EAS327: ELECTRONICS for Environmental Instrumentation

Objectives. Cover some important terminology and make you familiar with some of the simplest of electronic circuits that you may use in lab projects or that clarify important physical concepts. This material will also stress the exact and useful analogies between electronic systems and other physical systems. After studying this material you ought to be able to:

A1.1 Electrical Safety

The power-supply schematic gives the electrical setup usual in household and building wiring. The transformer is usually outside, and the centre-tap of the secondary winding, the "neutral," is grounded via a good connexion to earth, eg. an iron stake. The centre-tap neutral runs directly inside, and is (or should be) connected to the wide-slash pin of every outlet (but not uncommonly, household plugs are miswired). A and B are connected to the inside outlets through (in each case) a fuse or circuit breaker. The fuse/breaker is designed to sense excessive current, and break the circuit either by burning out due to excessive Joule heating ( i2R; fuse) or by some other reversible means (breaker).

The short-slash pins of the power outlets connect either to the A or to the B side, in roughly equal proportion. Any outlet having a "U-ground" should run a connexion from that pin to a local ground, usually the steel plumbing (but sometimes one sees outlets offering a U-ground that is not actually connected). Thus, for a properly-connected, grounded outlet, two of the three pins are actually grounds, - one local, one remote (there may be a small potential difference between these two grounds: a multimeter hooked between them might show a small non-zero voltage, and certainly some resistance (the wiring, at least).

The potential difference (voltage) between A and B varies sinusoidally

VAB = A sin ( 2 π f t )

where f=60 [Hz = s-1] is the "line frequency" and A is the amplitude. It is usual to quantify mains voltages in terms of the "root mean square" voltage (σ). For example, the mean square voltage across A-B is:

< VAB2 > =σ2 = 1/T 0∫ T A2 sin2 ( 2 π f t ) dt = A2/2

where the voltage (squared) has been averaged (integrated) over interval T=1/f (ie. the period of one cycle). Thus σ= A/2½. Now it is probably familiar that in North America the r.m.s. voltage of mains power is σ=120 Volts (nominally), so we have A= 2½ * 120 = 170 Volts. Some major appliances connect across AB (eg. stove; clothes drier).

Precautions in electrical/electronic work:

A1.2 OHM's LAW - the "current-voltage characteristic" for a resistor

The electrical potential at any point is defined to be the electrical energy [Joules] per unit charge [Coulombs] of any charge at that point. It is measured in volts, so a volt is the same as a [Joule Coulomb-1]. Current is measured in amperes, and 1 ampere = 1 Coulomb s-1.

A resistor has the function of "resisting" the flow of electrical charge, and is characterised by the value of its "resistance" R, [unit: Ohms]. See Figure 1 (definitional figure for Ohm's Law) representing a resistor, across which an electrical potential difference (or voltage) is applied, and through which consequently a current flows. Note that the diagram defines a sign convention for the direction of positive current flow. The potential difference across the resistor is V = V1 - V2, and Ohm's Law says that:

i = (V1 -V2)/R = ΔV/R [ amperes = volts/ohms]

where ΔV is just a shorthand symbol for (V1 -V2), the voltage drop across the resistor. The form of this equation is characteristic of many transport laws of physics:

Flow or Flux = Driving Force / Resistance.

In this case the flow is a flow of charge (current=rate of transfer of charge), and the driving force is an imposed electrical potential difference (voltage).

It is also useful to know that in terms of energy, a resistor is a "dissipative" circuit element; any current i through a resistor R causes "Joule Heating" at a rate P (power):

P = i2 R [Joules s-1, Watts]

In view of Ohm's Law, we can substitute for either i or R to obtain the alternative expressions:

P = i2 R = ΔV2 / R = ΔV i

Thus knowing the resistance of the block heater in your car, you can calculate the power consumed and the rate of release of heat energy.

