51.1.3. Operating Point Calculation

If we see that this I it is given as 1 mA and β is 100. So, we may approximate that collector current I_C = I_E and which is same as 1 mA, so this is the DC current. And then the voltage at the base it is given as 6 V and the emitter voltage here it can be obtained by considering 6 V here and V_BE all of 0.6 V. So, we can say at the emitter node we do have 6 ‒ 0.6 that is 5.4 V. Now, if we have this 1 mA of current at the emitter, approximately the collector current is also said by that, then the drop across this R_C it is 3 kΩ × 1 mA. So, that gives us this voltage it is V_C = V_dd - V_RC, V on the other hand it is 10 V.

In this chunk, we calculate the operating point of the transistor. The current through the transistor is given as 1 mA, and the current gain β is 100. From this information, we conclude that the collector current I_C is equal to the emitter current I_E, which also equals 1 mA. The base voltage is 6 V and the base-to-emitter voltage drop (V_BE) is approximately 0.6 V. Therefore, the emitter voltage (V_E) can be calculated as V_E = 6 V - 0.6 V = 5.4 V. Subsequently, we analyze the voltage across the collector resistor (R_C) when the collector terminal is connected to a supply voltage V_dd of 10 V. The voltage drop across R_C due to the emitter current gives us V_C at the collector terminal that drives the transistor to operate in its active region.

Think of a water tank system where you need to maintain a water level at a specific height (like V_C). The water flowing into the tank (like I_E and I_C) remains constant at a rate of 1 mA. The higher you want the water level, the more you need to adjust the opening at the outlet (like R_C) to maintain that balance. This analogy helps visualize how controlling the input affects the output in a circuit.

So now, the operating point it is we obtained. In fact, we can find what is the corresponding V_CE voltage which is 7 ‒ 5.4 here. If we see that this I_B is 1 mA and β is 100, so we can say g_m = 26 mV. And then the other parameter it is r_o it is expression is r_o = V_A / I_C where Early voltage V_A is given here it is 50 V and I_C is 1 mA. So, that gives us r_o = 50 kΩ.

After calculating the operating point, we can move to calculating the small signal parameters of the transistor. The voltage V_CE can be calculated as V_CE = 7 V - 5.4 V. The transconductance, g_m, can typically be approximated if the thermal voltage is provided (about 26 mV). For resistance due to Early effect, we can compute the output resistance r_o using the formula r_o = V_A / I_C, where V_A is the Early voltage (50 V) and I_C (collector current) is found as 1 mA. This calculation yields an output resistance of 50 kΩ.

Consider a garden hose where water pressure represents the voltage and the flow rate represents the current. g_m signifies how effectively pressure change (voltage) influences flow (current). The output resistance (r_o) can be imagined as the flexibility of the hose; a more rigid hose has a higher resistance to change in pressure and flow, affecting how much water you can push through based on the pressure.

So, the voltage gain starting from the emitter terminal to the collector terminal if I call that voltage gain it is A_V. The voltage gain can be approximated as A_V = g_m * (r_o || R_C). By plugging in the values of g_m, r_o and taking into account R_C = 3 kΩ, we will get A_V ≈ 100.

We now focus on calculating the voltage gain (A_V) of the common base amplifier, which is determined by the transconductance (g_m) and the output resistance. The expression for voltage gain in this context approximates to A_V = g_m multiplied by the parallel combination of the output resistance (r_o) and load resistance R_C. By substituting our known values, we discover that A_V approaches 100, which indicates that the voltage level at the collector can be a hundred times greater than that at the emitter.

Imagine using a megaphone to amplify your voice. The input sound (signal) level, when passed through the megaphone's bell (the effective amplifier), comes out much louder at the other end compared to what went in. Here, the gain factor (A_V) of the megaphone would reflect how many times louder the output is compared to the input.

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