BJT Amplifier – Understanding Input Impedance

Let's put some part reference designators on your diagram.

Always number all the parts. Then it is easy to discuss the diagram. Instead of "the emitter resistor of the second transistor" we just say R5.

• C1: This is a coupling capacitor which allows the AC signal to pass but blocks DC. It protects the microphone's coil from receiving a DC current from the amplifier's bias circuit and protects the amplifier's bias circuit from being disturbed by the impedance of the microphone. C1 transmits the voltage fluctuations from the microphone, superimposing them upon the bias voltage between R1 and R2.

• R1 and R2: These resistors form a voltage divider, establishing a voltage-divider bias for the base of transistor Q1. From a 9V power supply, R2 will develop about 1V. That's enough to forward bias the base junction of Q1, turning the transistor on.

• Q1: This BJT is the heart of the first amplification stage, a common-emitter (CE) voltage amplifier. Its job is to transform variations in the base current caused by the microphone voltage variations arriving over C1 into current variations through the collector-emitter circuit R3, R4 and C2.

• R3: This is the load resistor for the CE voltage amplification stage. Variations in current controlled by Q1 cause R3 to develop a voltage. This voltage is the output of the Q1 stage, directly conveyed to the base of Q2. The voltage is inverted with respect to the microphone signal. When the signal swings positive, more current flows through R3, developing a greater voltage drop. The top of R3 is pinned to the 9V power rail, so more voltage drop means that the bottom of R3 swings more negative.

• R4: This emitter resistor provides feedback to stabilize the DC bias of Q1. The bias provided by R1 and R2 turns on Q1 using a voltage of about 1V, mentioned above. This causes current to flow through the transistor. This current causes a voltage in R4. The transistor "rides" on this voltage. So the voltage opposes the 1V of bias. According to some rule of thumb calculations, R4 will develop about 0.3V, which is the voltage that is left over when we take the 1V bias voltage between R1 and R2, and subtract the base-emitter voltage drop of 0.7V. This 0.3V over 1500 ohms means that about 0.2 mA of collector current will flow through the transistor, at quiescence. This bias current also flows through the 10K R3 resistor, where it gives rise to a voltage of 2V. So the output of Q1 is biased approximately 2V below the 9V power rail.

• C2: This capacitor bypasses the R4 resistor for AC signals. The R4 resistor has the effect of feedback. The amplified current passes through R4 and develops a voltage, and Q1 rides on top of this voltage. The voltage being amplified is the difference between the input and the emitter. So R4 provides negative feedback, which reduces gain. By introducing C2, we get rid of this feedback for AC signals. AC signals do not experience negative feedback, and so the gain is much higher for those signals. R3 and R4 provide a stable DC bias for Q1, and C2 "cheats" around it, creating a higher gain for AC, so that the amplifier has a wider swing around the bias point (which, recall, is about 2V below the power rail). A lot of voltage gain is needed because microphones put out a rather small signal, and all the amplification is being done by a single stage.

• Q2: This transistor is set up as a current amplifying emitter-follower stage. Note that there is no load resistor similar to R3 in the previous stage. Instead, the output is taken from the top of the emitter resistor R5.

• R5: What happens here is that the top of resistor R5 follows the voltage applied to the base of Q2. It is simply that voltage, minus 0.7V. As the voltage at the base swings, the voltage at the top of resistor R5 goes through the same swing. This voltage is applied to the speaker through C3.

• C3: Another blocking capacitor. It prevents DC from flowing into the speaker, which would damage the speaker and also cause a lot more bias current to flow through Q2, since the speaker's impedance is a lot lower than that of R5.

• C2: This is a power-supply decoupling capacitor. In several places in the circuit, AC signals return to the power supply either through the 9V rail or through the common return (ground). These currents can develop a voltage across the internal impedance of the power supply. C2 provides a short circuit for these AC signals. Without power supply decoupling, current variations in Q2 could feed back into the Q1 stage, giving rise to oscillations. C2 also helps to keep stray noise from the power supply, such as power supply ripple, from affecting the circuit. Another way to look at it is that the capacitor provides current in response to sudden demands by Q2.

The Q2 stage is needed because, even though it does not amplify voltage, it amplifies power. It does that because it is able to deliver more current than Q1. Q1 has load resistor R3, which gives it a rather high output impedance. If the speaker were connected to the Q1 stage output, hardly any sound would come out of it because the Q1 stage cannot maintain its voltage into just an 8 ohm load. Q2 has no collector resistor, and so the output impedance is low. Current flucutations flow freely from the power supply, through the transistor's collector and across C3 to the speaker.

The Q1 stage is needed because a current driving stage like the one built around Q2 doesn't have any voltage gain. The Q2 stage alone could take the voltage from the microphone and put it across the speaker. Now it would be better than connecting the microphone directly to the speaker, because the microphone would be isolated from driving the low impedance of the speaker. But, in spite of that, would simply not be loud enough. Getting a reasonably loud sound out of the speaker requires a much higher voltage level.

The jobs of amplifying voltage, and then amplifying the current which enables that voltage to be put across a low-impedance load such as a speaker, are best implemented separately.

I found this on the internet: -

It shows collector current versus CE voltage for several base currents. The graphs indicate the natural slope resistance of the collector. For instance, when \$I_B\$ = 40 uA, I estimate that a collector current range of 8 mA to 8.75 mA corresponds to a CE range of 1V to 8V.

This means the slope resistance is \$\dfrac{7\space V}{0.75\space mA}\$ = 9333 ohms.

This slope resistance, in parallel with the collector resistor (Rc) dictates what the output resistance of the amplifier is. Re will have a small effect on this of course.

If you are serious about finding out, the next step would be simulation.