transformer primary voltage load current ambient temperature.

Transcription

1 INTODUCTION In the previous lesson it was shown how the use of a smoothing capacitor in a d.c. power supply can produce a relatively steady d.c. output voltage for any particular load conditions. The output voltage will, however, vary with varying load conditions. If for example the load were to draw more current then the average value of the output voltage will fall and, moreover, its ripple content will increase. Also, if the supply voltage were to vary then this too will be reflected in a variation of the output voltage. This type of supply is called an unregulated power supply. In an unregulated power supply the d.c. output voltage alters if there are changes in any of the following parameters: transformer primary voltage load current ambient temperature. The supply voltage for many electronic devices requires a constant d.c. voltage that is independent of the above factors. To power such devices a regulated power supply is required. A regulated power supply contains a voltage regulation stage that keeps the output voltage more constant despite changes in load, supply voltage and temperature.

2 2 YOU AIMS On completion of this lesson you should be able to: explain as analyse the action of a Zener diode voltage regulator explain the enhanced action of a transistor stabiliser circuit analyse the action of a voltage stabiliser employing an operational amplifier explain and analyse the features of a three-pin integrated circuit voltage regulator.

3 3 STUDY ADICE eference is made to the classic feedback equation derivation of this equation is given below. θ OUT θ IN G + GH. The θ IN x + G θ OUT H FIG. With referenced to the block diagram of FIGUE : θ Gx and x θ Hθ OUT IN OUT Thus: ( ) θ G θ Hθ OUT IN OUT θ Gθ GHθ OUT IN OUT θ + GHθ OUT θ + OUT OUT ( GH ) Gθ Gθ IN IN θout θ IN ( G + GH )

4 4 EGULATED POWE SUPPLY Most pieces electronic equipment will require a power supply to deliver a constant d.c. supply voltage that is independent of both the variations in supply voltage and the load; that is to say, they require a regulated supply. FIGUE 2 shows the block diagram of a regulated voltage power supply. Mains Transformer ectifier Smoothing oltage regulator Load FIG. 2 The forth stage of a regulated power supply is a voltage regulation stage that keeps the output voltage more constant. The use of a regulator stage has the added advantage of significantly reducing the ripple voltage from a smoothing circuit. This allows the smoothing capacitor to be of smaller value. This is beneficial, not just because large-valued capacitors are, of course, bigger in size and more expensive but also because large values of capacitance give rise to large values of peak current. This places greater demands upon other components, particular the rectifying diodes, whose ratings must be such as to meet these surges in current. We begin the lesson by investigating the simplest form of regulation the Zener diode voltage stabilizer circuit.

5 5 ZENE DIODE STABILISE If the reverse bias voltage across an ordinary diode is increased sufficiently, the diode will break down and be permanently damaged. There are, however, special diodes called Zener diodes which are specially designed to operate in the 'break down' mode. Such diodes make use of high doping levels to achieve artificially low reverse breakdown voltages. The Zener diode symbol is shown in FIGUE 3 and typical characteristics for a 9. Zener diode are given in FIGUE 4. FIG Forward Bias egion /volts I everse Bias egion I/mA FIG. 4

6 6 As FIGUE 4 shows, the potential drop across the diode in the reverse breakdown mode is almost constant over a wide range of currents. FIGUE 5 shows a practical current. The output voltage is equal to the Zener voltage and is, therefore, constant over a wide range of input voltages. The minimum limit of the input voltage range is automatically determined by the reverse breakdown voltage. In practice, the input voltage is maintained several volts above the breakdown value to allow for variations in load current and supply voltage. With reference to FIGUE 5, if the load is removed, all the current, I, will flow through the diode. The power dissipated by the diode will be given by P Z I Z. Therefore, when I L 0, then I Z I, and the maximum power becomes: () Pmax Z I... The Zener diode, therefore, must be capable of dissipating its no-load power and this imposes an upper limit on the input voltage. I IN (from rectifier circuit) I Z I L Z OUT L FIG. 5

7 7 Although the resistance of a Zener diode operating under reverse breakdown conditions is very low, it is nevertheless finite. This means that the Zener output voltage will vary with load current according to the value of what is termed the 'dynamic slope resistance' of the diode. With reference to FIGUE 4: dynamic slope resistance, rs... 2 I ( ) Example For the circuit given in FIGUE 5, IN 20 and kω. The reverse bias voltage for the Zener diode is 9. and the minimum reverse current requirement is 2 ma. If the dynamic resistance of the diode is 25 Ω, determine: (a) the maximum power dissipation in the diode (b) the change in the no load output voltage when the input voltage falls to 2 (c) the change in output voltage when the load current increases by 5 ma, assuming that the supply current I is constant. In solving this type of problem you may find it helpful to consider the Zener diode as a battery of open-circuit emf Z and internal resistance Z. This is shown in the equivalent circuit of FIGUE 6.