But how does one prove this expression for the rate of power dissipation? As charge flows into the resistor it has potential V1 [J Coulomb-1] and as it leaves it has lower potential V2. Thus the rate of flow of electrical energy to the resistor is iV1 while the rate of flow away is iV2. The difference is the amount of energy that has disappeared - been transformed to heat, as the equation above claims.

A1.3 KIRCHOFF'S LAW (conservation of charge)

The net current to a junction point of several circuit branches must vanish:

Σj ij = 0

The notation in this equation means: take the sum over the "dummy" subscript "j" which is used to label the nodes and the currents ij flowing in them. In the diagram, i ranges from (1,2...5), so, the equation simply says

i1 + i2 + i3 + i4 + i5 = 0

Note that a sign convention must be established and adhered to.

A1.4 CIRCUIT GROUNDS & CIRCUIT SCHEMATICS

"Earth," "earth ground," "power-line ground." A point in a circuit of low (in principle zero) resistance to earth. Eg. a 2m metal stake driven into the soil.

"Common," "floating ground," "signal reference", "floating reference point." The "reference point" with respect to which all voltages are measured: need not be power-line ground.

There are many minor variances in the style of circuit diagrams (the circuit shown is called a "half-bridge," or "voltage divider." Note that there is no voltage drop along a perfect conductor (any connecting line in a circuit diagram).

A1.5 THE CAPACITOR and its current-voltage characteristic

After the resistor, the next most commonly used "passive component" of electronic circuitry is the capacitor, whose property is capacitance C [unit: Farad], the ability to reversibly store charge and energy. Again, we recognise a potential drop (voltage) across a capacitor, and define a direction for positive current flow. Let q [Coulombs] be the charge stored on the capacitor. Then i = dq/dt, current is rate of change of charge. The current-voltage characteristic of the capacitor can be written in several equivalent forms:

V1 - V2 = q/C = 1/C ∫ i dt

or differentiating w.r.t. time:

d/dt (V1 - V2) = i/C

A capacitor may be regarded as an open circuit in so far as d.c. (steady) current is concerned: when a steady voltage V is suddenly applied by a voltage source across an initially discharged capacitor, current will flow, resulting in a buildup of charge until the voltage drop across the capacitor equals V: then current ceases. Contrariwise, a capacitor is a short circuit for very high frequency a.c.

A1.6 THE "Half-Bridge" or "VOLTAGE DIVIDER"

Refer again to the circuit diagram for a half-bridge. Provided no current is drawn at the Vo ("V - out") node, one can show using Ohm's Law that:

Vo = V R2 / (R1+ R2)

A1.7 The "Full-Bridge" or "Wheatstone Bridge"

The Wheatstone Bridge is made up of two half-bridges connected in parallel, and produces a differential voltage output Vo from the difference between the mid-point voltages of it's two respective "arms" (ie. dividers). This voltage is often called the error-voltage of the bridge. From the rule for the half-bridge, we have:

Vo = V [ R4 / (R3+ R4) - R2 / (R1+ R2) ]

and a little rearrangement gives us

Vo = (R1R4-R2R3 ) / [(R1+ R2) (R3+ R4) ]

from which we see that the bridge is "balanced" (ie. Vo vanishes) when:

R1R4 = R2R3

The Wheatstone Bridge is an invaluable circuit for the detection of very small imbalances in numerous types of system. For example, thermostatic systems (temperature control circuits) often utilise a Wheatstone Bridge, one arm of which contains a resistive sensor element whose resistance R=R(T) responds to the temperature T to be controlled. Let the desired value of T be Top. By providing a feedback system driven by the "error voltage" Vo we may take control the heating/cooling device so as to drive T back towards the target value Top. This is achieved by letting one of the other resistors, say R1, be tunable; and one tunes it such that the bridge will be perfectly in balance (Vo = 0) when T=Top.

