electronics for chemical instrumentation

8/3/2019 Electronics for Chemical Instrumentation

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Chemistry 30333 Electronics for Chemical Instrumentation 1

Electronics for Chemical Instrumentation

1. Basic Circuit Components1.1 Voltage. A central idea in all of science is the concept of potential energy difference

between two physical (or chemical!) states. In electronics, we will exclusively address potentialenergy differences in terms of electrical potential (voltage), V, measured in volts (V). Becausethe voltage is a potential energy difference, it is always measured between two circuit points.

Frequently, one of the points is implied to be the ubiquitous (and maddeningly named) groundpotential (see below). Furthermore, given the equivalence of work and energy, it should not be

surprising that we can identify the voltage with the work done in moving a given charge througha given potential difference. For example,

E=QV (1.1)

where the energy,E, is related to the product of charge, Q, and voltage. The SI units are defined

such that 1 Joule (J) of work is required to move 1 Coulomb (C) of charge through 1 volt ofpotential difference. To give some sense of proportion, the charge on a single electron is 1.6 x

10-19

C. Another commonly used measure of charge is constituted by a mole of electrons, forwhich we have,

Q mole( ) =QeNA = 1.6 x 10

-19 C( ) 6.02 x 1023 mo-1( ) = 9.64 x 104 C mo-1 =1 Faraday

(1.2)

We will make extensive use of the Faraday in electrochemistry, where it is useful to relate the

quantity of charge consumed or produced by an electrochemical reaction in terms of the reactantsor products that are expressed in moles.

Given the emphasis on voltages as potential

differences, wouldnt it be nice if there were acommon reference energy, that could be used to

relate different points in a circuit to a single referencevalue. Fortunately for us, there is the

aforementioned eponymous ground potential. Theground point in a circuit is physical location with a

potential that is negligibly different from theelectrical potential of the earth. Almost always,

ground potential in an instrumentation circuit isestablished through the 3rd prong of a grounded

outlet which is ultimately connected (through low-resistance thick copper wires) to a water pipe or

similar structure in intimate contact with the earth. InFig. 1.1, the ground is established through the thick

copper wires twisted together with the red connectorin the center. The physical basis for using the earth

as a reference potential derives from the relative

Figure 1.1. Exploded drawing of a three-

terminal duplex electrical outlet (U.S.).

The ground lead is connected to the round

terminal at the bottom of each receptacle.

Adapted fromhttp://www.thisoldhouse.com/toh/article/0,,216662,00.html


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Thus, the prescription for exploiting the phasor notation to make your life simple is:

1. Express the AC voltage of interest in terms of the first term in the numerator above. Forexample, V(t) = V0 cos twould be expressed as V0exp(it). Note carefully that the

exact equivalence is given in eqn. 1.8; the term used here is just the 1st

term in the

numerator of the exact expression.

2. Perform all the signal manipulation calculations you need to perform in the phasornotation, using the simplified expression above.

3. When you are ready to recover the exact answer, just write down the complex conjugateof the expression obtained at the end of step 2, sum them and divide by 2.

Again, we stress that these manipulations are for computational convenience only. No additionalinformation is required or added by performing the computations with phasor notation.

1.2 Current. While voltage is fundamental, the point of electronic circuitry is frequently toproduce and manipulate electrical currents,I, which can be defined as the time rate of change ofcharge at a given point in a circuit, i.e.

I=dQ

dt(1.9)

Another way to think about current is as a flux of charge. Current is measured in the SI systemin units of Amperes, with 1 Ampere being equal to a charge flux of 1 C s-1. Aside from the

fundamental physical origins of the two quantities voltage being a potential energy differenceand current a charge flux many other things about them are similar. For example, current can

be either positive or negative, corresponding to the direction of electron flow in a circuit; currentcan be either DC or AC (now the notation actually makes sense); and AC currents can be

handled using the same phasor approach we developed for voltages above. For example,I(t) = I0cos tcan be represented byI0 exp (t) and handled according to steps 1-3 above.

1.3 Passive Circuit Elements. The whole point of electrical circuits is to combine physical

elements that behave with unique current-voltage relationships,I = f(V), in such a way as tomanipulate signals in interesting and useful ways. A very commonI-Vrelationship that applies

to a specific class of electrical circuit elements is given by the famous Ohms Law,

V = IR (1.10)

whereR, the resistance measured in ohms, , is aproportionality constant that specifies the voltage needed to

produce a given magnitude current. Undoubtedly you alreadyknow that a voltage of 1 V will produce a current of 1 A in a 1

resistance. Physical resistors come in values in the range ~1108, and with this knowledge we are ready to constructour first circuit, Ckt. 1.

Circuit 1


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What questions can reasonably be answered about Ckt. 1? We might reasonably want to know:

(1) how much total current flows from supply to ground and (2) how is the current dividedbetween the two current paths,R1 andR2. In order to answer the questions, we need to introduce

two very useful laws.

Kirchoffs Law: This law deals with circuit nodes, i.e. points at which one or moreincoming and outgoing current paths intersect and says that the sum of all incoming

currents must equal the sum of all outgoing currents, i.e.

Iin = Iout (1.11)

Parallel Elements: This law deals with parallel circuit elements and states that thevoltage dropped across parallel elements must be equal.