8 8 A B I IN C D I Z I L Diode equivalent circuit Z r OUT L Z G F E FIG. 6 Solution (a) Maximum power is dissipated at no load, i.e. I L 0. Applying Kirchhoff's voltage law to loop ABCFG: IN I + IZ Z + Z Substituting I Z I: I + IN ( ) + Z Z I IN + Z Z

9 9 Substituting IN 20, Z 9., 000 Ω, Z 25 Ω: I ma P max I Z mW. (b) From part (a) no load current at IN 20 is 0.6 ma. + I OUT Z Z Z OUT 937. For IN 2 : I I IN Z ma Z + I OUT Z Z Z OUT 97. Drop in no load voltage

10 0 (c) Applying Kirchhoff's current law: I I + I Z L Therefore, if I L increases by 5 ma, I Z decreases by 5 ma. Change in ( change in ) I OUT Z Z ( ) eduction in OUT TANSISTO STABILISES Example illustrates the effective stabilising action of the Zener diode circuit. However, unless expensive high-power Zener diodes are employed, this type of circuit can only supply small load currents. An alternative circuit, FIGUE 6, which largely overcomes this limitation, incorporates a transistor in conjunction with a Zener diode. The transistor controls the flow of current through the load resistor, from the smoothed output of a rectifier, by means of the voltage applied between the base and emitter. The base voltage of the transistor is maintained at a fixed voltage by the Zener diode. This type of stabiliser can deliver a much greater load current than the simple Zener circuit. It is also capable of maintaining the output voltage, L, under varying load current conditions.

11 Transistor C E I L D B BE IN (from rectifier) L L Z A F FIG. 7 To appreciate the operation of the circuit we need to refer to FIGUE 8 showing the characteristics of a typical transistor. I B I C Operating range Increasing I B BE (a) input characteristics CE (b) output characteristics FIG. 8

12 2 The output characteristics indicate that for any particular value of base current, I B, the collector current, I C is almost constant over a wide range of collectoremitter voltage, CE. Applying Kirchhoff's voltage law to loop ABEDF of the circuit given in FIGUE 7: Z BE L 0 + Z L BE... ( 3) Also, from Ohm's law I... ( 4) L L L Substituting Equation (4) into Equation (3): Z ILL + BE... ( 5) Now consider the effect of reducing the load resistance, L. Because I L is almost constant for any given I B, the I L L term will tend to reduce and because the Zener voltage, Z, is also almost constant, BE must increase to balance the equation. eference to FIGUE 8(a) shows that a small change in BE within the normal operating range will initiate a rapid rise in transistor base current I B. This increase in I B drives the transistor harder causing I C to increase in accordance with the output characteristics of FIGUE 8(b). Hence, the load current demand is met and the output voltage L is restored to its original value.

13 3 Can you describe the stabilisation process outlined above in terms of control terminology? apid changes occur in I B, and hence I C of the transistor, in response to small changes in the output current and voltage. The induced changes in I C tend to counteract the initiating changes and restore the required output quantities. The circuit, therefore, operates as a closed-loop feedback control system.

14 4 OPEATIONAL AMPLIFIE OLTAGE EGULATO The performance of the transistor voltage stabiliser can be much improved by the inclusion of an operational amplifier in the base circuit of the transistor. FIGUE 9 shows a representative circuit. I L IN EF + EO 2 Sensing ciruit L L SENSE 3 FIG. 9 The operational amplifier is used to amplifier the difference between the reference voltage EF, as derived from the Zener diode, and the sensing voltage, SENSE, to produce an error signal EO. Thus A ( ) EO EF SENSE where A is the gain of the amplifier. The sensing voltage is some fraction of the output voltage, L. In this example the sensing voltage has been derived from the output voltage by means of a potentiometer. The regulatory action of the op-amp stabiliser can be explained as follows:

15 5 Case : Line voltage IN increases IL increases L increases SENSE increases decreases EO EO k EF SENSE Transistor collector current, IL decreases L decreases This process continues until EO > 0, so that L is restored to its regulated value. ( ) The reverse action occurs if IN decreases.

