68.1.12. Performance Summary of CE Amplifier with Active Load

So, here we do have the different parameters of the two transistors for Q β is 100, for Q β is 200, just for a change we are taking different value 1 2 of β. And V of transistor-1 it is we are approximating it is 0.6 V it should be mod BE(on) EB actually V of transistor-2 also we are assuming 0.6. The early voltage of transistor-1 it is 100 V. So, likewise let you consider early voltage for transistor-2 also 100 V and then the capacitances namely base to emitter C for transistor-1 it is 10 pF and likewise for π1 transistor-2 it is 10 pF and then on the other hand base to collector C it is 5 pF for both µ Q and Q .

In this chunk, we define the key parameters for two transistors (Q1 and Q2) used in the CE amplifier circuit. The beta (β) values indicate how much current is amplified through each transistor. For this example, Q1 has β = 100, and Q2 has β = 200. The base-emitter voltage (V_BE) for both transistors is assumed to be 0.6V, a common value for silicon transistors when in active operation. The early voltage, which reflects the output impedance of the transistors, is set at 100V for both, indicating good performance in terms of current gain. The capacitances represent the parasitic capacitance effects in the circuit, which can influence the frequency response of the amplifier, affecting both stability and bandwidth.

Think of each transistor as a water faucet: the β value is like the pressure enhancement from the faucet. A higher β means that a smaller input flow (current) can produce a much larger output flow. The base-emitter voltage (0.6V) can be seen as the minimum 'pressure' needed for the faucet to begin pouring water. The early voltage acts like the capacity of the system to handle higher water flow without overflowing, while the capacitances are like the resistance of the pipes; they affect how quickly water can flow through the system.

So, end of it what we are getting here it is collector current of transistor-1 assuming the device it is in active region of operation it is it is β × I and I it is supply voltage minus V of transistor-1 divided by R and B B BE(on) B1 that is 100 multiplied by. So, this is 12, this is 0.6 so, that is 11.4 and then R it is 570 k. So, that gives us this part it is 20 µA × 100. So, that gives us 2 mA.

This chunk discusses the collector current (I_C1) for transistor Q1 while in active mode. We calculate the current using the formula I_C1 = β × I_B, where I_B is the base current. The base current (I_B) can be calculated by taking the supply voltage (12V), subtracting the base-emitter voltage (0.6V), and dividing by the base resistor value (570kΩ). After performing the calculations, we find that the collector current for Q1 is 2mA. This signifies how much amplified current is passing through the transistor when it is properly biased and functioning.

Imagine filling a water tank with a hose (the supply voltage). The base-emitter voltage acts like a valve that's slightly closed, reducing the amount of water that can flow through. The resistor (570kΩ) is similar to a section of narrow pipe, which also limits the water flow. The overall current (2mA or water flow) is determined by the pressure in the tank (supply voltage) minus the pressure lost due to the closed valve (base-emitter voltage).

To get the DC voltage at the output node say V what we can do? we can compare β I × ( ) of transistor-1 = β I of transistor-2 × ( ) of transistor-2. Now this part and this part we have seen here they are equal. So, now, just by equating this factor what we can get here it is since the early voltage of the 2 transistors they are equal. So, that gives us V = V and also we know that V + V = V . So, this is V of transistor-2, this is V of transistor-1 that is V which is 12 V and from that we can say that both of them are equal and they are equal to 6 V.

This part focuses on deriving the output DC voltage (V_OUT) at the output node of the amplifier. By equating the output currents through both transistors, we establish the relationship between the voltages across them. Given similar early voltages, we can conclude that the voltages across both transistors must be equal when the currents are equal. By analyzing the circuit's total supply voltage (12V), we determine that each transistor (both in a balanced state) will have an output voltage of 6V.

Imagine you have two identical water tanks connected by a hose (the transistors). If both tanks are filled evenly (the collector currents are balanced), and the total pressure from the water supply is evenly divided, then each tank will effectively have the same water level (output voltage). So, with a supply pressure of 12 units, each would hold 6 units of water at equilibrium.

So, in summary what we can say that the cutoff frequency it is getting reduced gain got increased and output resistance also got increased of course, the input capacitance also got increased. So, to summarize the performance of this CE amplifier with active load probably we can see in the next slide and then we can compare performance of CE amplifier having passive load. So, here we do have the table to write different parameters namely yeah. So, we do have voltage gain in case circuit load CE amplifier load if it is active load voltage gain what we said is 1923.

In this concluding part, we summarize the findings on performance metrics for the CE amplifier with an active load compared to one with a passive load. Active load designs typically yield increased voltage gain, a higher output resistance, and more input capacitance. However, this often results in reduced bandwidth or cutoff frequency. Here, for the active load configuration, the voltage gain is documented at 1923—significantly higher compared to passive load configurations.

Think of the amplifier as a sport engine—when you make it more powerful (using an active load), it can run faster (higher gain) but also consumes more fuel (increased input capacitance), and the speed limit on most roads (cutoff frequency) is still an unavoidable factor you have to consider, leading to the conclusion that with greater power, one must also be prepared for the consequences.

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