We can now usefully apply these two laws to the analysis of the above questions. First,

Kirchoffs Law says that the (positive) current entering the top node will be divided into twooutgoing currents,I1 flowing throughR1 andI2 flowing throughR2. However, it does not tell us

the magnitudes ofI1 andI2. To determine that we need to apply Ohms Law (eqn. 1.10). Indoing this, clearly the applied voltage is 5 V, but what do we use for R, since there are two

resistors in the network. Here we utilize a trick to replace the pair of resistors with a singleequivalent resistance, determined by the (hopefully) familiar formula,

1

Requiv=

1

Rii=1

n

(parallel)

Requiv = Ri (series)i=1

n

(1.12)

In our case,

1

Requiv=

1

Rii=1

2

=1

1000+

1

2000;

Requiv = 666

Now, we simply apply eqn. 1.10, which yields,

Itotal =V

Requiv=

5V

666= 7.5 mA

Clearly, 8.33 mA is the total current entering the top node of Ckt. 1, but how much flows intoeach branch of the circuit. We could set up the simultaneous equations forI1andI2, but there is

an easier way. We simply recognize that the total voltage dropped across each resistor must bethe same, namely 5 V. Then we can apply Ohms Law to each branch in sequence, i.e.


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I1=

V

R1

=

5 V

1k= 5mA;

I2=

V

R2

=

5 V

2k= 2.5mA

Notice that our answers are internally consistent, i.e.I1 +I2 =Itotal. Furthermore, note that thefraction of the current flowing in each branch is inversely proportional to the resistance in that

branch a general observation for impedance (i.e. resistance) networks. We went through thisexample in significant detail, because all resistor network problems can ultimately be handled

using combinations of these simple steps.

The second commonly used passive circuit element is the capacitor. As the name implies,capacitors are charge storage devices, and the value of capacitance, C, measured in Farads,

denotes the ability of the particular component to store charge. Capacitors are governed by,

Q =CV (1.13)

and, because the current is the time derivative of charge we can relate the current through acapacitor to a time-varying voltage,

I=dQ

dt= C

dV

dt(1.14)

Right away we can see that capacitors will behave in fundamentally different ways from resistorswhen voltages are applied across them, because theI=f(V) is

different. In general, when a voltage is initially applied across acapacitor, the capacitor will begin to charge, meaning that

charge separation begins to occur. Eventually the capacitor willbecome fully charged. The time it takes for this to happen is

determined by the value of the capacitance and the total resistancethrough which the capacitive current must flow. As an example,

consider Ckt. 2, where the capacitor and resistor are in series. Attime t< 0, the applied voltage, Vin = 0, so the output voltage, Vout

= 0 also. Upon application of a step-input voltage, Vin at t= 0, allof the voltage immediately appears across the resistor, because the

capacitor has not had time to rearrange any of its charge. We begin by recognizing that all of thecurrent flowing in the circuit has to be the same through both elements, i.e. none flows through

the external impedance of whatever instrument is being used to measure Vout. Thus,

Vout= IR = RC

dV

dt (1.15)

Circuit 2


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We arrived at eqn. 1.15 through direct application of eqn. 1.14, but the equation is somewhatambiguous, because it does not tell us what the value of V is. Clearly this must be the voltage

across the terminals of the capacitor, i.e.Vin Vout. Thus, the correct equation is,

Vout= IR = RC

d

dt

VinV

out( )(1.16)

Because the input voltage does not change, its time derivative is zero, so we have,

Vout= RC

dVout

dt (1.17)

This is a simple boundary value problem of the type you have seen before. Rearranging and

directly integrating yields,

dVout

Vout =

1

RC dt;

lnVout=

t

RC+K

(1.18)

whereKis a constant of integration. Exponentiating both sides yields,

Vout= e

Ket/RC

=V0e t/RC

(1.19)

where we have re-labeled the constant of integration as V0. To finish up we simply re-use the

initial condition that at t= 0, Vout= Vin, finally giving us,

Vout=V

inet/RC

(1.20)

Apparently the behavior of the simple RC circuit is an exponential charging curve, where the

speed of the charging is governed by the value ofRC, which has units of time, i.e. expressingRin ohms and Cin Farads produces a product with units of seconds. A practical note

capacitance values found in typical electrical circuits in instrumentation have values in the range10

-12F < C< 10

-6F and are either denoted in units of pF (1 pF = 10

-12F) orF (1 F = 10-6 F).

Thus, a value of 10-9

F will be specified as 1000 pF, never as 1 nF.

Thus, if we had chosen, C= 1000 pF andR = 1 k, theRCtimeconstant would be (10-9 F) x (103) = 10-6 s = 1 s. If we choosean arbitrary value of 5RCas the time it takes the voltage to reachits steady state value, then 5 s after application ofVin the voltage

would appear entirely across the capacitor, and Vout ~ 0. Nowhaving gone through this example in detail you should be able to

derive (or at least convince yourself) that Ckt. 3 would begoverned by,Circuit 3


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Vout=V

in1 e

t/RC( )(1.21)

Note that the physical behavior of the elements in the two circuits is exactly the same, i.e. all of

the voltage initially is dropped across the resistor, then as time passes more of the voltage isdropped across the capacitor until at some later time (t> 5RC) all of the voltage appears acrossthe capacitor.

We can extend this analysis for both circuits by considering a square wave input and then use thebehavior in response to step changes in applied potential to develop the behavior in each half-

cycle. The behavior for Ckt. 2 is shown in Fig. 1.3. The applied potential ( ) is a squarewave with frequencyf= 2.77 kHz (T= 360 s). When 1/RC>>f( ), the full

exponential decay of the output voltage, indicated in eqn. 1.20 is apparent. One can easily seethat as theRCtime constant is made smaller (1/RClarger), the limiting output would be a series

of sharp spikes nearly coinciding with the voltage steps positive spikes for positive-goingchanges in the voltage and negative spikes for falling voltage edges. Alternatively, when 1/RC


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The same analysis can be applied to Ckt. 3, as shown in Fig. 1.4. The applied potential is again asquare wave with frequencyf= 2.77 kHz (T= 360 s). When 1/RC>>f(red), the full

exponential rise of the output voltage, indicated in eqn. 1.20 is apparent. One can easily see that

as theRCtime constant is made smaller (1/RClarger), the limiting output becomes a better andbetter reproduction ofVin. Alternatively, when 1/RC 1/RC) inputs, but to produce a faithful replica of the input for low (f


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frequency cutoffs to be of more than limited utility. The second point also refers to the Bodeplot and introduces a very central,

but frequently misunderstoodconcept, namely the frequency

bandwidth, . Figure 1.5

illustrates how this information iscommonly conveyed. They-axis is

the transmittance, T, expressed inunits of dB.

