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Take another look at the non-inverting amplifier.
A non-inverting amplifier with equal feedback resistors and 2 volts at the non-inverting input. |
The voltage applied to the non-inverting input is +2 V. The left end of R1 is tied to ground (0 volts) and the right end is tied to the inverting input. The op-amp will cause the inverting input to have the same voltage as the non-inverting input, which is +2 V. As long as the voltage at the non-inverting input is +2 V, R1 will have 2 volts across it. With a resistance of 1 k and 2 V across it, R1 will have 2 mA of current going through it. Where does this current come from? It comes from the output of the op-amp then through R2 (no current can come from the inverting input). R1 and R2 are in series, so they have the same current. Ultimately, the current through R2 is determined by the voltage at the non-inverting input and the value of R1. It doesn't matter how much resistance R2 has. As long as the output voltage of the op-amp doesn't hit one if its limits, R2 will have 2 mA flowing through it.
Let's say we want 50 mA of current to flow through some device (that device would be called the load since it is the ultimate consumer of the power). All we have to do is replace R2 with that device. Next, we have to choose a voltage for the non-inverting input combined with a resistor for R1 that results in 50 mA through R1. That 50 mA will first go through the load.
This circuit puts 50 mV across a 1-ohm resistor to get 50 mA to flow through the load. |
A typical op-amp cannot deliver 50 mA. That problem can be solved by simply adding a transistor to the op-amp's output.
If the op-amp cannot source 50mA, a transistor can be added to the circuit. The output of the op-amp will automatically compensate for the transistor. |
A practical circuit using a current controller would be a NiCd battery charger. A typical AA NiCd cell requires 50 mA of current to charge in 16 hours. In the above example, 50mV is applied to the non-inverting input. This puts 50mV across the 1-ohm resistor which results in 50mA flowing through it. Replace the load with one or more NiCd cells in series (positive end toward the op-amp output) and 50mA will flow through the battery, charging it at the proper 16-hour rate.
A nickel-cadmium battery charger. This is the same circuit as above, just drawn in a different style that some people prefer. R2 has been replaced with a NiCd battery. |
50mA is a little more current than a typical op-amp can supply. If more current is needed, a transistor can be put in the circuit as above. The specifications of the transistor are unimportant except that it must be able to handle whatever current is delivered to the load. The op-amp will make the base voltage whatever it takes to make the inverting input equal the non-inverting input. The current through the load will still be determined by the voltage at the non-inverting input and the 1-ohm resistor.
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