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Linear Voltage Regulators + + R IZ the Schematic Below Shows a Pretty Darn Good Design for a + V V =V RL Linear Regulator

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Linear Voltage Regulators + + R IZ the Schematic Below Shows a Pretty Darn Good Design for a + V V =V RL Linear Regulator 8/26/2010 LinearVoltageRegulators(overview).doc 1/20 8/26/2010 LinearVoltageRegulators(overview).doc 2/20 I IS L Linear Voltage Regulators + + R IZ The schematic below shows a pretty darn good design for a + V V =V RL linear regulator. It has good regulation, high output power, - S ZL and acceptable efficiency (for a linear regulator, that is!). − − I S ⎛⎞ R2 Remember, a zener diode is just a pn junction diode. It has R VLZK=+V ⎜⎟1 ⎝⎠R1 three operating modes: forward biased, reverse biased, and + + breakdown. + + 0 VS With most junction diodes, we try to avoid breakdown. - VZZK=V − I However, breakdown is the typical operating mode of zener − − L + diode! Im≈ 10 A Z R2 Recall that a zener diode in breakdown (but only in R1 RL VL breakdown!) has the following characteristics: − V V I > 0 ZZK Z Q: Yikes! Can’t we start with something simpler? where VZK is the zener breakdown voltage of the diode (a device parameter!). A: Sure. The simplest linear regulator is the shunt regulator. We simply add a zener diode to the previous “voltage division” In other words, a zener diode in breakdown behaves a lot like regulator: a voltage source with a value VoZK=V . Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 3/20 8/26/2010 LinearVoltageRegulators(overview).doc 4/20 > + IZ 0 + Iz Thus, the load current (and then some) must flow through Zener resistor R. + Vz CVD V Z V Breakdown Q: Yikes! The load current might be quite high—wouldn’t this ZK − Model cause quite the power dissipation in resistor R ? _ _ A: Absolutely! Moreover, if the load current is small, then all of IS must flow through the zener diode—again creating a Meaning that the regulator (if the zener is in breakdown) can power dissipation nightmare! be modeled as: Q: Is there some solution to this? I IS L + + A: Yes! We can insert a buffer between the zener diode and R IZ > 0 + the load. + V ZK V =V IS VS LZK RL - − − − R + + V + IL The regulated load voltage is VLZK=V ! + VS + 0 + - VZZK=V − − This shunt regulator provides both good line regulation, and V − − RL VLZK=V good load regulation. However, it is particularly inefficient! ImZ ≈ 10 A Q: Why is that? − A: From KCL it is evident that IIISZL=+. Moreover, since IZ and IL are both positive, we conclude that IIS > L . Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 5/20 8/26/2010 LinearVoltageRegulators(overview).doc 6/20 Recall that the circuit: 12V i =0 in 6 mA vin + Now let’s attach a 2K load to the 1K = voutv in output of this circuit; note the − output voltage drops to 6 V, and + the current increases to 6 mA. 2K 2K v =6V out − Q: But I thought the output voltage 3 mA 3 mA is known as a voltage follower. It is essentially an amplifier of this was 8.0 V; why is it now 6.0 V?? with a voltage gain of one (i.e., Avo = 1.0, so vout=v in ). A: 8.0 V was the open-circuit output voltage—the output is Q: A voltage gain of one! What good is that?? 8.0 V only when an open circuit is attached to it! A: The important aspect of a voltage follower is that its input As this example shows, attaching a load to the output will resistance is very large—the input terminals of an op-amp cause the output voltage to change. One way to prevent this draw almost no current. is with a voltage follower: 12V 12V Perhaps an example would help. 4 mA 1K 4 mA Op-Amp Consider this voltage divider, with an 1K supplies 0 = not shown open-circuit output voltage of 8.0 iin 0 vout = 8V volts. 