A standard dc current-limiting power supply shown in Fig. 2.29a provides 0–18 V at 3 A to a load. The voltage drop, VR, across a resistor, R, is used as a current-sensing device, fed back to the power supply and used to limit the current I. That is, if the load is adjusted so that the current to

Question

A standard dc current-limiting power supply shown in Fig. 2.29a provides 0–18 V at 3 A to a load. The voltage drop, [latex]V_{R}[/latex], across a resistor, R, is used as a current-sensing device, fed back to the power supply and used to limit the current I. That is, if the load is adjusted so that the current tries to exceed 3 A, the power supply will act to limit the current to that value. The feedback voltage, [latex]V_{R}[/latex], should typically not exceed 600 mV. If we have a box of standard 0.1-Ω , 5-W resistors, let us determine the configuration of these resistors that will provide [latex]V_{R}[/latex] = 600 mV when the current is 3 A.

Accepted Answer

Using Ohm’s law, the value of R should be

R = [latex]\frac{V_{R}}{I}[/latex]
= [latex]\frac{0.6}{3}[/latex]
= 0.2 Ω

Therefore, two 0.1-Ω resistors connected in series, as shown in Fig. 2.29b, will provide the proper feedback voltage. Suppose, however, that the power supply current is to be limited to 9 A. The resistance required in this case to produce [latex]V_{R}[/latex] = 600 mV is

R = [latex]\frac{0.6}{9}[/latex]
= 0.0667 Ω

We must now determine how to interconnect the 0.1-Ω resistor to obtain R = 0.0667 Ω. Since the desired resistance is less than the components available (i.e., 0.1-Ω ), we must connect the resistors in some type of parallel configuration. Since all the resistors are of equal value, note that three of them connected in parallel would provide a resistance of one-third their value, or 0.0333 Ω. Then two such combinations connected in series, as shown in Fig. 2.29c, would produce the proper resistance.

Finally, we must check to ensure that the configurations in Figs. 2.29b and c have not exceeded the power rating of the resistors. In the first case, the current I = 3 A is present in each of the two series resistors. Therefore, the power absorbed in each resistor is

P = [latex]I^{2}[/latex]R
= [latex](3)^{2}[/latex](0.1)
= 0.9 W

which is well within the 5-W rating of the resistors.
In the second case, the current I = 9 A. The resistor configuration for R in this case is a series combination of two sets of three parallel resistors of equal value. Using current division, we know that the current I will split equally among the three parallel paths and, hence, the current in each resistor will be 3 A. Therefore, once again, the power absorbed by each resistor is within its power rating.

Question 2.21: A standard dc current-limiting power supply shown in Fig. 2....

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