T dB( ) = 20log10V

out

Vin (1.23)

This is done to make the display ofwidely varying quantities practical.

Note that frequently the relevant

measured quantities are powers,rather than voltages, and since powerscales asP~ V

2, the factor of 20 is

replaced by 10 in eqn. 1.23. Thus a transmittance value of -20 dB corresponds to Vout= Vin/10.Also note that frequencies < 1 s-1 are denoted as the passband, while frequencies > 1 s-1

are denoted as the stopband. The bandwidth is simply the width of the pass band in frequencyspace. In our example here, the bandwidth contains all frequencies between 0 and 1 s

-1, so

= 1 s-1. Of course, you will also often see frequency displayed in units of circular, rather than

angular, frequency. The two are related by = /2, so = /2. The bandwidth, although

it is a frequently cited quantity, is a fuzzy quantity in the sense that the

circuit response does not drop abruptly to zero outside the passband, but rather

falls off in a gentler fashion, as shown in Fig. 1.5.

The last issue to raise with regard to capacitors and their behavior

concerns the equivalent of resistance, called capacitive reactance, or moregenerically, impedance. The basic idea can be gained by considering the

simple circuit shown in Ckt. 4. Here the input voltage is a pure sinusoidalvoltage, so we can use the phasor notation to write,

V(t) =V0

cos t+ ( ); or

V(t) = V0e

ite

i(1.24)

so if we use eqn. 1.14,I = C(dV/dt),

I(t) = CV0 sin t+ ( ); or

I(t) = iCV0e

i t+( )= iCV(t)

(1.25)

Figure 1.5. Bode plot for the frequency response of an idealizedButterworth filter with a cutoff (3 dB) frequency of 1 s-1.

Circuit 4


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Thus, if we recognize a generalized form of Ohms Law, V=IZ, whereZis a generalized form ofresistance, called impedance (generally) or capacitive reactance. For the case of a capacitor, as

shown in Ckt. 4, the capacitive reactance is given by comparing eqn. 1.25 to the generalizedform of Ohms Law, to yield,

Z() = 1iC

(1.26)

As we might guess from the form of the impedance in eqn. 1.26, the way in which combinations

of capacitors are accounted for is different than it is for resistors,

1

Cequiv

=

1

Cii=1

n

(series)

Cequiv = Ci (parallel)i=1

n

(1.27)

One final word about simple passive components. Less frequently used are inductors, which are

nothing more than small solenoids, coupling electric and magnetic fields. These passivecomponents are governed by,

Z() = iL (1.28)

whereL is the inductance in Henrys (H). Inductors are governed by anf(V) relationship givenby,

V = LdI

dt (1.29)

where a voltage of 1 V across an inductance of 1 H corresponds to a rate of change in the current

of 1 A s-1

.

Before leaving our consideration of circuits built onpurely passive elements, we introduce one last, extremely

important and ubiquitous, idea the voltage divider.Consider the circuit shown in Ckt. 5. Lets illustrate the

principle with a specific example; suppose Vin=10 V.

Then we can immediately calculate the total currentflowing because there is only one current path, and thecurrent will be given by the input voltage and the total

resistance,I= Vin/(R1+R2 +R3 +R4) = 10 V/10 k = 1mA. Now that we know the total current we cancalculate the voltage at all of the intermediate points in

the circuit by calculating the voltage dropped across each individual resistor. For example, thevoltage across R1 is given by V= (1 mA)(1 k) = 1V. Thus, the voltage between R1 and R2 is

Circuit 5


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V12 = Vin - V= 10 V 1V = 9V. Now, the voltage dropped across R2 is (1 mA)(2 k) = 2 V.Therefore, the voltage V23 = V12 2 V = 7 V. The voltages dropped across R3 and R4 aresubsequently given by (1 mA)(3 k) = 3 V and (1 mA)(4 k) = 4 V, respectively, giving V34 =V23 3 V = 4 V. Thus, the simple circuit gives a series of voltages at the intermediate points thatare determined by the relative magnitude of the resistances. For example, if we had chosenR1=

R2 =R3 =R4, then independent of the specific values chosen, the voltages dropped across each ofthe resistors would be the same, and the voltage would drop in four equal increments from the 10

V input to ground. Apparently then, the voltage divider provides a series of voltages at theintermediate points of the circuit that are proportional to the ratio

of the remaining resistances to Vin. The voltage divider in its manyguises is probably the most commonly encountered circuit element.

