Till now we completed the basic but major part of Analog Electronics. I hope you really enjoyed my posts and my courses as well. In this post, we will see the concept of current mirror circuits and it's application. I had seen most of the students struggling with the significance of the current mirror circuit, so I will try my best to explain the concept and the significance, i.e. why we are using the current mirror circuit and when we need it. Before starting the concept let me ask you one question, let's suppose I have one load resistance assume resistance as RL = 10Kohms, and to that load, I need to provide a fixed amount of current, assume the value of current as I = 1mA. So what process do you follow to provide fix amount of current? So you will tell me it is very simple, we know the value of current i.e. I = 1mA, and the value of resistance RL = 10Kohms based on this information we can easily calculate the voltage. Therefore Voltage(V) = I * RL = 1mA * 10Kohms = 10V So