16 6 Case 2 : Load varies Say L increases IL decreases SENSE decreases increases EO E O k EF SENSE Transistor collector current, IL increases L increases This process continues until EO > 0, so that L is restored to its regulated value. The reverse action occurs if L decreases. ( ) egulatory action viewed as a control system The op-amp regulator works as a classic control system with negative feedback and it is worthwhile diversion to see how this is so. FIGUE 0 represents the circuit of FIGUE 9 in block diagram form.

17 7 oltage-controlled current source I L SENSE EO IN Control circuitry EF OUT L FIG. 0 In the block diagram, the error signal is used to control a voltage-controlled current source (CCS). This source has a transfer characteristic like that of FIGUE, opposite. The greater the error voltage, the greater the current delivered to the load. We have seen that in practice this controlled source can be a transistor and its characteristic should be compared with that of FIGUE 8(b). I FIG. EO The bock diagram of FIGUE 0 can now be represented by the control system of FIGUE 2. Here a fraction of the output voltage L is sampled and fed back to the input (as SENSE k L. At the input SENSE is subtracted from the reference voltage EF and then amplified to produce an error voltage: A ( ) EO EF SENSE

18 8 The error voltage drives the CCS that has a gain of G amps/volt (i.e. G is a conductance and represents the slope of the graph of FIGUE ). The output of the CCS is a current I L that drives the load to produce the output voltage L. Op-amp CCS Load + A EF EO G L L SENSE k FIG. 2 Sensing Circuit The transfer characteristic of the complete control circuit can be derived as follows: SENSE kl... 6 A A k... 7 ( ) ( ) EO EF SENSE EF L I G GA k... 8 ( ) L EO EF L L LIL L GA EF kl... GA GAk L L EF L L ( ) ( ) ( ) ( ).. 9 ( ) Making L the subject of the equation: L GA L GAk EF L ( )

19 9 The above equation is now the form of the classic feedback equation θ IN + G θ OUT θ OUT θ IN G + GH H Dividing the HS of equation (0) by L GA: L EF k GA L ( ) Typically, A might have a value of 0 5 and G might have a value of S. So, for a 00 Ω load, / L GA The value of k, however, is likely to be between 0. and. Thus, providing these conditions hold, to good approximation equation () can be replaced by: ( ) L k EF The circuit of FIGUE 3 is used to derive the sensing voltage of an op-amp voltage stabiliser having a Zener voltage of 5. Determine the stabiliser s output voltage if 0 kω and 2 5 kω. SENSE 2 L FIG.3

20 20 k Thus, k 3 L EF volts 2 THEE-PIN INTEGATED CICUIT OLTAGE EGULATOS Three pin integrated circuit voltage regulators have all the circuitry in one integrated circuit package (IC). These IC regulators are cheap, simple to use, robust and require only a few external components. The basic functional block diagram for a three-pin voltage regulator is shown in FIGUE 4. oltage regulator In Controlling element passs transistor Current limiter and thermal cut-out Out I eference voltage + EO IN OUT I 2 2 Common FIG. 4

21 2 Assuming an ideal op-amp of infinite input resistance, then I I 2. The voltage is then given by: 2 OUT + For a sensible output then EF. Thus; 2 EF OUT OUT EF Thus the desired value of the output voltage is a function of the reference voltage and the ratio of resistor values. Practical voltage regulators will incorporate some form of circuit protection against current and thermal overload: Current limitation to prevent damage to the regulator if the load tries to draw too much current, if for example a short circuit is accidentally applied to the regulator s output.. Thermal cut-out to prevent overheating. This causes the regulator to shut down if its internal temperature approaches a damaging level. We now analyse these two forms of protection in a little more detail.

