R2 must be 1.20 volts so the current through R2 is 1.20/22,000, or 0.054 mA. Therefore, R1 must drop the rest of the supply voltage, which is 12 - 1.20 = 10.8 volts. The current flowing through R1 is a combination of the voltage-divider current plus the base current.

The base current is equal to the collector current divided by beta. It is found from:

ISo the total current through R1 is 0.054mA + 0.0128mA = 0.067mA, and R1 = 10.8/0.067mA = 160,000 ohms (160K)._{beta}= 1.28/100 = 0.0128 mA

Resistor R1 is the most critical resistor in the circuit. To ensure maximum voltage swing, it should bring the quiescent collector voltage to one half the supply voltage. After building the circuit, the value of R1 may have to be varied slightly to achieve that voltage swing.

We now have a circuit we can test.

__Interfacing:__Connecting the circuit to the outside world will require capacitor coupling. That serves to isolate the AC signal from any DC bias voltages. Figure 3 shows our complete circuit with input and output coupling capacitors. The values of those capacitors were calculated using C = 1/(3.2 x ƒ x R), where C equals the capacitor value in farads, ƒ equals the frequency at which response will be down 1dB, and R equals the impedance on the load side of the capacitor.

To calculate the value of C1, the amplifier's input impedance (15K) is used for R. To calculate the value of C2, the input impedance of the next stage (50K) is used for R.

The value of C1 can now be calculated for a drop of 1dB at 20 Hz: C1 = 1/(3.2 x 20 x 15000) = .00000104 farad = 1.0 uF. The value of C2 = 1/(3.2 x 20 x 50000) = .00000031 farad = 0.33uF.

To increase the gain of the stage, you could bypass R

_{e}with a capacitor, as shown in Fig. 4. Nothing comes for free, however. The price you pay for increase gain is lower input impedance, which will vary widely with beta. If that variation is not a problem, a significant gain increase can be realized by adding the bypass capacitor. Our original circuit has a gain of 10; if the emitter is bypassed the gain becomes R

_{c}/003/I

_{e}= 4700/(0.03/0.00129) = 4700/23 = 200 (approx).

The value of the bypass capacitor in farads is calculated from the formula C = 1/(6.2 x ƒ x R). Again ƒ is the low-frequency limit in Hz, and R is the dynamic emitter resistance (0.031/I

_{e}). In our example, if we stick to a 20-Hz lower limit we have C = 1/[6.2 x 20 x (0.03/0.00129)] = .000344 farads = 344 uF. A 350uF unit can be used.

__Thoughts:__A few thoughts on components before we finish: using 5% resistors allows closer adherence to the calculated values. Because of their temperature stability and low leakage specifications, silicon rather than germanium transistors are preferable for this type of circuit.

Finally, you've no doubt noticed that we have yet to specify a specific transistor. That's because for this type of application it really doesn't matter! Almost any small signal device will do fine.

Ricardo A monroy B

C.I. 17646658

EES

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