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I forgot to add Q₂'s loading on Q₁, which would reduce the gain, Av, to about -5.26. (It's loading changes the gain by (β⋅1kΩ) ⁄ (β⋅1kΩ + 10kΩ) ≈ .952)
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a) The first stage is a common emitter voltage amplifier with combination bias (voltage divider bias).
The second stage is an emitter follower. It is direct coupled to the first stage collector so its bias is taken from the DC output of the first stage.
b) The emitter of Q2 is 7.5V so its base is 7.5V + 0.7V for Vbe = 8.2V. The collector of Q1 is the same voltage, so the current is according to ohms law, I = (15V - 8.2V) / 10k = 0.68mA.
The voltage at the emitter is V = 0.68mA * 1.8K = 1.224V (ignoring the base current which is just 1/200 of the collector current, from the beta et al if you want to use it).
The base is ~0.7V higher for the Vbe = 1.924V. Now calculate the resistor for the voltage divider. Use the first link, which explains how. The combination bias type is selected as "voltage divider" in the drop down list. The result is close to 33K for Vbe = 630mV. The base current is ~3.4uA. You can work this through yourself, or search for other similar examples. The second link is an example.
How to get Vbe...
The critical parameter not given here is the base emitter voltage Vbe. The voltage is like 600mV at 25°C, and the base emitter voltage is modified from this by temperature and the amount of base emitter current flow (due to resistance of the material etc), just as the forward voltage of a diode. The collector emitter also influences this slightly. A voltage used by most people is 0.7V. This usually guarantees saturation as a logic switch, so is not very accurate here. The third link shows data for a similar transistor to the one in the questions. The Vbe saturation voltage (enough base current for collector to be saturated) is 0.7V typical, and can range from 550mV to 770mV when not saturated, depending on collector current and voltage slightly. Figure 2 in the data sheet shows with the value of collector current in the example this is more like 630mV. This in turn would change with temperature. Figure 4 shows this is ~2.22mV per degree change from 25°C at the collector current in this example. Figures 8 and 10 are similar diagrams for slightly different transistors. My examples with 0.7V above would be more accurate with 0.63V.
c) the quiescent base current is the collector current divided by the current gain (beta).
d) Input resistance (for AC) consists of:
the base emitter equivalent resistance = 630mV / 3.4uA = 185K
The two divider resistors. All are equivalent to being in parallel with the input and ground to an AC signal. There can be effects from feedback due to the emitter resistor too. Assume it is bypassed to AC signal by the emitter capacitor so it can be ignored.
e) The voltage gain of the first stage is R_Load / R_emitter = 10K / 1.8K = 5.555. Assume Ce which would modify R_emitter for AC signals is not connected. The second stage gain is almost 1, approx. 0.999. Thus overall gain is 5.555 and in dB that is 20(log(ratio)) = ~15dB. The gain is explained in the second link too.
The second stage is an emitter follower. It is direct coupled to the first stage collector so its bias is taken from the DC output of the first stage.
b) The emitter of Q2 is 7.5V so its base is 7.5V + 0.7V for Vbe = 8.2V. The collector of Q1 is the same voltage, so the current is according to ohms law, I = (15V - 8.2V) / 10k = 0.68mA.
The voltage at the emitter is V = 0.68mA * 1.8K = 1.224V (ignoring the base current which is just 1/200 of the collector current, from the beta et al if you want to use it).
The base is ~0.7V higher for the Vbe = 1.924V. Now calculate the resistor for the voltage divider. Use the first link, which explains how. The combination bias type is selected as "voltage divider" in the drop down list. The result is close to 33K for Vbe = 630mV. The base current is ~3.4uA. You can work this through yourself, or search for other similar examples. The second link is an example.
How to get Vbe...
The critical parameter not given here is the base emitter voltage Vbe. The voltage is like 600mV at 25°C, and the base emitter voltage is modified from this by temperature and the amount of base emitter current flow (due to resistance of the material etc), just as the forward voltage of a diode. The collector emitter also influences this slightly. A voltage used by most people is 0.7V. This usually guarantees saturation as a logic switch, so is not very accurate here. The third link shows data for a similar transistor to the one in the questions. The Vbe saturation voltage (enough base current for collector to be saturated) is 0.7V typical, and can range from 550mV to 770mV when not saturated, depending on collector current and voltage slightly. Figure 2 in the data sheet shows with the value of collector current in the example this is more like 630mV. This in turn would change with temperature. Figure 4 shows this is ~2.22mV per degree change from 25°C at the collector current in this example. Figures 8 and 10 are similar diagrams for slightly different transistors. My examples with 0.7V above would be more accurate with 0.63V.
c) the quiescent base current is the collector current divided by the current gain (beta).
d) Input resistance (for AC) consists of:
the base emitter equivalent resistance = 630mV / 3.4uA = 185K
The two divider resistors. All are equivalent to being in parallel with the input and ground to an AC signal. There can be effects from feedback due to the emitter resistor too. Assume it is bypassed to AC signal by the emitter capacitor so it can be ignored.
e) The voltage gain of the first stage is R_Load / R_emitter = 10K / 1.8K = 5.555. Assume Ce which would modify R_emitter for AC signals is not connected. The second stage gain is almost 1, approx. 0.999. Thus overall gain is 5.555 and in dB that is 20(log(ratio)) = ~15dB. The gain is explained in the second link too.