Question

The problem is about op-amp voltage amplifiers and signal combining. 1.a You have two DC voltage supplies ±9 V, unlimited supply of standard resistors, two 741 opamps and a single voltage signal of v1​(t)=2sin(2π⋅600t)V. Create two new signals v2​(t)=4.5sin(2π⋅600t)V and v3​(t)=−0.5sin(2π⋅600t)V. Each of the new signals, showing up at each op-amp output, is applied to a load resistor: v2​(t) is connected to a resistor RL2​ whose other side is in ground, and v3​(t) is connected to RL3​. How small is each of the loads, RL2​ and RL3​, allowed to be before op-amp current limit problems arise? Demonstrate.

Answer 1

To avoid exceeding the 25 mA current limit of the 741 opamps, the smallest allowed load resistor for the v2(t) signal is 180Ω and for the v3(t) signal is 20Ω.

Step-by-step explanation:

To create two new signals v2(t)=4.5sin(2π∙600t)V and v3(t)=-0.5sin(2π∙600t)V from an initial signal v1(t)=2sin(2π∙600t)V, we would need to setup two operational amplifier (op-amp) circuits. The 741 opamps can be used for this task, and typical configurations would be a non-inverting amplifier for v2(t) with a gain of 2.25, and an inverting amplifier for v3(t) with a gain of -0.25.

The current limit for a 741 opamp is typically around 25 mA. The load resistors RL2 and RL3 must be chosen such that the opamps do not exceed this limit. For signal v2(t), the peak output voltage is 4.5 V, so using Ohm's law (I = V/R), to keep the current below 25 mA, the smallest allowed value for RL2 would be RL2 ≥ 4.5V / 0.025A = 180Ω. Similarly, for v3(t), with a peak of 0.5 V, the smallest RL3 would be RL3 ≥ 0.5V / 0.025A = 20Ω.