22 22 Over Current Limitation FIGUE 5 shows a common way of applying over-current limitation to the output of a regulator. (This method is also widely used in other circuits such as for example audio power amplifiers and operational amplifiers.) BE T I L B T 2 + EF IN L L FIG. 5 The current limiting circuit consists of just two extra components, a second transistor, T 2, and a resistor B. The full-load current I L flows through B to produce a voltage BE. This voltage, as its subscript implies, is applied across the base-emitter terminals of T 2. Providing BE remains below the turn-on voltage of the transistor (about 0.7 volts) then T 2 remains switched off and it has no effect upon the regulator. If, however, BE exceeds the turn-on voltage then T 2 switches on. By this we mean the transistor s emitter is effectively connected to its collector, as depicted in FIGUE 6, where the output of the regulator is shown shorted to ground. In this application the transistor is modelled as a simple switch the operation of which is controlled by the base-emitter voltage. When the switch of T 2 closes current is diverted from the base of T. The effect of this

23 23 is to increase the collector-emitter resistance of T which in turn will reduce the load current to a value just sufficient to sustain BE at its turn-on value. BE >0.7 T I L B T2 + FIG. 6 FIGUE 7 shows graphically the action of the current overload circuit. So long as BE remains below 0.7 volts, the load can vary but the load voltage will remain at its regulated value. When, however, the load current is large enough to cause BE to exceed 0.7 volts then regulation is lost as the load voltage falls. The load current is limited to a maximum shortcircuit value, I SC. L FIG.7 BE >0.7 I SC I L The regulator of FIGUE 5 has a maximum input voltage of 9 and the output current is limited to 2 A. Calculate: the required value of resistor B the power dissipated in the regulator when passing a current of A the power dissipated in the regulator under short-circuit conditions.

24 24 B Ω P ( ) I ( 9 5) 4 W IN OUT P ( ) I ( 9 5) 2 8 IN OUT W The above example shows the disadvantage of over-current limiting: a high power is dissipated in the device. Under fault conditions this level of power dissipation could be maintained for some time and damage could be caused by overheating before the fault is rectified. To overcome this many regulators employ the more sophisticated method of over-load protection now discussed below. Current fold-back limiting An idealised graph of load voltage against load current of a regulator with current fold-back protection is given in FIGUE 8. The graph shows that regulation occurs, i.e. the load voltage remains constant under a varying load, up to the point of maximum rated current. Beyond this point both the load voltage and current fall, so that the short-circuit value of fault current is less than the maximum rated value. L I SC FIG.8 I MAX I L FIGUE 8 also reveals how the technique gets its name; the graph looks like it has been folded along the oblique line, rather like the corner of the page of a dogeared book.

25 25 FIGUE 9 shows a representative regulator circuit with current fold-back protection. The circuit is similar to that of FIGUE 5 but the base of transistor T 2 is connected to a potential divider formed by A and B rather than directly to the emitter of T. T 6 I L 5 6 T EF IN L L O 4 2 FIG. 9 The voltage across the base emitter junction of T 2, BE, is given by 6 I L 6 BE B 5 T 2 Note that for T 2 to turn on, then BE 6 5 > 0.7 volts. Now 6 I L 6 and so I BE L 6 5

26 26 Also, by potential divider action, 5 O So long as the regulator has regulatory action, then is constant and so, therefore, is 5. Under this condition (i.e. regulatory action), BE is determined by the value of load current I L. So long as BE I L 6 5 is less than 0.7 volts then T 2 remains turned off and the current fold-back circuit is inoperative. O When, however, the load current increases to such a value that BE I L 6 5 exceeds 0.7 volts, then T 2 will turn on. In doing so it robs base current from T, so tending to turn it off. This has the effect of both reducing I L and causing to fall so regulation is lost. As falls so will 5 and as BE O I L 6 5 it can still exceed 0.7 volts, despite the fact that the load current is falling. It is this action the enables the current fold-back. The sequence of actions is illustrated opposite: O

27 27 I L BE 2 increases > 0.7 T turns on T tends to turn off I O L 5 falls falls falls I >. BE T remains on I 2 L L 'folds back' Design Equations for Current Fold-back The resistor ratio 4 5 Given the desired values of load voltage, L, short circuit load current I SC and maximum load current I MAX, then the ratio of 4 to 5 can be found from: 4 5 L ISC 0.7 I MAX I SC

28 28 The resistors 4 and 5 The potentiometer voltage formed by 5 / 4 should not be loaded by T 2. That is to say, the voltage should be independent of the base current of T 2. This can be achieved (approximately) by making the current through 5 much greater than the base current, say at least ten times greater. But 4 and 5 should not be set to unnecessary low values as this will drain current from the load. Moreover the consequential I 2 heating within the integrated circuit is highly undesirable. Once the desired ratio of 4 to 5 has been found from above, suitable optimal values can be determined from the following argument: The maximum output current of the op-amp is about 25 ma (the opamp will also have short circuit protection and this is its short circuit value). Assume all this current flows into the collector of T 2 (worse case scenario ( wcs )). I Assume that T 2 has a d.c. current gain of C h (a FE 00 IB conservatively low value). If the output of the regulator is short-circuit ( wcs ) then the base current of T 2 is about 25 ma µa. Set the current through 5 to ten times this value to give 2.5 ma.