A specific realization of the voltage divider that is particularly

useful in chemical instrumentation is the Wheatstone bridge,shown in Ckt. 6. First, a note about schematic reading the arrow

over the resistor Rb indicates that this resistor is variable, i.e. it can

be changed during circuit operation, either manually or, morecommonly today, automatically as part of a control circuit. Theanalysis of the circuit starts with the fact that the two voltage

dividers R1-Rb and R2-Rx are in parallel, so Vin must be droppedacross both pairs, i.e. I1(R1 +Rb) =I2(R2 +Rx), and Rb is adjusted to

make Vbal= 0, i.e.I2Rx =I1Rb andI1R1 =I2R2. We can solve thefirst equation forI2 =I1(R1 +Rb)/(R2 +Rx). After some algebra, we finally obtain,

Rx=

R2R

b

R1

(1.30)

Rarely do we actually use the bridge to actually measure the value of Rx. More commonly, wemeasure the amount that Rb must be changed in order to keep Vbal= 0, i.e. the null condition. As

a specific example the commonly encountered thermal conductivity detector in gaschromatography places a heated filament with resistanceRx(T)in the effluent stream of the GC.

Small changes in the gas composition cause the thermal conductivity to change which changesthe temperature and, therefore, the resistance,Rx, to change. Rb is continually altered to maintain

the null condition, Vbal= 0, and the signal required to do this is logged as the GC detector outputvoltage.

1.4 Active Circuit Elements. Although there are a number useful functions that can be

accomplished with combinations of passive components only, to realize the full potential of

modern electronics, we need to take advantage of so-called active components. The first elementthat we introduce is the diode, represented by the triangle-bar symbol in Ckt. 7. First, letsexplain the symbol. Diodes, as the name implies are bipolar devices with an anode (excess

positive carriers, i.e. holes) and a cathode (excess negative carriers, i.e. electrons). Themnemonic to remember how the diode is oriented is The arrow points north, where north

stands for the n-type material (i.e. the material with excess electrons). Under forward biasconditions, this is the electrode that serves as the cathode. An alternative way to think about it is

Circuit 6


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that a negative voltage applied to the cathode will repel the electrons in the n-type region, drivingthem over the barrier and into the p-type material, thereby supporting an electrical current.

TheI=f(V) relationship of the diode

can be expressed mathematically, but it

is much simpler to just refer to thegraph of theI-Vcharacteristic, anexample of which is shown in Fig. 1.6.

First, note that they-axes areasymmetric, with reverse bias currents

in the low range, while forwardbias currents are orders of magnitudelarger (mA). Second, there is a large

asymmetry between the forward andreverse-bias voltage axes. While the

forward bias current turns on at a

voltage near 0.7V (Si diodes) or 0.3 V(Ge diodes), the large increase inreverse current occurs at much larger

voltages, 75-100 V is a typical value.In fact, there are specific diodes which

are meant to be run in reverse biasmode, called Zener diodes. Third,

there is a sharp increase in the forwardbias current past the threshold or turn-

on voltage, meaning that the forward bias voltage is held very close to the threshold voltage dueto current limitations under normal circumstances. As an example

of this phenomenon, consider Ckt. 7 withR = 1 kand Vin = 10 V.Even if there were no resistance at all in the diode, the maximumcurrent would be 10 mA. However, this current occurs in thesteeply rising portion of theI-Vcharacteristic and naturally limits

the voltage dropped across the diode, e.g. the 1N4148 Si diodewould limit at ~ 0.7 V forward bias voltage under these conditions,

and the remainder of the voltage would be dropped across the loadresistance,R.

By far the

dominant application of diodes exploits their

unidirectional conduction properties. Thecircuit shown in Ckt. 8 illustrates the point. Asquare wave input, symmetric about 0 V, is

differentiated by the C-R1 combination toproduce a series of alternating positive- and

negative-going spikes at the input of the diode,D1. Assuming that the spikes are larger than

0.7 V, the forward spikes (less the 0.7 V

Circuit 7

Figure 1.6. I-Vcharacteristic of a typical Si diode.Bothx- andy-axes are asymmetric. Adapted from

koala.ece.stevens-tech.edu/~utureli/EE359/lecture/uncompressed/lecture2.ppt.

Circuit 8


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forward bias voltage) are passed on to the output and dropped across output resistor,R2. Ofcourse, we could have chosen to turn the diode around and passed only the negative-going

spikes.

A slightly more complicated variant of the diode is the bipolar transistor,

essentially the circuit element on which the electronics revolution wasbased. These devices are signified by the symbol shown in Fig. 1.7 andalways have 3 terminals emitter, base and collector. These 3 terminals

are eithern-p-n orp-n-p. In other words the base is composed of Si dopedwith the opposite polarity material as the emitter and collector. Standard

circuit drawing conventions always place the arrow at the base-emitterjunction, and again we use the mnemonic arrow points north to determine

which terminal is n-type (frequently in modern practice the circle isomitted). Clearly the device shown is an npn transistor. Basically, we will

be concerned with an extremely limited set of transistor behavior,determined by a simple set of rules: (1) The collector must be more

positive than the emitter for an npn (reverse this for apnp); (2) The base-emitter and basecollector junctions act like diodes normally the base-emitter is conducting and the base-

collector is not; (3) The circuit must be designed not to exceed maximum values ofIC,IB andVCE; and (4) If rules 1-3 are obeyed then the current flowing in the collector arm is determined

by

IC= h

FEIB

(1.31)

where hFEis the transistor gain, typically of the orderhFE~ 100. Essentially the simple bipolartransistor is a current amplifier, where the current flowing in the collector arm is a large multiple

of the current in the base.

This property has an enormous number of useful consequences, of which we will examine onlyone, illustrated in Ckt. 9, the emitter follower. First, as long as Vin

stays below about 10.7 V, the base-collector junction is reverse-biased, as required by rule 1. Then, forward biasing the base-emitter

junction requires ~0.7 V, so the emitter voltage is the base voltageless the forward bias requirement, i.e.Vout= VE= VB 0.7 V.

WheneverVB changes, the output voltage faithfully follows thechange, minus the 0.7 V forward bias.

