WEBVTT NOTE Power Supplies and Voltage Regulators 1 00:00:00.000 --> 00:00:01.987 Two ways to regulate. 2 00:00:01.987 --> 00:00:08.846 Question. How does a linear electronic voltage regulator work? 3 00:00:09.746 --> 00:00:17.266 Answer. The conduction of a control element is varied to maintain a constant output voltage. 4 00:00:17.866 --> 00:00:22.866 Question. How does a switchmode voltage regulator work? 5 00:00:23.766 --> 00:00:29.566 Answer. By varying the duty cycle of pulses input to a filter. 6 00:00:30.166 --> 00:00:38.230 A linear regulator burns off the difference in a pass element, quiet, simple, and inefficient. 7 00:00:38.230 --> 00:00:49.726 A switching regulator chops the input and filters the result, adjusting the duty cycle to set the output, efficient, small, and noisy. 8 00:00:49.726 --> 00:00:58.855 Question. Why is a switching type power supply less expensive and lighter than an equivalent linear power supply? 9 00:00:59.755 --> 00:01:10.161 Answer. The high frequency inverter design uses much smaller transformers and filter components for an equivalent power output. 10 00:01:10.761 --> 00:01:19.342 The same argument as the General lesson: magnetics shrink with frequency. Series versus shunt. 11 00:01:19.342 --> 00:01:26.102 Question. Which of the following describes a three-terminal voltage regulator? 12 00:01:27.002 --> 00:01:30.442 Answer. A series regulator. 13 00:01:31.042 --> 00:01:41.545 Question. Which of the following types of linear voltage regulator operates by loading the unregulated voltage source? 14 00:01:42.445 --> 00:01:45.308 Answer. A shunt regulator. 15 00:01:45.908 --> 00:01:56.425 A series regulator puts the control element in series with the load and varies its resistance, the familiar 78xx three-terminal part. 16 00:01:56.425 --> 00:02:12.873 A shunt regulator sits across the load and draws whatever current the load does not, keeping the total constant. Simpler, but it dissipates most when the load draws least, so it is only sensible at low power. 17 00:02:12.873 --> 00:02:16.748 A Zener used as a regulator is a shunt regulator. 18 00:02:16.748 --> 00:02:22.868 Question. What device is used as a stable voltage reference? 19 00:02:23.768 --> 00:02:27.168 Answer. A Zener diode. 20 00:02:27.768 --> 00:02:37.680 Its reverse breakdown voltage is sharply defined and stable, which is exactly what a reference needs. Reading figure E7-2. 21 00:02:37.680 --> 00:02:46.860 This question refers to one of the exam's circuit diagrams, which you will need to look at in the written lesson. 22 00:02:46.860 --> 00:02:53.140 Question. What type of circuit is shown in Figure E7-2? 23 00:02:54.040 --> 00:02:57.760 Answer. Linear voltage regulator. 24 00:02:58.360 --> 00:03:06.791 Question. What is the purpose of Q1 in the circuit shown in Figure E7-2? 25 00:03:07.691 --> 00:03:13.691 Answer. It controls the current to keep the output voltage constant. 26 00:03:14.291 --> 00:03:23.131 Q1 is the pass transistor, the control element in series with the load, exactly as the definition describes. 27 00:03:23.131 --> 00:03:31.213 Question. What is the purpose of C2 in the circuit shown in Figure E7-2? 28 00:03:32.113 --> 00:03:38.433 Answer. It bypasses rectifier output ripple around D1. 29 00:03:39.033 --> 00:03:52.694 D1 is the Zener reference, and any ripple appearing across it would be amplified straight through to the output. C2 shunts that ripple to ground, keeping the reference clean. 30 00:03:52.694 --> 00:04:02.273 This is why a good linear regulator has a capacitor across its reference and a poor one does not. Dissipation and dropout. 31 00:04:02.273 --> 00:04:09.673 Question. Which of the following calculates power dissipated by a series linear voltage regulator? 32 00:04:10.573 --> 00:04:18.493 Answer. Voltage difference from input to output multiplied by output current. 