29 29 O I5 and assuming O is the regulated value (4.7 in the circuit of FIGUE 20) then O 5 4 I 5 Find 4 from above and hence 5. Series esistor 6 Having determined the required ratio 4 5 then 6 can be determined from: I SC 4 The plot of load voltage against load current of FIGUE 20 is from a PSpice simulation of the regulator circuit of FIGUE 9. In this circuit the current fold-back resistors were given values of 6.65 Ω, Ω and 4 2 kω.

30 30 FIG. 20 Thermal cut-out Semiconductor devices can be irreparable damaged by overheating. In a linear regulator by far the principle source of heat is caused by the I 2 loss in the pass transistor. FIGUE 2 shows the principle of operation of the thermal shutdown circuit (TSC). A temperature sensitive device within the TSC monitors the internal temperature of the integrated circuit and when this temperature exceeds its maximum rated value the base current of the pass transistor, T, is diverted, so causing T to turn off.

31 3 T IN + EF L L Thermal shutdown circuit FIG. 2 Integrated circuits are optimised for the fabrication of transistors or diodes and it is therefore convenient to use one of these devices as a temperature sensitive device. The volt-drop across a forward biased diode or across the base-emitter junction of a transistor is temperature sensitive, having a negative temperature coefficient of about 2 m/ C, and is moreover linear over the temperature range of interest. FIGUE 22 shows the change in BE of a discrete transistor over the temperature range of 0 to 200 C. For this transistor (a BC07), the corresponding change in voltage is about 320 m. FIG. 22

32 32 Determine the temperature sensitivity of the BC07 used in the circuit of FIGUE 22 in m per C. Sensitivity k T BE 6. m/ C FIGUE 23 shows how a transistor is used in a TSC. The temperature-sensing transistor T 3 is placed in good thermal contact with the pass transistor T. The base of T 3 is set to a constant voltage, derived from the reference voltage by the potentiometer, of about B 0.4. (We should, at this juncture, distinguish between B, a property of the circuit and BE, the volt-drop across the forward-biased base-emitter junction of the transistor.) At normal operating temperatures T will be off as BE > B. However, as the temperature within the circuit increases, say due to increased current flow through T, the BE of T 3 falls and at a sufficiently high (but abnormal) temperature B becomes greater than BE. T 3 will then switch on and divert the base current of T to ground, so turning T off. The regulator is now effectively shut down and will remain so until the temperature within the circuit falls below it designed maximum operating value.

33 33 T OUT EF HEAT FLOW + 2 TSC 0.4 T 3 FIG. 23 Estimate the maximum operating temperature of the circuit of FIGUE 22. Assume T 3 has BE 0.7 at 20ºC and that BE of T 3 has a negative temperature coefficient of.6 m per ºC.

34 34 BE T BE where k k 5. m/ C 300 T 88 C. 6 T MAX C Comment: We see below that we have made a certain assumption in this calculation. FIGUE 24(a) shows a PSpice circuit model of a regulator with the TSC highlighted. The bias points have been shown for a temperature of 0 C. At this temperature the base voltage of Q2 sits at 77 m. FIG. 24(a) The temperature response of the circuit, FIGUE 24(b), shows that the TSC operates at a temperature just over 50 C, which corresponds to a Q2 base voltage of about 600 m.

35 35 FIG. 24(b) The quiescent value of Q2 base voltage given in FIGUE 24(a) is a lot higher than we would expect, e.g.the characteristic of FIGUE 22 would predict a voltage of about 540 m. Can you explain why this is so?

36 36 FIGUE 24(b) shows that one cause of the anomaly is caused by the reference voltage falling with temperature. This is because the reference source, a Zener diode, also has a negative temperature coefficient of about.6 m/ C. Thus at 50 C the reference voltage will fall by about 240 m. A second cause is that the base I voltage B of Q2 has two C components. One is BE, the voltage across the forward-biased B Q2 p-n junction. The other is the B E r E voltage E that is dropped across B the emitter resistance r E. This resistance is due to the bulk of BE E E resistive material in the emitter B that does not form part of the baseemitter junction. The voltage E is given by I E r E I C r E, i.e. the emitter voltage is directly proportional to the collector current. Typically r E is a few ohms and if I C about 0 ma then E will be of the order of 20 m.

