We wish to design an efficient electric heater that operates from a 24-V dc source and creates heat by driving a 1-Ω resistive heating element. For the temperature range of interest, the power absorbed by the heating element should be between 100 and 400W. An experienced engineer has suggested

Question

We wish to design an efficient electric heater that operates from a 24-V dc source and creates heat by driving a 1-Ω resistive heating element. For the temperature range of interest, the power absorbed by the heating element should be between 100 and 400 W. An experienced engineer has suggested

Accepted Answer

In the first case, the required network is shown in Fig. 7.44. The variable resistance element is called a rheostat. Potentiometers are variable resistors that are intended for low power (i.e., less than 1 W) operation. Rheostats, on the other hand, are devices used at much higher power levels.
We know from previous work that the heating element voltage is

[latex]V_{o} = V_{S} \frac{R_{he}}{R_{he} + R_{adj}}[/latex] 7.34

Changing [latex]R_{adj}[/latex] will change the voltage across, and the power dissipated by, the heating element. The power can be expressed as

[latex]P_{o} = \frac{V^{2}_{o}}{R_{he}} = V^{2}_{S} \frac{R_{he}}{ (R_{he} + R_{adj})^{2}}[/latex] 7.35

By substituting the maximum and minimum values of the output power into Eq. (7.35), we can determine the range of resistance required for the rheostat:

[latex]R_{adj, min} = \sqrt{\frac{V^{2}_{S}R_{he}}{P_{o, max}}} - R_{he} = \sqrt{\frac{(24^{2})(1)}{400}}[/latex] - 1 = 0.2 Ω 7.36

[latex]R_{adj, max} = \sqrt{\frac{V^{2}_{S}R_{he}}{P_{o, min}}} - R_{he} = \sqrt{\frac{(24^{2})(1)}{100}}[/latex] - 1 = 1.4 Ω

So, a 2-Ω rheostat should work just fine. But what about the efficiency of our design? How much power is lost in the rheostat? The rheostat power can be expressed as

[latex]P_{adj} = \frac{(V_{S} - V_{o})^{2}}{R_{adj}} = V^{2}_{S} \frac{R_{adj}}{(R_{he} + R_{adj})^{2}}[/latex] 7.37

We know from our studies of maximum power transfer that the value of [latex]R_{adj}[/latex] that causes maximum power loss, and thus the worst-case efficiency for the circuit, occurs when [latex]R_{adj} = R_{he}[/latex] = 1 Ω. Obviously, the resistances consume the same power, and the efficiency is only 50%.
Now that we understand the capability of this voltage-divider technique, let’s explore the alternative solution. At this point, it would at least appear that the use of a switched inductor is a viable alternative since this element consumes no power. So, if we could set up a current in an inductor, switch it into the heating element, and repeat this operation fast enough, the heating element would respond to the average power delivered to it and maintain a constant temperature.
Consider the circuit in Fig. 7.45, where the switch moves back and forth, energizing the inductor with current, and then directing that current to the heating element. Let’s examine this concept to determine its effectiveness. We begin by assuming that the inductor current is zero and the switch has just moved to position 1. The inductor current will begin to grow linearly in accordance with the fundamental equation

[latex]i_{L}(t) =\frac{1}{L} \int v_{L}(t) dt =\frac{1}{L} \int V_{S}(t) dt = \frac{V_{S}}{L} t[/latex] 7.38

Note that [latex]V_{S}[/latex]/L is the slope of the linear growth. Since [latex]V_{S}[/latex] is set at 24 V, we can control the slope with our selection for L. The inductor current increases until the switch moves at time t = [latex]t_{1}[/latex], at which point the peak current is

[latex]I_{peak} = \frac{V_{S}}{L}t_{1}[/latex] 7.39

This inductor current will discharge exponentially through the heating element according to the equation

[latex]i_{L}(t^{\prime}) = I_{peak}e^{-t^{\prime}/ au}[/latex] 7.40

where [latex]t^{\prime}[/latex] is zero when the switch moves to position 2 and [latex] au = L/R_{he}[/latex]. If the switch is maintained in position 2 for about 5 time constants, the inductor current will essentially reach zero and the switch can return to position 1 under the initial condition—zero inductor current. A sketch of the inductor current over a single switching cycle is shown inFig. 7.46. Repeated switching cycles will transfer power to the heating element. If the switching period is much shorter than the element’s thermal time constant—a measure of how quickly the element heats up—then the element’s temperature will be determined by the average power. This is a concept we don’t understand at this point. However, we will present Average Power in Chapter 9, and this example provides at least some motivation for its examination. Nevertheless, we should recognize two things: the load current is just the exponential decaying portion of the inductor current, and the initial value of that exponential is [latex]I_{peak}[/latex], as defined in Eq. (7.39). Increasing [latex]I_{peak}[/latex] will increase the element’s power and temperature. As Eq. (7.39) indicates, this is easily done by controlling [latex]t_{1}[/latex]! It is impossible to proceed with the design until we can accurately predict the average power at the load.

Returning to our original concern, have we improved the efficiency at all? Note that there are no power-consuming components in our new circuit other than the heating element itself. Therefore, ignoring resistance in the inductor and the switch, we find that our solution is 100% efficient! In actuality, efficiencies approaching 95% are attainable. This is a drastic improvement over the other alternative, which employs a rheostat.

Question 7.18: We wish to design an efficient electric heater that operates...

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