Page 130 - Op Amps Design, Applications, and Troubleshooting

Current Amplifier 113

The value of current (ij) can also be computed by Ohm's Law as

Substituting this into the equation for load current produces

Factoring f/ gives us the equation for I L:

In this form, it is easy to see that we do indeed have a current amplifier. That is,
the input current (f/) is multiplied by a constant (Rz/Ri + 1) to produce the output
or load current. The constant is the current gain of the circuit, shown below.

In the case of the circuit in Figure 2.30, the current gain is calculated as
shown:

It is especially important to note that the value of load current is independent of
the value of load resistor. That is, the op amp circuit is acting as a current
source.
Although many current sources are essentially EC (e.g., transducers), there
may be an application requiring current amplification at higher frequencies. As
the frequency of operation is increased, the actual current gain will begin to
decrease from the low frequency value calculated. This effect is caused by the
reduction in open-loop op amp gain as the input frequency is increased. The higher
the value of current gain (A/), the more significant the effects of op amp voltage
gain variations.

Load Currant. The input current (ij) for the circuit in Figure 2.30 is indicated to
be in the range of 10-50 microarnps. The output current can be found by transpos-
ing the basic current gain equation, Equation (2.1).