Question 6.3: A sinusoidal generator with open−circuit voltage EG = 1 V ha......
A sinusoidal generator with open-circuit voltage E_G=1\text{ V} has internal resistance R_G=50 \; \Omega. The generator is connected with a small-signal load R_L=-25 \; \Omega (provided, e.g., by a device with a local negative differential resistance). Analyze the power transfer within the circuit.
The current flowing in the circuit (positive out of the generator and positive entering the load) is:
I=\frac{E_G}{R_G+R_L}=\frac{1}{50-25}=40\text{ mA}.
The voltage on the internal generator resistance will be V_G=R_G I = 50·0.04 = 2 V while the voltage on the load resistance will be V_L=R_L I = -25·0.04 = -1 V. Thus, the power transfer within the circuit is as follows: the generator supplies a power P_G=E_G I=1 \cdot 0.04=40\text{ mW}; the internal generator resistance absorbs a power P_{R_G}=2 \cdot 0.04=80\text{ mW}, and the load resistance absorbs a negative power P_L=-1 \cdot 0.04=-40\text{ mW}, i.e., supplies 40 mW. In other words the circuit is formally stable because the loop resistance is not zero (it is in fact positive), but the load supplies power rather than absorbing it. (From the energetic standpoint the power supplied by the negative resistance is actually converted from the DC bias of the device providing the negative differential resistance.) If the load is the input of an amplifier, clearly this is not the operation corresponding to amplification as we commonly understand – it is rather the principle of the so-called reflection amplifiers.