|
The primary advantage of switching-mode power supplies is they can accomplish power conversion and regulation at 100% efficiency -- given ideal parts. All power loss is due to less than ideal parts and the power loss in the control circuitry. In this section we explore some of the switching-mode power supplies that can be constructed using only one each of the parts from our simple parts list.
One of the simpler switching-mode power supplies is one that is used to control a heater. For example, an inexpensive space heater I have applies full household ac voltage to a heater element when the temperature is less than that set by a simple thermostat and turns the ac power off when the temperature is above the set point. The heater turns on and off every several minutes to keep the room temperature constant. Closer to our example might be a heater in a crystal-oscillator oven used to keep the temperature of the crystal within narrow limits. For this circuit, we only need to use the switch from our list of components. The schematic is shown in Figure 3-1. 
Figure 3-1: Heater Schematic
As with our previous examples, Vin is 12 Vdc and the load resistor R2 is 0.25 ohms. The objective is to open and close the switch so that the average voltage across R2 is 5 Vdc. The waveform of the voltage across R2 is shown in Figure 3-2. 
Figure 3-2: Heater Waveform
This is the first waveform we have encountered in this tutorial. In your design of switching-mode power supplies you will have to work with many waveforms and calculate properties such as period, frequency, duty cycle, harmonics, average and rms (root-mean-square) values -- both with and without the dc component, etc. Unless you really like solving definite integrals and doing Fourier analysis, you will want a handbook that gives you characteristics of common pulse and periodic waveforms. In Figure 3-2, the period, T, of the waveform is Ton + Toff, and by definition, the reciprocal of the period is the frequency. For example, a period of 50 us results in a frequency of 20 kHz. The ratio of Ton/(Ton + Toff) is called the Duty cycle, D, -- a parameter much used in switching-mode power supply calculations. The average value of the waveform over a period is shown by a dotted line. For those who have had calculus you will remember the average value of a function is (1/T)*(Integral zero to T of the function). In this case you have to evaluate the integral from 0 to Ton and from Ton to T and add them. Or you can look it up in a handbook and see the average is Vin*D. Again, those with calculus know to get the RMS (Root-Mean-Square) of a function you first square it and then take the mean or average of the result as before. You then take the square root to get the answer. Or you can look it up in a handbook and see the RMS is Vin*SQRT(D). Example:
Vin = 12 Vdc
Vo = 5 Vdc (average)
D = (5 V)/(12 V) = 0.417
Vrms=(12 V)*SQRT(0.417) = (12 V)*0.645 = 7.75 Vrms.
Iaverage = (0.417)*(12 V)/(0.25 ohm) = 20 A
Pin = (12 V)*(20 A) = 240 W These are the answers we got in our series and shunt regulator examples and are no surprise. We might expect that the power in R2 is the same as before, (5 V)*(20 A) = 100 W, but that doesn't make sense, since we are taking 240 W from the source and there is nowhere it can go except into R2. Recalling that power in a resistor is (Vrms)*(Vrms)/R, and the rms voltage is 7.75 Vrms, then we get (7.75 Vrms)(7.75 Vrms)/(0.25 ohms) = 240 W in R2, and everything balances. Notice that the rms value of the waveform is higher than the average value. This is true in all duty cycle controlled switching-mode power supplies. Also notice that all the power taken from the source is delivered to the load assuming ideal components. This will be true in all the examples discussed in this section, however, there is one notable exception. The exception is if you try to switch directly into a capacitor, which we discuss next. Editor: Jerrold Foutz |