37 37 NOTES

38 38 SELF-ASSESSMENT QUESTIONS. With the aid of a sketch graph explain why a Zener diode is a suitable device for the provision of a stabilised d.c. supply. State what factors determine the range of input voltage over which a Zener diode stabiliser is operational. 2 A Zener diode stabiliser circuit as shown in FIGUE 5 of the lesson is used to provide a 9 d.c. supply from a rectified input of 20. The reverse bias voltage for the diode is 9., 200 Ω, L 2.4 kω and the 'dynamic resistance' of the diode is 25 Ω. Apply Kirchhoff's laws to the equivalent circuit of FIGUE 6 to produce two equations in I L and I Z. Hence determine: (a) the load current (b) the output voltage. Compare the answer to part (a) with the calculated value of load current assuming the output voltage to be a constant FIGUE 25 shows a three-pin voltage regulator (the LT0833) used in a circuit to increase its output voltage to 5. It nominal reference voltage, EF, is.25. (a) Show that, assuming no current is taken from the ADJ pin, that O EF + 2 (b) If kω, find a suitable value for 2.

39 39 FIG Justify the choice of values for resistors 4, 5 and 6, from the information given in FIGUE 20.

40 40 ANSWES TO SELF-ASSESSMENT QUESTIONS Forward Bias egion /volts 2 I everse Bias egion I/mA FIG. 4 (eproduced) The supply voltage equals the pd across the Zener diode and is, therefore, constant over a wide range of input voltages. The characteristics given in FIGUE 4 also show that the pd across the diode in reverse 'breakdown' mode is almost constant over a wide range of currents. The input voltage range must lie between a value slightly above the P reverse breakdown voltage and a value somewhat below m where P m is I the power rating of the diode and I is the 'no load' current.

41 4 2. I IN (from rectifier circuit) I Z I L Z OUT L FIG. 5 (eproduced) Applying Kirchhoff's voltage law to loop FCDE: Z + IZZ ILL Substituting Z 9., Z 25 Ω, L 2400 Ω: I 2400I Z Or 2400I 25I L Z L () Applying Kirchhoff's voltage law to loop ABCFG: IN I + IZZ + Z

42 42 Substituting in 20, 200 Ω, Z 25 Ω, Z 9. : I + 25I Z + 9. But by Kirchhoff's current law: I I + I I + I 25I 9. Or 200I + 225I Z L Z L ( ) + + Z L Z ( ) (a) We now need to solve Equations () and (2) simultaneously. 2400I 25I L Z 200IL + 225IZ () ( ) Equation () [2 Equation (2)]: 2425I 2. 7 Z I Z 524. ma Substituting in Equation (): 3 ( ) 2400I L I L 434. ma

43 43 (b) + I OUT Z Z Z ( ) Assuming a constant output voltage of 9 : I I L L OUT L ma compared with the actual value of 4.34 ma. 3. (a) Assuming the same current flows in both resistors; EF O EF O EF ( ) 2 EF + EF O 2 EF + O 2 O EF + 2 O EF + 2

44 44 (b) Ω Ω to the nearest preferred value. 4. From the graph of FIGUE 20, L 4.7, I SC 500 ma and I MAX 940 ma. 4 5 L ISC 0.7 I MAX I SC I SC Ω O I Ω

45 Ω Ω Comment: These values compare reasonable well with the circuit values. The cause of the discrepancies is that the circuit was designed for a maximum current of A and a short circuit current of 0.5 A. In this question, whilst we have assumed 0.5 A for the latter, we have taken the maximum current as 940 ma.

46 46 SUMMAY We have shown how the Zener diode can be used as a voltage reference device when used its reverse-biased mode. The simple Zener-diode voltage regulator consists of just two components the diode and a series resistor. Such a simple circuit is perfectly adequate for many applications where only low load currents (a few milliamps) are required. For applications requiring larger load currents (00 of ma or a few amps) we have seen how the Zener diode can be used as the controlling devices of a series element a bipolar transistor. Finally we have seen how the use of an operational amplifier can much improve the regulator s performance. Integrated circuit regulators can consist of just three pins and are extremely simple to use. Such regulators can incorporate more refined features such as current overload protection and thermal shutdown.