{Parenthetical but

important! remark.This seems to be a fairly

trivial accomplishment, so what is the point of theemitter follower in a phrase its load management.

In chemical instrumentation we often encountermeasurement components that have extreme values

of impedance; a good example is an indicatingelectrode used in potentiometry which can have an

Figure 1.7.

Symbol fornpntransistor.

Circuit 9

Circuit 10


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equivalent input impedance near the Grange. Lets think about the problem presented in Ckt.10, where Vm is a voltage measuring device, e.g. a voltmeter. Up to this point we have beenthinking about voltage measurements as an ideal process, by which we mean that the process

of measuring the voltage does not perturb the circuit in any way. However, this is never strictlytrue. Just to the side of the voltmeter is a small resistor, Rm, which is meant to convey the idea

that the voltmeter has an internal resistance, called the input impedance.

If we want to measure the voltage being dropped across resistor R2, then we want the inputimpedance,Zin, to be as high as possible. Lets see why. Suppose Vin = 10 V, andR1 =R2 = 1

M. Then, the ideal voltage dropped across R2 is just 5V, as given by the voltage dividerequation. Suppose that we are using a Keithley 2000 6-1/2 digit voltmeter with an input

impedance rated at 10 G. Now, as the arrows indicate, current splits at the circuit branch pointwith part going through each arm. Now we can exercise a straightforward (but tedious) voltage

divider analysis by (1) calculating the equivalent resistance,Requiv, of the R2-Zin parallelcombination; (2) calculate the total currentItotal= Vin/(R1 +Requiv); (3) solve for the current spit in

the lower half of the circuit,Im = (R2/R2 +Rm)Itotal; (4) finally calculate the measured voltage

using Vm =ImZin. When we go through this for the case stated, we find thatItotal = 5.000 25 A,and Vm = 4.999 75 V. Since the ideal voltage would be 5.000 00 V, the error is (5.000 00 4.999 75)/5V = 0.005%.

Now consider the case where we are trying to measure the voltage of an indicating electrode

presenting an output impedance of 1 G. The same process leads toItotal= 5.2381 nA, and Vm =4.7619 V, for a relative error of 0.24 V/5 V = 4.8%. Now the fractional error we are willing totolerate will depend on the application to which the voltage measurement is put, but 5% is too

large for most chemical determinations. The solution of course is to use a voltage measuringdevice with a much higher input impedance. Fortunately for us such instruments exist; for

example there are special electrometers built exactly for this type of application withZin values ~

2 x 1014

. End of parenthetical remark}

With this background we can now analyze the behavior of the emitter follower with respect to

load management. Previously we noted that Vout= VE= VB 0.7 V. Clearly, any change in thebase voltage will immediately be recognized at the emitter, i.e.VB = VE. he change inemitter current is then IE= VB/R. Now using eqn. 1.31 andIE= IC+IB, we getIE= hFEIB +IB

= (1 + hFE)IB. Rearranging,

IB=

1

1+ hFE( )

IE;

IB =1

1+ hFE( )

IE

(1.32)

and combining with the expression for change in emitter current yields,

IB=

VB

1+ hFE( )R

(1.33)


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Finally we are ready to recognize the input impedance of the emitter follower as,

Zin=

VB

IB

= 1+ hFE( )R hFER (1.34)

Typically the value of the transistor gain, hFE~ 100, so the input impedance is the actual loadresistance R multiplied by ~100, making the load much easier to drive it draws a factor of 100

less current than the load would by itself.

Now lets turn the problem around and use the emitter follower in Ckt. 9 on the output end of adriving circuit. The output impedance will be governed by the quantity, VE/IE. Using the

relations, VB = VEandIE= IC+IB, we can derive an expression for the output impedance.

Suppose the base voltage is being determined by another circuit with characteristic voltage andimpedance, Vsource andZsource. Then,

IB

=

VB

Zsource (1.35)

and

IE= 1+ h

FE( )IB =1+ h

FE( )VBZ

source

=

1+ hFE( )VE

Zsource

(1.36)

so

Zout=

VE

IE

=

Zsource

1+ hFE

(1.37)

Thus, the output impedance is lowered by the same factor that the input impedance was raised.If we think back to the example in the parenthetical remark, we could run the output of our 1 Gindicator electrode through an emitter follower with hFE~ 100 and produce a voltage source withan equivalent output impedance of ~10 M, thereby making the measurement problem muchsimpler and reducing the measurement error.

It is hard to overemphasize the importance of loading in electrical circuitry and, by extension, in

chemical instrumentation. Together with ground loops (another pernicious problem that we willnot be able to address) in this class, they represent one of the biggest problems to the design of

highly functional circuitry for chemical instruments. We always want to design for high inputimpedance and low output impedance. The terminology we use is that poorly designed (or just

intrinsically high output impedance) source circuits will load the measurement device, leading toinaccurate measurements.

Finally we note that there are an enormous number of different uses for and types of transistors.We have specifically addressed one application of bipolar transistors, and mainly as a way to

introduce the important topic of loading. The other major type of transistor function is provided


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by so-called field-effect transistors, or FETs, where the application of a gate voltage, VG, controlsthe passage of current from a source to a drain contact,IDS. These devices are extraordinarily

useful; in fact the majority of contemporary digital logic is based on the complementary metaloxide semiconductor (CMOS) version of FETs.

2. Integrated Circuits and Instrumentation Systems

2.1 Operational Amplifiers.

If we had to rely on individual components to assemble the electronic functions we have al cometo now and love, then your personal music device would be the size of a room. What powered

the electronics revolution was the development of the integrated circuit, or IC. In an IC all of thecircuit components are fabricated into a single piece of silicon, utilizing suite of technologies

ranging from photolithography for pattern transfer to high temperature diffusion to alter theconductance properties of exposed portions of the silicon wafer. For our purposes we need not

concern ourselves with how ICs are made but can treat them as functional elements to beincorporated into chemical instrumentation to produce modules of great power and utility.

