 # Widlar Op-Amp Current Source for Grounded Load The simple current source is imperfect as the load affects the current. The voltage drop VL across the load (suppose that it is a varying resistor R) is harmful as it enervates the excitation voltage V thus decreasing the current I.

According to the basic idea above, we change continuously the voltage Vvar by following the voltage drop VL across the load, in order to compensate it. In this way, Vvar acts as a following voltage source keeping constant voltage difference Vvar - VL, which produces a constant current. Graphically presented, when the load increases its resistance RL, the IV curve rotates clockwise round the zero of the coordinate system. As the voltage source Vvar reacts to this intervention by increasing the voltage, its IV-characteristic moves horizontally to right remaining parallel to itself.

Vice versa, if the load decreases its resistance RL, the IV curve rotates contraclockwise round the zero. Now, the voltage source Vvar reacts to this intervention by decreasing the voltage; so, its IV-characteristic moves horizontally to the left remaining again parallel to itself.

In the two cases, the working point A slides horizontally over the dynamic IV-characteristic of the constant current source obtained.

Implementation: Using an op-amp follower as a following voltage source. We may see this idea in the Widlar's bilateral current source. In this clever circuit, the following voltage Vvar is produced by summing two voltages - the reference voltage VREF and the load voltage drop VL.

As a result, the potential of the point B follows the potential of the point A, the reference voltageVREF is applied across the resistor R (VR = Vvar - VL = VREF + VL - VL = VREF) and a steady current I = VREF/R flows through the load. Well, let's now see how this idea is implemented into a circuit. Viewpoint 1. The input voltage VREF applied to the inverting amplifier R1-R2-OA creates initial excitation voltage at point B.

The voltage drop variations across the load (point A) are attenuated (R3+R4)/R3 times by the voltage divider R3-R4 and then amplified (R1+R2)/R1 times by the non-inverting amplifier R1-R2-OA. As R1=R3 and R2=R4 the equivalent ratio is K = 1; so, the potential of the point B follows the potential of the point A.

Viewpoint 2. The resistors R1, R2, R3, R4 and the op-amp OA form a differential amplifier. It acts as a summer needed, as the two input voltages VREF and VL have different polarities.