33 00:04:19.093 --> 00:04:33.835 P equals (V in minus V out ) times I out The pass element carries the full load current with the full voltage difference across it, and P = E times I. 34 00:04:33.835 --> 00:04:43.467 A 5 V regulator fed from 15 V at 1 A dissipates (15−5) times 1 = 10 watts, twice what it delivers. 35 00:04:43.467 --> 00:04:59.585 That single calculation explains every oversized heat sink you have ever seen, and why switching regulators displaced linear ones for anything drawing real current. 36 00:04:59.585 --> 00:05:05.185 Question. What is the dropout voltage of a linear voltage regulator? 37 00:05:06.085 --> 00:05:12.942 Answer. Minimum input-to-output voltage required to maintain regulation. 38 00:05:13.542 --> 00:05:20.713 Fall below it and the regulator stops regulating and simply follows the input. 39 00:05:20.713 --> 00:05:38.734 A standard 78xx needs about 2 V; a low-dropout part manages a few hundred millivolts, which matters when running a 12 V regulator from a battery that sags to 12.5. High-voltage supply refinements. 40 00:05:38.734 --> 00:05:44.342 Two features specific to the supplies inside tube amplifiers. 41 00:05:44.342 --> 00:06:00.189 Question. What is the purpose of connecting equal-value resistors across power supply filter capacitors connected in series? A. Equalize the voltage across each capacitor. B. 42 00:06:00.189 --> 00:06:11.482 Discharge the capacitors when voltage is removed. C. Provide a minimum load on the supply. D. All these choices are correct. 43 00:06:12.382 --> 00:06:15.902 Answer. All these choices are correct. 44 00:06:16.502 --> 00:06:32.557 Capacitors in series divide voltage according to their leakage, not their capacitance, so without equalising resistors one may take far more than its rated share and fail, cascading into the others. 45 00:06:32.557 --> 00:06:37.422 The same resistors double as bleeders and as a minimum load. 46 00:06:37.422 --> 00:06:44.942 Question. What is the purpose of a step-start circuit in a high-voltage power supply? 47 00:06:45.842 --> 00:06:51.695 Answer. To allow the filter capacitors to charge gradually. 48 00:06:52.295 --> 00:06:57.782 At switch-on, uncharged filter capacitors look like a short circuit. 49 00:06:57.782 --> 00:07:15.455 A step-start inserts a resistor for a moment, then shorts it out once the capacitors have partly charged, sparing the rectifiers, the transformer, and the mains breaker from a very large inrush. Batteries and inverters. 50 00:07:15.455 --> 00:07:20.153 Question. How is battery operating time calculated? 51 00:07:21.053 --> 00:07:26.866 Answer. Capacity in amp-hours divided by average current. 52 00:07:27.466 --> 00:07:32.953 The General figure, restated. Average, not peak. 53 00:07:32.953 --> 00:07:39.993 Question. What is the purpose of an inverter connected to a solar panel output? 54 00:07:40.893 --> 00:07:46.627 Answer. Convert the panel's output from DC to AC. 55 00:07:47.227 --> 00:08:05.026 An inverter makes AC from DC. It is not a charge controller and not a regulator, three different devices for three different jobs, and the solar questions in the General track ask about the other two. Check yourself. 56 00:08:05.026 --> 00:08:17.882 A linear regulator delivers 12 V at 3 A from an 18 V input. Dissipation? Why do series filter capacitors in a high-voltage supply need equalising resistors? 57 00:08:17.882 --> 00:08:26.287 Your 12 V regulator loses regulation when the battery sags to 12.8 V. What specification have you hit? 58 00:08:32.287 --> 00:08:47.008 (18 − 12) times 3 = 18 watts in the pass element, half again what it delivers. Series capacitors divide voltage by leakage, not capacitance, so one can exceed its rating and fail. 59 00:08:47.008 --> 00:09:03.949 The resistors equalise the division, bleed the charge, and provide a minimum load. Dropout voltage, the minimum input-to-output difference needed to regulate. A low-dropout regulator would still be working.