On Wed, 2 Jul 1997 18:24:28 +0200 EJA van Es writes: >Hi! > >Anyone know how to build a constant current source using Op-amps. I >know >how to do it with transistors, but I want to try it with op-amps. > If you can put up with a floating load, just put the load between the output of an op-amp and the inverting input, then put a resistor from the inverting input to ground. Apply a voltage to the non-inverting input. Since the inputs will "follow each other", the inverting input will be at about the same voltage, causing I=V/R through the resistor to ground. Since an ideal op amp has no input current, this current also goes through the load. You can also ground the noninverting input and apply a voltage to the end of the resistor that was grounded. The "happy op amp" will keep the two inputs together, keeping the inverting input at ground. The current through the resistor and the load will be I=V/R, since the voltage at one end of the resistor is zero. This is the basis of an inverting integrator. The controlled current flows through the capacitor (the load). Since the inverting input is at ground, the op amp output voltage is the same as the capacitor voltage. Another circuit with a grounded load is described below (my ascii art isn't very good). Imagine two equal value resistors in series. Connect the load from the junction of the two resistors to ground. Connect the reference voltage to one end of the series resistors. Connect the junction of the resistors (and the load) to the input of a noninverting amp with a gain of 2. Connect the output of the amp to the other end of the two resistor network. Clever circuit! Try a few ohm's law analysis runs on it. Start with a load of zero ohms to see that the current is Vin/R. Then assume the circuit works and use ohm's law to find the voltage at the top of the load (assuming a resistor), then determine the voltage at the output of the amp (twice the load voltage). Apply Kirchoff's current law. It works! Harold