An ideal current-controlled voltage source has an output voltage that (1) is equal to a constant (k) times the magnitude of an independent current: Vo = kl., and (2) is independent of the load connected to it. Here, the constant k has the units of ohms. A current-controlled voltage source can be thought of as a current-to-voltage converter, since output voltage is proportional to input current. It is useful in applications where current measurements are required, because it is generally more convenient to measure voltages.
A very simple current-controlled voltage source. Since no current flows into the – input, the controlling current lilt is the same is the current in feedback resistor R. Since v- is virtual ground,
Once again, the fact that the amplifier has zero output resistance implies that the output voltage will be independent of load.
Figure 14-15 shows a noninverting, current-controlled voltage source in which the controlling current has a return path to ground. Since V/ = liltRs, we have
Current-Controlled Current Sources
An ideal current-controlled current source if) one that supplies a current whose magnitude (1) equals a fixed constant (k) time. the value of an independent controlling current: I.. == k l., and (2) is independent of the load to w••ich the current is supplied. Note that k is dimensionless, . ince it is the ratio of two currents. Figure 14-16 shows a current-controlled curr nt sour ~e Nith floatrng load RI Since; into the – input, the current in Rl must equal ijn• Since vis at virtual ground, the voltage V2 is
An electronic differentiator produces an output waveform whose value at any instant of time is equal to the rate of change of the input at that point in time. In many respects, differentiation is just the opposite, or inverse, operation of integration. In fact, integration is also called “antidifferentiation.” If we were to pass a signal through an ideal integrator in cascade with an ideal differentia tor, the final output would be exactly the same as the original input signal.
Demonstrates the operation of an ideal electronic differentiator. In this example, the input is the ramp voltage Vjll = Et. The rate of change, or slope
Of this ramp is a constant E volt/second. (For every second that passes, the signal increases by an additional E volt.) Since the rate of change of the input is constant, we see that the output of the diffcrcntiator is the constant de level E volts. The standard symbol used to represent differentiation of a voltage u is du/dt. (This should not be interpreted as a fraction. The tit in the denominator simply means that we are finding the rate of change of v with respect to time, I.) In the example shown in Figure 14-25, we would write
Note that the derivative of a constant (de level) is zero, since a constant does not change with time and therefore has zero rate of change. Figure 14-26 shows how an ideal diflcrcutiator is constructed using an operational amplifier. Note that we now have a capacitive input and a resistive feedbackagain, just the opposite of an integrator. It can be shown that the output of, this differentiator is
Thus, the output voltage is the (inverted) derivative of the input, multiplied by the constant RfC. If the ramp voltage in Figure 14-25 were applied to the input of this differentiator, the output would be a negative de level. Readers not familiar with calculus can skip the prescn parugreph, in which we show how the circuit of Figure 14-26 performs differentiation. Since the current into the -: terminal is 0, we have, from Kirchhoff’s current law,
Equation 14-57 shows that when the input is sinusoidal, the amplitude of the output of II differentiator is directly proportional to frequency (once again, just the opposite of an integrator). Note also that the output lags the input by 90°, regardless of frequency. (In practice, non inverting differentia tor circuits are used to introduce plase a systern.) ne gull. ul the differentiator.