![]() I've provided a schematic on how to realize this. A third amplifier can then generate a static 10V signal which when used as a reference to the 0-20V signal gives a voltage range of +/-10V. To generate a +/-10V signal we can cheat a little and modify our amplifier circuit to amplify by a factor of 4 to give 0-20V output. This can be achieved easily by amplifying the 0-5V signal we had form out 0-20mA cct by a factor of 2 to generate a 0-10V output signal. In addition to that we'd like a 0-10V or +/-10V output. Looking at the datasheet for the LM324 we can see it's capable of driving 30mA and can therefore be used as the basis of our simple current source without an additional drive transistor. We can therefore construct a circuit that converts a voltage to a current. ![]() If V(R1) = 5V then I(R1) = 5/250 = 20mA and since RL forms a series cct (no current flow into (-) terminal) with this it must also have 20mA flowing through it. The only place for this to come from is the op amps output therefore the op amp sources enough current to ensure the negative terminal is at 5V. Since the voltages at an opamps input are equal when the positive terminal is 5V the negative terminal must be likewise. The above schematic details a voltage to current converter. ![]()
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