8.0V + 4 mA 2K 4 mA 2K + − 2K vout =8V 4 mA − Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 7/20 8/26/2010 LinearVoltageRegulators(overview).doc 8/20 Q: But the output of the voltage divider is not connected to Note here that the load current IL no longer passes through an “open-circuit” in this case—it’s connected to the input of resistor R, or the zener diode. Thus, only a small amount of the voltage follower. Why then does the output voltage not current is required (e.g., ImAz ≈ 10 ). This will bias the zener change? into breakdown, and create a reference voltage Vref =VZK for the voltage follower to sense. A: Ah, but the output of the voltage divider is attached to an VS open circuit! Remember, the input resistance of the op-amp is very high—essentially an open circuit. − A Reference VVSZK R R Voltage This is why the voltage follower works; it “senses” the voltage Circuit on its input, but does not draw any current from the circuit V =V (V >V ) ref ZK SZK + it is sensing—it senses directly the open-circuit output voltage, and then places that voltage across the load. V =V ZZK − Thus, the current through the load does not have to come from the original circuit—the load current instead is provided by the op-amp (through its power supply). Q: Say I want a different regulated voltage then VLZK=V , must I find a new zener diode? An application of this then is with our regulator circuit: V S A: You can, but an easier way would be simply to change the gain of the voltage follower. Consider this circuit: Op-Amp supplies R not shown vin + v out I + L − + 0 + R2 V =V − ZZK − V =V RL LZK ImZ ≈ 10 A R1 − Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 9/20 8/26/2010 LinearVoltageRegulators(overview).doc 10/20 Recall that an op-amp is a differential amplifier, such that: Thus, if the output of the op-amp is not saturated, the condition v+−≅ v must be true! vout =−Avd ()+− v So, returning to this circuit, where v+ and v− are the voltages on the inverting (+) and non- we find from voltage division: vin + inverting (-) input terminals, and Ad is the differential gain of R1 vout v =v the op-amp. + out ()RR12+ − R Recall that this differential gain is typically very, very From KVL: 2 7 large—for example, Ad = 10 or more. v− =vin Q: I don’t understand. After all, a gigantic Ad would R1 mean that the output v =−Av() v would likewise o d +− Because of virtual short: be very large—thus saturating the amplifier—unless of course the inputs were approximately equal v−+=v (v12≅ v ). And of what use is a differential amplifier if the Combining these equations, we find: inputs must be approximately equal?? v =⇒=+vvvRR121 in out()RR12+ out in () R 1 A: Op-amps are generally not implemented by themselves! Instead, they typically are but one component of many in a Thus, the gain of this non-inverting amplifier is: feedback amplifier (see the above circuit as an example). vRo 2 Avo ==+1 In these applications, we will indeed find that v+−≅ v -- but we vRi 1 will also find that this is a desirable condition! Q: Hey wait! I thought you said the gain was something really 7 The condition v+ ≅ v− is known as a virtual short. If this is not huge, like 10 ?? true, the output of the op-amp will be saturated. Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 11/20 8/26/2010 LinearVoltageRegulators(overview).doc 12/20 A: The differential gain of the op-amp (i.e., Ad ) is really Q: Yikes! A voltage regulator will likely need to deliver more huge, but do not confuse an op-amp with the amplifier. Our load current than that; is there some solution to this amplifier consists of an op-amp and two resistors. An op-amp problem? is but one element of this feedback amplifier—an amplifier whose gain is a mere 1 + R2 ! A: Absolutely! We can add a pass transistor to the circuit— R1 in this way the load current will not need to pass through (and thus limited by) an op-amp. Combining these elements, we now have a regulator whose Collector voltage is adjustable by the proper selection of resistors R1 Typically, this pass transistor is a npn i and R2: V + C S Bipolar Junction Transisitor (BJT). vCB iB + - VV− SZK Recall that a npn BJT is constructed vCE R R Base + Op-Amp such that it has two pn junctions. We - v 0 supplies BE not shown call these junctions the Collector- - + IL iE + Base Junction (CBJ) and the + Emitter-Base Junction (EBJ). Emitter V =V ZZK − − R2 R2 RL V =+V 1 0 LZK()R1 − VZK VVLZK + = R1 RR12 R 1 Note in the circuit above that the load current is delivered by the op-amp. This can actually be a bit of a problem, as the current at the output of an op-amp is typically limited to a rather moderate amount(e.g., 25 mA).
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