A1.8 The VOLTAGE SOURCE

The diagram gives a representation or schematic of a "voltage source". Rs is the called the "source resistance" or "internal resistance," and is necessarily non-zero for any real device (ie. realizable voltage source). Sometimes one speaks of an "ideal voltage source" for which Rs=0. Such a (notional) device has the property that it imposes the specified voltage drop across its terminals irrespective of the type of load applied and the current drawn. Such a device would be capable of providing infinite power, and so is an electrical equivalent of the mechanical perpetual motion machine.

This "equivalent circuit" may represent anything from a simple battery to a very complex laboratory power supply.

Now apply a "load," having resistance RL, across the source. The larger RL the smaller the load (RL= infinity corresponds to open circuit, ie. no load). The addition of the load resistor yields us a voltage-divider circuit. By the voltage-divider rule, the voltage applied across the load is:

VL = Vs RL/ (Rs+RL)

Unless RL >> Rs, VL is less than Vs. This means that if you want to test the voltage on your car battery, you should test it under load, ie. with the motor turning over.

Examples of common sensors/devices as voltage source: thermocouple; photovoltaic semiconductor.

A1.9 Voltage Receivers

Many environmental sensors provide an electronic output in the form of a voltage signal, and are correctly regarded (from the electronic viewpoint) as being voltage sources, thus characterised by having a calibration (the link between Vs and the value of some environmental variable, eg. temperature T ) and a source resistance Rs. To use (ie. read, interrogate) such a sensor, we require to connect it to a "voltage receiver."

A voltage receiver is a device with two voltage-input terminals, often labelled V+ (or "hi" or "+"), and V- ("lo" or "-"). Ideally, the voltage receiver "responds to" (measures, or amplifies) only the difference (V+ - V-). For example, an A/D (analog-digital) converter will produce a number, representing the input (V+ - V-); and (if we neglect "common-mode error," which is discussed below) that response may be written as:

N = α (V+ - V-) + Vos

Here Vos is a small (but usually non-zero) error called the "input offset voltage." Input offset voltage may be measured by short-circuiting the input (ie. by setting V+ = V-) and noting the reading (N=N0). Usually Vos may be eliminated, or nulled out, by adjusting a "zero" potentiometer, until N=N0=0. And provided the receiver is correctly calibrated in terms of its sensitivity (or span), the slope-constant α will be unity. If not, most receivers, in addition to providing for the nulling-out of zero-offsets (Vos=0), also provide means to adjust the span (ie. sensitivity), so as to ensure that α=1.

Characteristics of voltage-receivers:

Matching source resistance and input resistance: By referring to the schematic for a voltage source connected to a voltage receiver, you should be able to see that in order for the receiver to "see" all of the signal (Vin = Vs), we require Rin >> Rs. Unless this "matching requirement" is fulfilled, "loading" of the source by the receiver reduces the signal actually detected. Hence as a general principle, any voltage receiver used to inspect, display, record or digitise a voltage signal should have an input resistance Rin greatly exceeding the internal resistance (output resistance, source resistance) of the source. Sometimes it is necessary to introduce a circuit called a "buffer," between the source and the receiver. Ideally, a buffer (in practise, some form of amplifier) has infinite input resistance, and zero output resistance (ie. its output stage functions as an ideal voltage source, with nominally zero source resistance).

Summing up, signal receivers and signal sources must be carefully matched with repect to source and input resistances... and care must be taken over the issue of receiver type and grounding... consider the implications of the alternative choices of single-ended or differential, and the implied grounding of the source circuitry. For example, the error voltage from a power-line grounded Wheatstone bridge will require to be measured using a differential receiver, or a floating single-ended receiver: but not a single-ended, grounded receiver. Why?

A1.10 Determining the internal resistance of a voltage source

Specifications of purchased voltage sources (receivers) will state the source (or input) resistance. But in case not - or you lost the manual - how can it be determined?

A1.11 The CURRENT SOURCE

The diagram gives a representation or schematic of a "current source". Again, Rs is the "source resistance" or "internal resistance," less than infinite for any real current source: sometimes one speaks of an "ideal current source" for which Rs = infinity . An ideal current source has the (notional) property that it drives the specified current across its terminals irrespective of the type of load applied and the corresponding voltage drop.