A particularly useful and ubiquitous

integrated circuit is the operationalamplifier (op amp), essentially a

multi-stage differential amplifierdesigned to exhibit extremely large

open-loop gain values,A ~ 106, very

high input impedance and very low

output impedance. Based on whatwe just learned about load

management, these sound likeintriguing properties. In addition, it

is usually utilized in combinationwith negative feedback to control

the output, but before we introducethis concept, lets take a look at

what the packages look like andhow they are represented. Figure

2.1 shows the circuit diagram for ageneric op amp and a photograph of

a set of individual ICs. We can seefrom the figure that the device is a

3-terminal device with twoadditional power supply inputs.

The 3 additional connections in the8-pin DIP are typically either used

for null offset or are left unconnected. The two input connections labeled V+ and V- are the non-inverting and inverting inputs, respectively. As shown in the configuration of Fig. 2.1, the op

amp output would be governed by the simple equation,

Figure 2.1. (Top) Photograph of

8 DIP (dual in-line package) modules. At the lower leftis the famous 741C op amp. (Bottom) Complete circuit

graphic representing an op amp. Typically the positive(VS+) and negative (VS-) supply voltages are not shown

in the circuit diagram, even though they must always bepresent for the device to function.


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Vout= A V

+V

( ) (2.1)

with the caveat that the maximum (minimum) voltage output is governed by VS+ (VS-). In other

words, the output voltage can never be greater than VS+ or less than VS-. Thus, ifA = 106

and|VS| = 10 V, then the measurable voltage difference between the (+) and (-) inputs would be 10

V. Any V greater than 10 V would result in the output limiting at eitherVS+ orVS-. Ofcourse, the specific voltage range depends both onA and the supply voltages, but the point is that

very little difference in the input voltages causes the output to slew to its positive or negativelimit, according to eqn. 2.1. Since voltages used in practical circuits would rarely need to be

discerned with this level of precision, the op amp wouldseem to have limited utility, outside of the (useful) function

of acting as a voltage comparator.

The development that saves the op amp and adds greatpower and flexibility to its operation is the feedback loop,

more specifically negative feedback, as shown in Ckt. 11.

The feedback loop shown here between the output and theinverting input produces behavior such that the op ampadjusts the output to keep the two inputs equal, i.e.V+ = V-,

or in the case of Ckt. 11, since the output is connected directly to the inverting input, Vout = V+,effectively forming a voltage follower. Again, just as in the case of the emitter follower, there is

no voltage gain, but as you might have guessed there is great capacity for load management; infact instead of simply increasing (decreasing) the input (output) impedance by hFE, the op amp

inherently has a very largeZin and a very lowZout. Thus, the voltage follower shown in Ckt. 11performs the very useful function of isolating the load from the drive circuitry.

In fact the concept of negative feedback is so powerful that it gives rise to two ideal properties

that can be used to evaluate the function of the vast majority of op amp circuits. They are:

1. The op amp adjusts the output to keep V+ = V-, and2. The op amp draws no current, i.e.Zin = .

To see how useful these properties are, lets take a look at afew op amp circuits using feedback. Ckt. 12 shows a very

simple design, that employs some common op amp circuitfeatures. First, the non-inverting input is grounded. Thus, the

inverting input, by property (1), is held very close to ground

potential. It is said to be at virtual ground. Property (2) saysthat the op amp will draw no current, so all of the currentIinpresent at the inverting input is shunted through the feedback

resistor producing an output voltage, Vout= -IinR, i.e. theoutput voltage is proportional to the input current, effectively

converting current to voltage, an extremely useful function forchemical instrumentation. To see why consider just two very different types of detectors, flame

ionization detector used in gas chromatography and photomultiplier tubes used in various types

Circuit 11

Circuit 12


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of optical spectrometers. Both of these detectors produce a current signal, however, thecomputers used to run the instruments and log the data can record only voltages (vide infra).

Thus, the current-to-voltage converter shown in Ckt. 12 is very handy.

A slightly altered version of this circuit is shown in Ckt. 13. This circuit can be analyzed in

essentially the same manner as was just used for Ckt. 12. The inverting input is held at virtualground, and it draws no current. That means that thecurrent coming into the input leg of the circuit, Iin =

Vin/Rin, is equal in magnitude to the current in the feedbackloop,If= -Vout/Rf. Equating the two yields,

Vout =

Rf

RinV

in (2.2)

Thus, the input voltage is inverted and amplified by the

ratioRf/Rin. We can extend this idea to make a summing

amplifier, if the input legs are arranged to sum at the (virtual ground) inverting input, as shownin Ckt. 14. Now things are starting to get very handy; by choosingR1 =R2 =R3 =Rf, one canreadily sum the three input voltages and obtain,

Vout = RfV1

R1

+

V2

R2

+

V3

R31

= V1 +V2 +V3( ) (2.3)

There are other useful combinations of the passive elements in the summing amplifier in Ckt. 14,e.g. choosingR3 = 2R2 = 4R1 produces an output

that is the weighted sum of the inputs, andsignificantly the weighting factors vary by powers

of 2, suggesting an application in converting thebinary digits (bits) of a computer representation of a

digital number to an analog voltage proportional tothat number, i.e. a digital-to-analog converter, or

D/A. This, in fact, is exactly how modernpotentiostats produce voltage programs needed for

voltammetry experiments the computer designsthe voltage program in software, and the digital

values are fed to the reference electrode-working electrode pair through a high bandwidth D/A.

We can, of course, put other circuit elements in the feedback loop. Replacing Rfof Ckt. 13 witha capacitor, C, produces a nearly ideal integrator. The current in the input leg, Vin/R must equal

the current in the feedback loop,If= -dQ/dt =-CdVout/dt. Integrating this last equation yields,

Vout=

1

RCV

in dt + const. (2.4)

Circuit 13

Circuit 14


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As you can imagine this is very handy. Frequently, it is necessary to vary some voltage in alinear fashion in time, i.e. to produce a ramp voltage. The ramp can produced quite nicely from

an op amp integrator and a constant Vin. Choosing theRCtime constant then sets the rate ofincrease (decrease) of the ramp.

2.2 Analog-to-Digital Conversion

It will com as no surprise that measurements are typically carried out so as to produce analog

(i.e. continuously varying) quantities, while digital computers need, well, digital information toprocess. Whole textbooks can be written about the various methods of analog-to-digital, or A/D,

conversion, so we have time only to hit the high spots. We begin with the orientation that noneof us in the chemical sciences are likely to build our own A/D, so we are faced with the task of

purchasing A/D converters designed by someone else intelligently. The three principal figures-of-merit that we need to consider are: (a) range, (b) precision and (c) speed. All three parameters

need to be carefully chosen and specified in order to optimize the performance/cost ratio.Generally we want enough performance to accomplish the measurement of interest, but extra

performance costs more, so we want to be circumspect.

Lets begin with a real example. Consider a flameionization detector in a gas chromatograph capable

of producing output currents in the range, 0 >Iout > -100 A being fed to a current-to-voltage converter

withRf= 50 k, thus converting the current tooutput voltages in the range, 0 5 V. This clearlysets the range needed for our application (logging the

output of the gas chromatograph detector into thecomputer for processing and display). Now the fun

begins. We have to decide what kind of precision weneed for both they- (voltage) andx- (time) axes. Tounderstand lets look at a generic time-dependent

signal, such as the one shown in Fig. 2.2. The figure makes the idea of sampling clear. We wantto obtain specific digital representations of the waveform at regularly spaced time intervals, so

that instead of being represented by the continuous curve in the figure, the waveform can berepresented as a set of (voltage, time) pairs.

The first problem is how frequently do we need to sample the waveform in order to obtain an

accurate representation. Remember that we can go very fast indeed if we are willing to spend alot of money, so there is a speed/cost trade-off that we have to consider in practice. Here we are

guided by a very important concept known as the Nyquist Sampling Theorem, which states:

Nyquist Sampling Theorem: In sampling an analog waveform the samplingfrequency must be at least twice the highest frequency component found in the

original waveform to avoid aliasing the results.

Thus, once we know the bandwidth of the signals to be sampled, we can determine the maximumfrequency needed for the A/D converter. But wait a minute you say, how do we know the

Figure 2.2. Arbitrary time-dependent

signal being sampled at regularintervals.


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maximum frequency associated with the data in Fig. 2.2, since it does not consist of regularperiodic signals? There are several ways to get at this. First, one could acquire similar data,

subject it to Fourier analysis and look for the highest significant amplitude components in thefrequency domain. Admittedly this involves some work. Second, we could determine (for

example by looking up the specs) the bandwidth of the detector being used. If the detector has a

bandwidth of 100 Hz, there is little point in paying for a 1MHz A/D, since very little of thebandwidth would actually be used. Finally, we could assess the signal directly and make ourchoice based on an analysis of the time-domain behavior. Admittedly this latter approach is a bit

seat-of-the-pants but it has the advantage of being simple, fast and cheap.

Going along with our example, lets assume that the flame ionization detector is specified at amaximum bandwidth of 100 Hz. We could choose anything with 200 Hz or higher for A/D.

Choosing a faster A/D would not hurt anything, but we would of course be paying for unusedbandwidth. However, if we chose anything less than 200 Hz, then our A/D converter would

introduce a deterministic error, known as aliasing, into the results.

The other issue we need to consider is the precision of the A/D, typically specified by thenumber of bits, e.g. a 16-bit A/D has the capability to provide readings over its range with a

precision of the range/216

= range/65,536. In our example above, that would provide us with (5V)/65,536 = 76.3 V wide data bins. In other words, two adjacent digital values, say 2154 and

2155 would represent voltages separated by 76.3 V. A/Ds with up to 24-bits of precision arereadily available commercially, so the choice is again one of price (you pay more for more bits)

vs. performance. Most commonly the choice here is made based on the noise characteristics ofthe measurement. For voltages measured across resistive load,R, with bandwidth, f, the noise

floor is determined by the Johnson noise given by,

VJ = 4kTRf (2.5)

where kis Boltzmanns constant and Tis the temperature in K. For the example considered here

with f= 100 Hz andR = 50 k, the Johnson noise floor is ~ 300 nV. However, a typical flameionization detector is much noisier then this thermal limit. Lets use a typical figure determinedfrom the instrument of 1 mV. Under these conditions, the 16-bit A/D is overkill, because it

would devote 2 mV/.076 mV ~ 26 units of measurement just to noise. Given these conditionswe could get by with a (much cheaper) 12-bit A/D with a precision of 1.2 mV in a 5 V range.

Alternatively, we do not want to do what Fig. 2.2 implies and introduce so-called digitizationnoise into the measurement. Notice just past halfway along the time axis the sampled data do not

do a particularly good job of representing the data. Thus, we are looking for a compromiseposition, where we sample with enough precision to represent the data well, but not so much as

to devote significant measurement capacity to noise.

Finally we note that there are several different ways to accomplish the actual A/D conversion.They include successive approximation converters, sampling converters, and flash converters to

name a few. A simplified circuit schematic for a very fast flash A/D converter is shown in Fig.2.3. Clearly the successive comparators are arranged to have non-inverting inputs of 1 V, 2 V, 3

V,.. etc. The analog input voltage is available for comparison at the inverting input of all 7comparators simultaneously. Thus, if the analog input is Mvolts, then all of the comparators


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Mwill be negative, while the rest will be positive. The boxlabeled Encoder Logic, then takes this information and

produces a 3-bit number. For example, ifM = 5.2 V, thenthe first 5 comparators would be negative, and the output

would be {101}. This type of A/D is very fast (hence the

name) but extremely inefficient, since it requires 1comparator for each distinguishable voltage level, e.g. an 8-bit flash A/D would require 256 comparators. We could go on

to explore the inner workings of other types of A/Ds, but forour purposes we do not care so much about the operational

details but rather about their operating characteristics.

2.3 Lock-In Amplifiers

Finally we end by taking a somewhat closer look at theconcept of electronic noise. We have already encountered the

concept of Johnson noise, the noise associated with thediffusive motion of electrons through resistive materials.

Another fundamental limit is placed on electron transmissionwhen the absolute number of electrons becomes small.

Called shot noise, this type of noise arises from the discretenature of electrons and has a magnitude given by,

Ishot = 2eIf (2.5)

where e = electron charge = 1.6 x 10-19

C,Iis the dc current magnitude and fis the

measurement bandwidth. Like Johnson noise, shot noise is defined as an rms (root-mean-square)

quantity, and also like Johnson noise it is Gaussian and white (frequency-independent). Shotnoise is significant only at relatively small currents. If the current is 1 A and the measurement

bandwidth 10 kHz, Ishot= 57 pA, or more to the point, Ishot/Idc = 0.0057%. However, ifIdc = 1

pA, then the shot noise is 57 fA, Ishot/Idc = 5.7%. The Gaussian nature of shot (or Johnson)

noise refers to its amplitude distribution, i.e.

p I, I+ dI( ) =exp I

2

2Ishot2

2 IshotdI (2.6)

which gives the probability of a given current fluctuation having an amplitude betweenIandI+dI. Perhaps more important for this discussion is the fact that both Johnson noise and shot noise

have flat frequency distributions, i.e.shot() = constant.

Both shot noise and Johnson noise are fundamental, meaning that under normal circumstances,they are unavoidable. However, there are other sources of noise which can swamp these

fundamental noise sources and these can (and must) be addressed by design. The most importantsource of noise is actually a whole family of noise sources, that is denoted by the common

Figure 2.3. Flash A/Dconverter. Adapted from

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic

/adc.html


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frequency spectrum and called 1/fnoise. The 1/ffrequency spectrum of this ubiquitous noisesource, not to mention the fact that both Johnson and shot noise decrease as the square root of

bandwidth suggest a genericapproach to low noise

measurements. What if we could

arrange to make measurementswhere the 1/fnoise is small, i.e. athigh frequencies, and over a limited

bandwidth, thereby making fsmall

as well. Then we could addressboth issues simultaneously. There

are a whole class of electronicmeasurement strategies that fit this

description and are knowngenerically as bandwidth-narrowing

technique. We will consider only

one such technique, lock-indetection.

The basic idea for lock-in detectionis shown in Fig. 2.4. To begin lets

assume that we are dealing with a time-periodic signal. If the signal is not naturally periodic,then we can always arrange to make it periodic by simply repeating the perturbation applied to

the chemical system in a periodic fashion to yield,

E(t) = EScos t+ ( ) (2.6)

This in turn can be mixed with (i.e. multiplied by) a reference square wave of the samefrequency that transitions at the zeroes of sin t,

Eref (t) =1; 0 < t< /

1; / < t< 2 /

(2.7)

If we low pass filter the resulting output, thereby averaging the result over times long compared

to the period, =RC >> 2/, the output will be,

Vout

=

1

2E

Scos t+ ( )

0

EScos t+ ( )

2

=

2ES

sin (2.8)

where the angle brackets, denote a time average. The operation conveyed mathematicallyin eqn. 2.8 is carried out between the red box, labeled PLL for phase-locked loop, and the

oscillating signal. The PLL is critical. It basically locks onto the frequency of the signal insuch a way that if a frequency difference, , develops, the voltage-controlled oscillator, VCO,

changes its frequency in such a way to drive the reference frequency in the direction to bring

back to zero. This is good, but what if the relative phases just happen to be at a particularly poor

Figure 2.4. Schematic block diagram of a simple lock-in amplifier. Adapted from www.ee.tamu.edu/People/

bios/hemmer_files/18-Lock-in-amplifiers.ppt


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value, e.g.= 0. Since we cannot control the phase of the experimental perturbation, in general,

it is necessary to add a phase adjustment to the lock-in amplifier. This is typically under operatorcontrol via a front-panel adjustment.

Finally, we also have to choose the low pass filter for the output again via a front panel

adjustment. This is also important, because the bandwidth of the low pass filter effectivelydetermines the pass band of the electronic filtering. Suppose we wish to perform an ac

voltammetry experiment, where the potential is modulated by a small amplitude sine wave at 1kHz added to the usual ramp voltage. We can de-modulate via a lock-in amplifier with say a =

1 s time constant, giving a bandwidth, 1/= 1/(1 s) = 1 Hz. This is equivalent to making a new

measurement covering the frequency range from 999.5 Hz to 1000.5 Hz, thereby discriminating

against all of the low frequency noise sources, such as 1/fnoise and power line bleedthrough atmultiples of 60 Hz.

electronics for chemical instrumentation

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