8/26/2010 LinearVoltageRegulators(overview).doc 1/20 8/26/2010 LinearVoltageRegulators(overview).doc 2/20

I IS L

Linear Regulators + + R IZ The schematic below shows a pretty darn good design for a + V V =V RL . It has good regulation, high output power, - S ZL and acceptable efficiency (for a linear regulator, that is!). − −

I S ⎛⎞ R2 Remember, a zener is just a junction diode. It has R VLZK=+V ⎜⎟1 pn ⎝⎠1 three operating modes: forward biased, reverse biased, and + R + breakdown. + + V 0 S - With most junction , 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 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 where VZK is the zener breakdown voltage of the diode (a device parameter!). A: Sure. The simplest linear regulator is the regulator. We simply add a zener diode to the previous “voltage division” In other words, a zener diode in breakdown behaves a lot like a with a value V =V . regulator: simpler oZK ?

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 R. + Vz CVD VZ VZK Breakdown Q: Yikes! The load current might be quite high—wouldn’t this cause quite the − Model 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 LZKRL - −

− − R + V +

+ IL The regulated load voltage is VLZK=V ! V + + 0 + S - V =V − This shunt regulator provides both good line regulation, and ZZK V − − − V =V good load regulation. However, it is particularly inefficient! RL LZK

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 = vout in v 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 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 in =v ). 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 , 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:change? But the output of the voltage 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 ref = divider is not connected to into breakdown, and create a reference voltage V 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 ref on its input, but does not draw any current from the circuit V =V (V >V ) ZK SZK + it is sensing—it senses directly the open-circuit output must I find a 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 ). Q: Say I want a different regulated voltage then VLZK=V ,

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

R supplies not shown vin +

vout + IL − + 0 + R2 V =V − ZZK − = RL VLZKV

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 =−Av()+− v d So, returning to this circuit, where v+ and v− are the on the inverting (+) and non- we find from voltage division:

vin + inverting (-) input terminals, and Ad is the differential gain of out R1 v v =v out the op-amp. + ()RR12+ − R Recall that this differential gain is typically very, very From KVL: 2 7 large—for example,be very Ad = large—thus10 or more.

of course the inputs were approximately v− =vin Q: I don’t understand. After all, a gigantic Ad would ( saturating R1 mean that the v =−Av() v would likewise o +− the amplifier—unless d Because of virtual short: output v =v ≅ ). −+ v 12v equal

And of what use is a differential amplifier if the Combining these equations, we find: inputs

must out in out in be approximately equal?? RR v =⇒=+vvv121 RR ()RR 12+ () 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). A vR vo ==+o 1 2 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 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.

the gain was something really

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

problem? A: The differential gain of the op-amp (i.e., Ad ) is really Q: Yikes! A wi 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 . An op-amp is but one element of this feedback amplifier—an amplifier ll likely need to deliver whose gain is a mere 1 + R2 ! A: Absolutely! We can add a pass 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 - supplies vBE 0 call these junctions the Collector- not shown - + IL iE + Base Junction (CBJ) and the + Emitter-Base Junction (EBJ). Emitter = VZZKV − LZK − R R 2 RL V =+V ()1 2 0 R1

− VZK VVLZK RR+ R= R1 12 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). Each of these junctions can either be forward biased or reverse biased, and therefore there are four possible operating mode of a npn BJT. Each mode has its own name:

Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 13/20 8/26/2010 LinearVoltageRegulators(overview).doc 14/20

MODE EBJ CBJ Active

Cutoff Reverse Reverse An npn BJT in active mode exhibits these properties:

Active Forward Reverse C VBE ≅ 07. V ii= β B VCE > 07. iB > 0 Reverse Active Reverse Forward A transistor is somewhat like a valve used to control liquid Saturation Forward Forward current. In this analogy we find:

The reverse active is typically not used—the three This …is (sort of) like… this fundamental bias states of an BJT are thus Cutoff, Active, and Saturation. Liquid current

Each of these modes can be described mathematically (or, at Voltage Pressure least approximately so). Collector Valve Input

Cutoff Emitter Valve Output

An npn BJT in cutoff exhibits these properties: Base Valve Control Knob Collector “Collector” Current === < > iiiB 0 VBE 0 VCB 0 iC C (liquid) E Saturation + + v “Base” P Pressure Bas e CE CE An npn BJT in saturation exhibits these properties: - -

C ii< β V ≅ 07. V V ≅ 02. V Emitter B BE CE “Emitter” Cutoff is analogous to the value being completely closed—no current will flow through the value, regardless of how much

pressure (vCE ) is applied.

Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 15/20 8/26/2010 LinearVoltageRegulators(overview).doc 16/20

Saturation is analogous to the value being completely open—it Note from KCL that: takes almost no pressure (v ) to get a lot of current to flow CE B II+=+ IVZK through the valve. C 1 L R

ZK Active mode is analogous to having the value partially open—it B V Typically we find that I << IC and R1 << IL , so that:

requires some pressure (vCE ) to get current to flow.

Moreover, this current can be increased by further opening IIC ≅ the valve (increasing base current i ) or decreased by further L B closing the valve (decreasing base current i ). B In other words, the load current no longer comes from the

op-amp. Instead, the load current is delivered directly from It is the active mode that we desire for the pass transistor the source VS (e.g., the output of the power supply filter). in our regulator circuit. The base of the npn is attached to

the op-amp output, the emitter to the load, and the collector This means that the load current can potentially be very big. to the source voltage: However, there is still one problem:

V V S S Æ This big load current must travel through the npn BJT! C VVSZK− II= β B R R As a result, we potentially have a major power dissipation

0 problem in the BJT. The power absorbed by the BJT is + + IB CE approximately: + =− BJT V VVSL CE E VZZK= V − − PVIVVI≅≅−()SLL − + IL 0 R2 LZK This is the major cause of the inefficiency associated with R R V =+V 1 2 L ()R1 this regulator. If we have designed our regulator well, then VZK VVLZK RR+ R= R1 the BJT should be the only device that dissipates much heat. 12 1

− In other words, the efficiency of this regulator is

approximately:

Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS should be? After all, it has 8/26/2010 LinearVoltageRegulators(overview).doc 17/20 8/26/2010 LinearVoltageRegulators(overview).doc 18/20 parameter P η = L Q: But how does the op-amp “know” what this base current PPBJT+ no “knowledge” of what the BJT L β is! where PVILL= L . True, but what it does know (indirectly) is what the load

We can also conclude from conservation of energy that the voltage is. The load voltage VL is divided by resistors R1 and power delivered by the input source is: R2 , this voltage is then “sensed” as the inverting input

voltage v− . ≅+ PPSL P BJT =−() + Remember, if the load voltage V is at its proper level, then VVIVI L =−+SLLLL v−+==VvZK —the virtual short is enforced! What is its VI VI VI = S L LL LL VI SL However, if the load voltage VL drops below its regulated a result that is hopefully evident when examining the circuit value (for example, if the load RL instantaneously decreases), above! then:

Q: So what about the current op-amp? v−+<=VvZK , causing… purpose now? the op-amp output voltage to increase, causing… The op-amp increases or decreaseat sthe its outputoutput ofvoltage the until the base current IB becomes the correct value to provide the the base current to increase, causing… required load current of IILB≅ β . the load current to increase, causing…

Æ In other words, the base current IB controls our valve—increasing or decreasing the base current the load voltage to increase, causing… (metaphorically) adjusts the valve “control knob” one way

or the other. v− to increase until we again arrive at the steady state condition wherein the virtual short is enforced

(v−+==VvZK ).

Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS 8/26/2010 LinearVoltageRegulators(overview).doc 19/20 8/26/2010 LinearVoltageRegulators(overview).doc 20/20

V V S S − C Burns told me to keep the VVSZK = β R IIB pressure at R for 14.3 psi Analogous toReference 0 some reason + pressure has , but Voltage + I the B better open the value a bit Analogous to + VCE =−VV dropped. SL more, I unregulated until V =V increases back to ZZK − − the pressure supply − + 14.3 psi 0 R IL 2 . V Analogous to RL L Analogous to Analogous v

nt” Feedback Of course, if v−+>=VvZK , the opposite set of reactions will “Curre Voltage occur, causing the regulated voltage to drop to its correct value. Note here that water pressure is analogous to voltage; water flow is This system demonstrates the beauty of feedback. You do analogous to electrical current. not need to know precisely how much to adjust the “knob”;

since you have a way of measuring (sensing) the value you wish

to control. Load (with change “current” Analogous to Æ You just keep “turning” until the quantity returns to the requirement!) desired value!

Jim Stiles The Univ. of Kansas Dept. of EECS Jim Stiles The Univ. of Kansas Dept. of EECS