Even with no load connected, current Is flows internally, through Rs. When a "load" having resistance RL is connected across the output, load current IL flows through the load, and I = IS + IL. It can be shown using Ohm's Law that:

IL = I / {1 + RL/RS}

Unless RL << Rs, the load current IL is less than I.

Example of current source: semiconductor phototransistor used as shortwave radiation sensor.

A1.12 RESPONSE OF A LINEAR SYSTEM

A1.13 SIGNIFICANCE OF THE TIME CONSTANT τ

Consider the accompanying schematic of a signal y(t), seen over an interval of time (t1 <= t <= t2) during which its mean value (average) was Y. We can "decompose" y(t) as:

y(t) = Y + y'(t)

where y'=y'(t) is called the "fluctuation." The fluctuation is simply the instantaneous deviation from the mean (Y).

Now suppose the fastest fluctuations in the signal have period Tmin and frequency fmax=1/Tmin. Our sensor has time constant τ.

If we want our sensor to "show," ie. reveal, or measure, the fastest fluctuations, then we must ensure that

τ < = Tmin /10

ie. make sure we use an instrument with a time-constant at least 10 times smaller than the fastest periods of interest.

But IF we only care to measure the average value of the signal, or the very slow changes, we can safely allow τ >> Tmin, a choice which will result in our "seeing" all changes with periods T>>τ.

This implies that we can use a "slow" sensor to achieve averaging of a variable signal, as an alternativeto use of a faster sensor, and subsequent digital averaging after the signal has been aquired (digitised, logged).

A1.14 THE LOW-PASS RC FILTER

We study the RC low-pass filter because many environmental sensors function on analogous principles, and secondarily because it is a widely-used device.

Suppose we regard this "thing" as a black box, to which we apply some input voltage Vs, and which produces an output Vo (we will assume no current is drawn from the output node). How does the output relate to the input?

Step-Response of RC lowpass filter

Suppose that for t<0, Vs=0, and the capacitor is fully discharged (ie. holds no charge). Then Vo, the voltage on the capacitor, is zero for t<0. Then at t=0, we switch the input Vs to some fixed, constant value, say Vs=V. That is, we "drive" the input with a step change (analogy: dropping a thermometer into boiling water). Can we calculate the output?

Yes. We use Ohm's Law:

Vs - Vo(t) = i(t) R

and we have the current-voltage characteristic for the capacitor

Vo (t) = q(t)/C = 1/C 0∫ T i(t) dt

Note that (t) denotes "time-dependence" (it does not mean multiply by t)! There are many ways to proceed from here, but suppose we differentiate the second equation, noting that C is constant. We have:

dVo/dt = i(t)/C

and if we substitute for i(t) from the first equation we have:

dVo/dt = (Vs - Vo(t))/ (RC)

This is a first-order ordinary differential equation, and its form is exactly that of the earlier-given general class; the class of instruments characterised by a falling rate of response to a new equilibrium. That is, the rate of increase of Vo depends on the difference between the "driving" voltage Vs and the response, Vo: and, that rate of response is "moderated" by the RC product, which is the product of the resistance to flow (in this case, of charge) and the capacity of the device to store the property of interest (in this case charge). If you do happen to have Calculus, you may know how to solve this differential equation (which term makes the equation inhomogeneous, ie. "forces" the evolution of the response? What is the auxiliary equation?). You will of course make use of the "initial condition," that Vo(0) = 0, ie. output voltage vanishes at t=0. The solution is:

Vo(t) = Vs [ 1 - exp ( - t / τ )]

where τ=RC is of course the (RC) "time constant" of the device. Quiz: what are the values of Vo/Vs at particular times t=(0, RC, 2RC, 3RC)?

Sinusoidal response of RC lowpass filter

Suppose now we "drive" the input of the device with a sinusoidal voltage V(t)=Ai sin(2πft), such as can be generated by a laboratory function-generator (voltage generator).

Assuming all start-up transients have decayed away, the output (proof is outside the scope of the course) is also a sinusoid, and of the same frequency f, but delayed in phase and with a reduced amplitude Ao

(Ao/Ai)2= 1/{ 1 + ( f/fo)2 }

where fo=1/(2π RC) = 1/(2π τ) is called the "half-power frequency." It is conventional to call (Ao/Ai)2 the "power gain" and represent it by the symbol G=G(f). Clearly if f=0, G=1. Thus, the power gain of the low-pass filter is unity at very low frequencies. Equally clearly, if f is infinite, G=0. So the power gain of the low-pass filter is zero at very high frequencies. You can easily see that at the particular frequency f=fo, ie. at the half-power frequency, G=½.

Exercise: Plot a graph of this frequency response function.

Useful mental picture: regard a capacitor as an "open circuit" for DC (that is, steady voltages and currents), and a short circuit for very high frequency (f>>fo) AC. Thus, after switching transients have "died out," the RC filter "shows" the DC input at its output. But if driven by very fast AC, it shows "nothing" at its output.

Fluid flow analogy:

Consider a huge reservoir R, filled to a (unchanging) depth d. At the base of this reservoir a fine tube, closed by a valve, can transfer water to another, much smaller reservoir R', whose cross sectional area A' is small, and whose depth is d'=d'(t). (Schematic of this flow system). Suppose that (if the valve is open) the flow rate of water from R to R' is

Q = k (d - d')

where k is some proportionality constant. Those of you familiar with fluid systems will understand that the hydrostatic pressure on each side of the valve is controlled by the depth (d or d'), ie. the pressures on each side of the valve are ρ g d and ρ g d'. The formula above assumes the flow rate is linearly related to the pressure difference across the valve, and we could make it look like Ohm's law by writing

Q = (d - d')/ (1/k)

where now (1/k) could be interpreted as the flow resistance between the two sides.

Now suppose that at t=0 the capillary R' is empty (ie. d'=0) and that at that instant we open the valve. What is the equation for the "response", ie. the depth in the capillary, d'(t)? I think you can easily prove that again, we will have "falling rate of response to a new equilibrium," where the final (equilibrium) depth in R' is just d'=d.

How does all this relate to environmental instrumentation? Well, suppose instead of charge or water transfer (ie. electronic or liquid flow) we consider heat transfer; instead of resistance to charge (or water) flow, resistance to heat flow. Instead of electrical or volumetric capacitance (ie. charge storage or water storage), heat storage (ie. heat capacity). The mathematics for describing the behaviour of a thermometer is identical to the maths done here.

A1.15 Noise Pickup and Sheilding

"Whenever low-level signals are involved, say in the mV or μ range, it is almost mandatory to use shielded connecting cables." (Benedict)

Two circuits which are not directly interconnected by a conductor may interact indirectly through electromagnetic coupling. Consider a measurement circuit, composed of a voltage source connected to a receiver by a careless layout of conducting wire. The figure also shows a second circuit, in which there is a time-varying current i=i(t). For an infinitely-long straight wire, at any point at radius r away from the wire the "magnetic induction" B can be represented by vectors oriented in circles about the wire (as shown), and the "strength" of the field is B(t) ∝ i/r, where i(t) is the current in the wire.

Now in the second (susceptible) circuit, a voltage source has been coupled rather carelessly to a voltage receiver, using two single conductors with a large open area A between them. By the "Faraday Law of Induction," if the magnetic field B from the circuit on the left intersects this loop, and if B is changing, then a voltage is induced in the secondary circuit. Specifically, if AT is the area of the loop projected perpendicular to the field B then,

Electro-motive force = EMF = "induced voltage" = - Rate of change in time of ( B AT )

or

EMF = - d/dt ( B AT) .

This voltage (lets call it vi , for "voltage induced") adds algebraically to any other voltage in the circuit and causes a current to flow in inverse proportion to the total resistance Rs+Rin about the loop.

To avoid this E/M coupling: