�1

Biosensors and Instrumentation: Tutorial 3 Solutions

1. A schematic cross section of an ion sensitive field effect transistor (ISFET) is shown in figure 1.

�Figure 1. Schematic cross section through an ISFET

1.1. In this device, how does the value of pH affect the potential at the oxide/solution interface and the effective threshold voltage of the ISFET?

The interface between the thin gate oxide over the channel region of the transistor and the solution will consist of hydroxyl (OH) groups, which can either accept or donate protons (H+ ions) from the solution. The balance of these charges at the surface will depend on the pH in the solution and will effectively change the surface potential of the oxide and therefore the threshold voltage of the transistor.

1.2. Equations [1] and [2] define the change the surface potential, and therefore VT, with pH for an ISFET.

[1]

where

! [2] 1.2.1. Define the term Cdl and describe how it will be affected by changes in the measured

solutionCdl is the capacitance of the electrical double layer between the gate oxide surface and the solution. It will depend on the surface potential and on the ionic concentration of the solution. If the ionic concentration increases then so will the capacitance.

1.2.2. Define the term βint and why it should be maximised

βint is the intrinsic buffer capacity of the oxide surface. It is a measure of the ability of the surface to accept or donate protons to solution. The buffer capacity should be maximised to increase the sensitivity by making α closer to unity.

p−Si

n−Si n−Si

DrainSource

SiO2

Encapsulation

Vref

ReferenceElectrode

Solution

€

ΔΨ = −2.3αRTFΔpHbulk

↵ =1

(2.3kT/q2)(Cdl/�int) + 1

�2

1.2.3. If α=1 what will be the sensitivity δΨ/δpH at 25 °C (298 K)?If α = 1 then the sensitivity will be:

�

1.2.4. What type of response does the result of 1.2.3 resemble?This is a Nernstian type response similar to a glass electrode pH sensor.

1.2.5. Suggest a gate oxide material which has high sensitivity with reasons for your choice Tantalum Pentoxide (Ta2O5) is a high sensitivity gate oxide material due to its very large buffer capacity (βint), which also makes it less sensitive to changes in Cdl.

1.3. Figure 2 shows an alternative circuit for biasing and measuring the output of an ISFET pH sensor. Define the ISFET current and voltage IDS and VDS.

Assuming the op-amp inputs are equal then the drain source voltage of the ISFET will be the same as the voltage across R2:

�The current through the ISFET will be the same as the current through resistor R3 and the voltage across R3 is the same as that across R1.

�

1.4.The source-drain current through a MOSFET (or ISFET) in the linear region of operation is given by the equation below. Given that the source drain voltage (VDS) and current (IDS) of an ISFET are kept constant by the op-amp feedback circuit shown in figure 3, what will be the dependence of VGS on VT?

�If IDS, VDS are constant we can rearrange to get:

�The first term is a constant so if VT increases then VGS will increase by the same amount.

1.5. Given that the effective value of VG is controlled by the reference electrode what will happen to the output voltage Vout if the threshold voltage increases due to a change in the pH of the measured solution?If the threshold voltage increases then the gate source voltage will have to increase by the same amount to balance this. As the gate voltage is effectively set by the reference

�

�pH= �2.3

kT

q= �59.2 mV pH�1

IDS = �

✓(VGS � VT )VDS �

V 2DS

2

◆

VGS =1

VDS

✓IDS

�+

V 2DS

2

◆+ VT

�3

electrode and is constant then the source voltage (which is equal to Vout) will decrease by the same amount as the threshold voltage.

1.6. If the ISFET used has a threshold voltage VT with a pH sensitivity of -58mV/pH at 298K and has been set up so that the output voltage Vout is -1V at pH = 7 what will the output be for a pH of 1 and 12? Design an amplifier to give an output with a pH dependence of 1V/pH. The output voltage Vout is effectively –VGS with some offset due to the fixed potential of the reference electrode. Either way the sensitivity of the output with pH has the same magnitude, but the opposite sign, to the sensitivity of the ISFET VT with pH. So we known that Vout is –1V at pH = 7 and the pH sensitivity is +58 mV/pH. So at a pH of 1 the output will be –1.35 V and at a pH of 12 the output will be –0.71 V. In order to get an output sensitivity of 1 V/pH the amplifier needs a gain factor of 17.24. If a non-inverting amplifier is used the gain is (1 + R2/R1) and a gain of 17.25 (close enough) can be obtained with R1 = 2.4 kΩ and R2 = 39 kΩ.

�Figure 2. Source-Drain Follower ISFET circuit

2. Figure 3 shows the ISFET amplifier circuit we looked at in the lecture. We’re now going to analyse this in a little more detail.

−

+

D

SISFET

+−

RRef. electrode

V

V1 in

2

out

R

R

3

�4

�Figure 3. ISFET Amplifier

2.1. The important voltages are numbered V1...V5 in figure 4. Define each of these in terms of the known voltages and currents in the circuit. Assume that R1 = R2 = R3 = R.Voltages are as follows:

�

�Which means that we can put this back into the equation for V1:

�

�

�

��

2.2. What is the voltage VDS and current IDS of the ISFET?�

�

2.3. Explain how the current IDS is set by controlling the input Vref.Feedback from the output of the comparator through Vout will control the voltage across the resistor RS which then controls the effective gate source voltage of the ISFET. This

D

SISFET

Ref. electrode

−

+

−

+

−

+

R

R R

R2

2 3

3

VoutR1

Vdd

R1

RDS

−

+

refVRout

RS

Iin

I f

V1

V2

V3

V4

V5

V1 = V2 + IinRDS

V2 = (If + Iin)RS

V1 = IfRS + IinRS + IinRDS

V5 = IinRDS + 2IDSR

VDS = V1 � V2 = V2 + IinRDS � V2 = IinRDS

IDS =V5 � IinRDS

2R=

V5 � VDS

2R

�5

keeps IDS at a constant value which keeps the output of the instrumentation amplifier (V5) at a value approximately equal to Vref. Therefore the drain source current will be:

�

2.4. The calibration routine suggests that the ISFET be placed in a buffer of a known pH (pH 7 typically) while the voltage Vref is adjusted to zero the output voltage Vout. What does this imply about the voltage V2 at this point? What about the output of the final op-amp comparator? At this bias position the voltage across Rout is zero so If is also zero. This means that the voltage V2 = Iin RS which is the source voltage of the transistor. The output of the final op-amp stage will also be equal to V2 (relative to ground) so that there is no voltage across Rout.

2.5. If the VT of the ISFET is dependent on pH with a sensitivity of –50 mV pH–1 choose appropriate values of RS and Rout to give an output voltage change of 1V pH–1. The output voltage is:

�So if we choose RS = 1 kΩ then Rout = 20 kΩ.

3. The “O2-FET” mentioned in lecture 11 can potentially measure both the pH and oxygen concentration in a solution. Please try to read reference [7] mentioned in the notes for lecture 11 before attempting this question. Figure 4 shows the response of an O2-FET used to investigate a cell culture. 3.1. When the circulating pump is operating the conditions in culture chamber are kept constant.

In the graph above, what is the difference between the two measurement times when the pump been turned off?The first period when the pump has been turned off only measures the acidification of the solution caused by the cells. In the second pump-off period the voltage applied to the noble metal electrodes causes an additional pH change due to transduction of the oxygen content.

3.2. Estimate the change in pH for the first measurement time if the ISFET has a sensitivity of 55 mV pH-1.The change in voltage is approximately 20 mV meaning that the solution becomes more acidic by ~0.36 pH.

3.3. There is an additional change in the ISFET output of 10 mV between the change in pH measured with the oxygen transducer switched off, and with it switched on. What is the source of this?The cells in the culture chamber are absorbing oxygen from the culture media to fuel their metabolism. This is respiration.

IDS =Vref � VDS

2R

Vout

= ��VT

Rout

RS

�6

�Figure 4. O2-FET measurements

3.4. Can you think of reasons why an ISFET is better suited to measuring short term changes in pH rather than long term measurement of absolute pH?As mentioned in the lectures, almost all ISFETs have a small constant drift in the measured pH with time, caused by charge trapping, oxidation of the surface etc. This means that the output voltage for a particular pH is constantly changing though this change is slow. So it will easily pick up short term changes but would need to be recalibrated regularly or have some drift correction applied to achieve absolute measurement.

4. Figure 5 shows a plot of membrane potential against time for a neuron being activated and delivering an action potential.

4.1. Describe the changes in the cell, in terms of ion channels and the flow of charge, that cause the action potential.Stimulus from another neuron, or some other source, depolarises the membrane of a neuron cell above the threshold level causing voltage gated sodium ion channels to open. The flood of Na+ ions quickly raise the potential until at some positive value potassium ion channels will also open and let K+ ions out of the cell. Near the peak of the AP pulse the sodium channels close again and the flow of positive ions out of the cells will begin to move the membrane potential again in the other direction. The membrane potential will typically overshoot the resting potential before the potassium channels close and the cell will return to a state where the membrane permeabilities for sodium and potassium are at levels found in the resting condition.

M. Lehmann et al. / Biosensors & Bioelectronics 16 (2001) 195–203 199

Fig. 7. Fig. 9.

not have a significant effect on the output signal of theISFET. It only produces a constant offset of less than 1mV, which does not influence the measurements. Thefollowing negative potential causes a basification signalof the ISFET due to the generation of hydroxyl ions inthe surrounding of the ISFET. The application of thebiphasic potentials is repeated seven times and producesthe ‘new’ acidification–respiration signal. When thepositive potential is applied, the ISFET measures theacidification (Fig. 7). However, while applying the neg-ative potential the ISFET measures both the acidifica-tion and respiration. For the evaluation of therespiration the following method is applied: substrac-tion of two absolute values of the basification peakswhen the potential is changed from positive to negativesignals. Due to possible side effects caused by fluidicmovement when the pump is turned off [see also (Sohnand Kim, 1996)], the two chosen values for our evalua-tion were the second and the last peak (Fig. 7).

The respiration of the cells before the addition ofTriton-X-100 was 10 mV (Fig. 7). After addition of

0.3% Triton-X-100, both the acidfication and respira-tion disappeared (Fig. 8).

The first potential steps before and after the Triton-X-100 addition have a comparable value of approxi-mately 40 mV (Fig. 7 and Fig. 8). This can be explainedby the same oxygen content in the media at the begin-ning of the pump–off period.

The aim of the second experiment was to show thatthe sensor is able to measure opposite trends in acidifi-cation and respiration which may occur in the course ofan experiment. In earlier experiments with iodoacetate(Brischwein, 1997; Lehmann et al., 2000), it was shownthat iodoacetate, a potent inhibitor of glycolysis, byblocking the activity of glycerinaldehyde-3-phosphate-dehydrogenase (a key enzyme of glycolysis), causes thiseffect. The second experiment was therefore made withiodoacetate. Fig. 9 depicts the whole measurement.Both the respiration and acidification are present. Bysubtracting the slope of the acidification from theacidification–respiration signal there is a respirationvalue of 3.35 mV on average (Fig. 10). The acidification

Fig. 8. Fig. 10.

V sou

rce

(vs.

Ref)

[mV

]

�7

�Figure 5. Schematic plot of an action potential in a typical neuron cell

4.2. In the Goldman equation, shown below, which parameters are principally affected by the the opening and closing of ion channels?

�The PX parameters describe the permeability of the membrane to the particular ion X which is controlled (assuming the cell is healthy) by the opening and closing of ion channels.

4.3. Suggest two possible ways in which a neuron could be excited into an action potential.Neurotransmitters released into the synaptic gap between an axon terminal and a dendrite terminal activate ligand gated ion terminals in the dendrite. If enough of these are opened the local membrane potential can be depolarised to the threshold level at which an action potential may be triggered. Another possibility is the use of external stimulation to generate an action potential; for example in a current clamp setup, a fixed current is forced into the cell through a glass needle containing an electrode and a conductive electrolyte. This can depolarise the cell until the threshold is reached and an action potential occurs. If the stimulation continues another action potential will be triggered, following the usual refractory period. There is a graph in the lecture that shows something similar to this.

4.4. An example of a current clamp circuit which can inject a current I into a cell is shown in figure 6. What natural process does this attempt to emulate?This emulates the opening of ion channels and therefore the flow of ion current into or out of a cell, which can excite an action potential.

4.5. Given that all resistors marked R are equal, and the voltage input (Vin) and output Vp are referenced to ground, what are the voltages at points 1-3? The op-amp which is driving point 1 is configured as a summing inverting amplifier so V1 = –(Vin + Vp), this feeds a second inverting amplifier so V2 = Vin + Vp. The voltage Vin is the input to the circuit while the voltage Vp is the output of the third op-amp which is configured as a buffer, therefore the voltage at point 3 is the “probe voltage” Vp.

Em

=RT

Fln

✓PK[K+]

out

+ PNa[Na+]out + PCl[Cl�]inPK[K+]

in

+ PNa[Na+]in + PCl[Cl�]out

◆

�8

�Figure 6. Current clamp circuit for neuro-electrophysiology

4.6. Choose a resistor value for Rout that will give a maximum current output of 1nA. Assume that the resistance of the probe is 1 GΩ, and the op-amps have ±15V power supplies. If the current is 1 nA then the maximum value of Vp will be ~1.04 V at the peak of an action potential. That is 1 V for the potential across the probe resistance and 40 mV for the membrane potential. That means the maximum input voltage Vin can be is 15 - 1.04 = 13.96 V. If we limit this to 10V to give a large margin then Rout should be 100 MΩ. This is obviously very large. If Rout = 100 kΩ then Vin for 1 nA will be 0.1 V. It would be easier to overload the current

4.7.When a step change of current is injected the output voltage Vp shown in figure 7 is recorded. Explain how this differs from the actual membrane voltage Vm and what the source of the error is. Can you suggest a way of correcting this?The measured probe voltage has an offset compared to the actual membrane potential due to the resistance of the micropipette probe which can be in the GΩ range. This voltage offset will have the value IRout. A way of correcting this is to subtract a voltage from the probe voltage which is proportional to the input current.

�Figure 7. Input current and output voltage traces from current clamp electrophysiology experiments

−

+−

+

−

+

V R

R

R

R

R

R

in

V pV p

V +Vin p

V in

I

I

out

Cell

1 2

3

3.2. Intracellular Recording Current Clamp

The Axon Guide - 2500-0102 Rev. C 31

Figure 3.3: The <Bridge BalanceA technique.

This technique is used to separate the membrane potential (Vm) from the total potential (Vp) recorded by the micropipette. The technique is schematically represented in A. A dif-ferential amplifier is used to subtract a scaled fraction of the current (I) from Vp. The scal-ing factor is the micropipette resistance (Rp). The traces in B illustrate the operation of the bridge circuit. When the current is stepped to a new value, there is a rapid voltage step on Vp due to the ohmic voltage drop across the micropipette. Since the micropipette is intracellular, changes in Vm are included in Vp. Thus the Vp trace shows an exponential rise to a new potential followed by some membrane potential activity. The bridge ampli-fier removes the instantaneous voltage step, leaving the Vm trace shown.

There are several ways to set the bridge balance. A commonly used technique is to apply brief repetitive pulses of current to the micropipette while it is immersed in the prepara-tion bath. The Bridge Balance control is advanced until the steady-state pulse response is eliminated (Figure 3.4). At this point, the circuit is balanced and the micropipette resis-tance can be read from the calibrated readout of the Bridge Balance control. The same technique can even be used after the micropipette has penetrated the cell. Details of how to achieve this are described in the Axoclamp 900A and MultiClamp 700B manuals.

�

+1V

VA

B

mp

I

I

Vp

( Vp Vm= + IR )p

Vm

�9

5. Figure 8 shows a circuit that could be used in patch clamp measurements of the ion current (Ich) through a single ion channel.

�Figure 8. Patch Clamp Circuit for Electrophysiology

5.1. Derive an expression for the output Vout.Assuming ideal amplifiers and stable feedback, the potential at the inverting input of the op-amp will be Vin. If we define the current flow out of the cell, through the patch clamp electrode and the feedback resistor Rf as Ich then the voltage at the op-amp output (V+) will be:

�Then the output of the the Differential Amplifer will be:

�

5.2. If Ich is in the pA range suggest a suitable value for Rf giving reasons for your answer.If we’re trying to measure a pico-ampere current then typically Rf will be in the GΩ range so that the voltage will be in the µV to mV range.

5.3. Given the above information why should the operational amplifier have FET based input terminals?We need to make sure that the input bias current of the op-amp is as small as possible as this will cause an error in the measurement.

5.4. Assume that the resting value of the transmembrane potential is -70mV and the patch clamp has been successfully located over a voltage gated potassium ion channel. Describe how this ion channel can be activated, and what the effect will be on the output of this circuit.

−

+Diff.Amp.

Cell

+

−

Bath Electrode

Vout

Vin

Rf

V+ = Vin � IchRf

Vout

= V+ � Vin

= –Ich

Rf

�10

Voltage gated potassium channels are activated when the membrane is strongly depolarised to near the maximum action potential value of +50 mV. When they open the positively charged potassium ions should flow out of the cell, though this is contingent on there being a similar concentration gradient between the cell and the patch clamp electrolyte as you would find in a normal physiological situation. Anyway, assuming this, the flow of positive ions out of the cell is equivalent to a flow of conventional current out of the cell, through the electrode and through the feedback resistor. Therefore the output of the differential amplifier will change by –ImRf

6. The quenching of a Ru based fluorescent molecule by oxygen is described by the “Stern-Volmer Equation” which describes the kinetics of the quenching process. This is as follows:

�I0 is the intensity when there is no quenching while I is the intensity at the given oxygen concentration [O2]. In a similar way, τ0 is the fluorescence lifetime with no quenching and τ is the lifetime at oxygen concentration [O2]. KSV is the Stern-Volmer constant which is a measure of the quenching efficiency and is often expanded to be the product of the baseline lifetime τ0 and kq, a rate coefficient for the quenching process such that:

�

6.1. The intensity data from an oxygen sensor made from a fibre optic cable tipped with an immobilised fluorescent material is shown in figure 9. Suggest why this data would make it difficult to use for a sensor and explain a possible source for the observed drift. It should be fairly obvious that there is a downward drift in the maximum and minimum values of the fluorescence intensity. This would be difficult to use in a sensor without calibrating the drift and applying a correction to the measurement data. The source of this drift is likely to be photobleaching of the fluorescent material by the illumination.

�Figure 9. Intensity data from a fluorescent oxygen sensor exposed to saturated atmospheres of O2 and N2

I0

I=

⌧0

⌧= 1 + KSV [O2]

I0I

=�0�

= 1 + kq�0[O2]

294 P.A.S. Jorge et al. / Sensors and Actuators B 103 (2004) 290–299

case where larger fiber core diameters are used. Indeed fora fiber with a 1mm core diameter light coupling efficienciesof ≈20% were achieved.In the experimental set-up of Fig. 1 the butt-coupling con-

figuration was used with a 550!m core fiber. In such situa-tion, an optical power of 70!Wwas available in each outputof the fiber coupler.Preliminary tests, in gaseous environments, were per-

formed using the coated glass slides in order to characterizethe sol gel thin films. The parameters Q and !τ were eval-uated as a function of the solution aging time, from 2 h tomore than 48 h at 20 ◦C. Both these parameters increasedwith increasing aging time. Values for Q parameter between30 and 40% were obtained. Values for!τ varied between 77and 176 ns. As stated in the literature, this behavior results

0 60 120 180 240 300 360 42040

60

80

100

120

140

160

180

(a)

100% N2100% N2

100% O2100% O2

Fluo

resc

ence

Inte

nsity

(mV)

time (s)

0 60 120 180 240 300 360 420-20

-18

-16

-14

-12

-10

-8

-6(b)

100% N2

100% O2

100% N2

100% O2

φ d (d

egre

es)

time (s)

Fig. 5. Response of fiber taper to O2/N2 saturation cycles: (a) fluorescence intensity response; (b) phase response.

from increased film porosity and easier oxygen diffusioninto the matrix [8]. This traduces in higher quenching ef-ficiency and increased oxygen sensitivity. The unquenchedlifetime, τ0, decreased with increased aging time, 630 and545 ns, for 2 and 24 h aging time, respectively. With a dipcoating speed of 3mm/s, film thickness between 600 and800 nm were obtained (measured with a perfilometer). Theresponse time of the sensing films when they undergoneO2/N2 saturation cycles varied between 9 and 11 s (framedto the 10 and 90% reference levels). In the slide configu-ration the detected fluorescence emission was very weak(Pd ≈ 0.36 nW), and due to a low signal to noise ratio (SNR)the phase signal was unstable. The resulting Stern–Volmerplots were non-linear indicating that the sensing complexhad a heterogeneous distribution inside the porous silica

�11

6.2. The lifetime, which is generally unaffected by issues causing drift in the intensity, is commonly measured through modulating the light input with some frequency (f) and measuring the phase shift in the resulting fluorescence. The phase shift (φ) is related to the lifetime (τ) by the following equation:

�Estimate the values of τ0 and τ[100% O2] from the data in figure 10, given that the input frequency is 75 kHz

�Figure 10. Phase data from a fluorescent oxygen sensor exposed to saturated atmospheres of O2 and N2

Estimations of the phase angle for 100% O2 and 100 N2 are approximately -9.5º and -18º respectively. The equation for the phase shift can be rearranged to give:

�So τ0 will be 0.69µs, and τ[100% O2] will be 0.355µs. The angle in the graph above is negative because it is a phase DELAY compared to the modulation of the excitation. Plugging that into the equation would obviously give you a negative lifetime which isn’t sensible so just take the absolute angle and use it in the calculation.

6.3. The data shown in figure 11 is phase and intensity data for step changes in the atmosphere being sensed. Estimate values of I0/I and τ0/τ for each oxygen concentration, assume that the results for [O2] = 0.4 % are effectively the same as for 0% oxygen. Why don’t you actually require the value of the input modulation frequency?When calculating τ0/τ you’ll have 1/(2πf) on both the top and bottom of the equation and they will cancel out so that the calculation will simply be:

�The tables below give all of the results for the estimations I’ve made from the graphs, don’t worry if yours are a little different. Below that I’ve plotted the results in figure T2.

tan[�] = 2⇡f⌧

294 P.A.S. Jorge et al. / Sensors and Actuators B 103 (2004) 290–299

case where larger fiber core diameters are used. Indeed fora fiber with a 1mm core diameter light coupling efficienciesof ≈20% were achieved.In the experimental set-up of Fig. 1 the butt-coupling con-

figuration was used with a 550!m core fiber. In such situa-tion, an optical power of 70!Wwas available in each outputof the fiber coupler.Preliminary tests, in gaseous environments, were per-

formed using the coated glass slides in order to characterizethe sol gel thin films. The parameters Q and !τ were eval-uated as a function of the solution aging time, from 2 h tomore than 48 h at 20 ◦C. Both these parameters increasedwith increasing aging time. Values for Q parameter between30 and 40% were obtained. Values for!τ varied between 77and 176 ns. As stated in the literature, this behavior results

0 60 120 180 240 300 360 42040

60

80

100

120

140

160

180

(a)

100% N2100% N2

100% O2100% O2

Fluo

resc

ence

Inte

nsity

(mV)

time (s)

0 60 120 180 240 300 360 420-20

-18

-16

-14

-12

-10

-8

-6(b)

100% N2

100% O2

100% N2

100% O2

φ d (d

egre

es)

time (s)

Fig. 5. Response of fiber taper to O2/N2 saturation cycles: (a) fluorescence intensity response; (b) phase response.

from increased film porosity and easier oxygen diffusioninto the matrix [8]. This traduces in higher quenching ef-ficiency and increased oxygen sensitivity. The unquenchedlifetime, τ0, decreased with increased aging time, 630 and545 ns, for 2 and 24 h aging time, respectively. With a dipcoating speed of 3mm/s, film thickness between 600 and800 nm were obtained (measured with a perfilometer). Theresponse time of the sensing films when they undergoneO2/N2 saturation cycles varied between 9 and 11 s (framedto the 10 and 90% reference levels). In the slide configu-ration the detected fluorescence emission was very weak(Pd ≈ 0.36 nW), and due to a low signal to noise ratio (SNR)the phase signal was unstable. The resulting Stern–Volmerplots were non-linear indicating that the sensing complexhad a heterogeneous distribution inside the porous silica

⌧ =tan[�]2⇡f

⌧0

⌧=

tan[�0]tan[�d]

�12

�

�Figure 11. Sensor response to step variations of O2 concentration: (a) phase response; (b) intensity response.

P.A.S. Jorge et al. / Sensors and Actuators B 103 (2004) 290–299 297

difference is expected to be maximum around 188 kHz,which is the ideal modulation frequency [5]. This way, an in-crease in modulation frequency is expected to improve sen-sor performance. In Fig. 5(a) and (b) the phase and the fluo-rescence intensity response of the sensing system to O2/N2saturation cycles can be observed. The phase signal showssome instability indicating the need for SNR improvement.In order to demonstrate the phase insensitivity to opticalpower drift a simple test was performed. With the sensinghead in a 21% O2 atmosphere, the optical power injectedinto the fibre system was changed up to 25%. Fig. 6 showsthe consequence of this variation in the intensity and phaseof the fluorescence signal. Although a significant change in

0 60 120 180 240 300 360 420 480-20

-18

-16

-14

-12

-10

-8(a)

80%

100%

39,1%

60%

20,6%

12,3%8%

0,4%

φ d (d

egre

es)

time (s)

0 60 120 180 240 300 360 420 4800,03

0,04

0,05

0,06

0,07

0,08

0,09

0,10

0,11(b)

100%80%

60%

39,1%

20,6%

12,3%8%

0,4%

Fluo

resc

ence

inte

nsity

(mV)

time (s)

Fig. 8. System response to step variations of O2 concentration level: (a) phase response; (b) fluorescence intensity response.

the intensity response occurs, the phase response remainsessentially unchanged. This confirms the ability of thephase detection scheme to avoid power fluctuations inducederrors.The Stern–Volmer plots obtained from phase measure-

ments with all configurations are clearly non-linear, indicat-ing that the dopant is not homogeneously distributed withinthe silica matrix. Instead, the ruthenium complex occupiesenvironments with different oxygen accessibilities. In thissituation, the standard Stern–Volmer equation (Eq. (1)),based on a single exponential decay, no longer describesaccurately the quenching behavior. Alternatively a dual ex-ponential model, corresponding to two different dominant

P.A.S. Jorge et al. / Sensors and Actuators B 103 (2004) 290–299 297

difference is expected to be maximum around 188 kHz,which is the ideal modulation frequency [5]. This way, an in-crease in modulation frequency is expected to improve sen-sor performance. In Fig. 5(a) and (b) the phase and the fluo-rescence intensity response of the sensing system to O2/N2saturation cycles can be observed. The phase signal showssome instability indicating the need for SNR improvement.In order to demonstrate the phase insensitivity to opticalpower drift a simple test was performed. With the sensinghead in a 21% O2 atmosphere, the optical power injectedinto the fibre system was changed up to 25%. Fig. 6 showsthe consequence of this variation in the intensity and phaseof the fluorescence signal. Although a significant change in

0 60 120 180 240 300 360 420 480-20

-18

-16

-14

-12

-10

-8(a)

80%

100%

39,1%

60%

20,6%

12,3%8%

0,4%

φ d (d

egre

es)

time (s)

0 60 120 180 240 300 360 420 4800,03

0,04

0,05

0,06

0,07

0,08

0,09

0,10

0,11(b)

100%80%

60%

39,1%

20,6%

12,3%8%

0,4%

Fluo

resc

ence

inte

nsity

(mV)

time (s)

Fig. 8. System response to step variations of O2 concentration level: (a) phase response; (b) fluorescence intensity response.

the intensity response occurs, the phase response remainsessentially unchanged. This confirms the ability of thephase detection scheme to avoid power fluctuations inducederrors.The Stern–Volmer plots obtained from phase measure-

ments with all configurations are clearly non-linear, indicat-ing that the dopant is not homogeneously distributed withinthe silica matrix. Instead, the ruthenium complex occupiesenvironments with different oxygen accessibilities. In thissituation, the standard Stern–Volmer equation (Eq. (1)),based on a single exponential decay, no longer describesaccurately the quenching behavior. Alternatively a dual ex-ponential model, corresponding to two different dominant

pO2 (%) Intensity (mV) Io/I

0.4 0.106 1

8 0.088 1.2

12.3 0.085 1.25

20.6 0.076 1.39

39.1 0.066 1.61

60 0.0565 1.88

80 0.0465 2.28

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�Figure T2. Plot of fluorescent intensity ratio (I0/I) and lifetime ratio (τ0/τ) against oxygen

concentration for a fluorescent quenching oxygen sensor based on an optical fibre

Neither set of results is terribly linear, and the paper in which they are taken from suggests that there are two different linear regions of operations over different oxygen concentration ranges. The suggestion is that this is a result of inhomogeneity in the glass matrix containing the fluorescent material which affects the diffusivity.

6.4. Explain the discrepancy between the two sets of measurement results. As with the first plot this measurement is subject to bleaching of the fluorophore as the experiment proceeds this leads to a lowering of the intensity which causes an increase in the intensity ratio I0/I.

100 0.04 2.65pO2 (%) Intensity (mV) Io/I

pO2 (%) φ (degrees) Tan(φ) τ0/τ0.4 18.4 0.33 1

8 15.7 0.28 1.18

12.3 15.2 0.27 1.22

20.6 13.8 0.25 1.35

39.1 12.6 0.22 1.49

60 11.4 0.2 1.65

80 10 0.18 1.89

100 8.8 0.15 2.15

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6.5. Describe a possible method of miniaturising the sensor and light source and integrating it with a microfluidic system, perhaps with the aim of measuring the oxygen partial pressure in blood samples.The laser light source could be replaced by an LED, while the light detection could be performed with a single photon avalanche photodiode detector which is ideal for the measurement of fluorescent lifetime. If the application is to measure the oxygenation of blood in small samples the detector and light source could be integrated with a microfluidic flow channel with fixed fluorophore material in contact with the blood. One possible issue with this system would be in trying to prevent cross talk between the source and the detector. Optical filters to stop the light from the LED reaching the SPAD while allowing the fluorescence emission through is very difficult. Can you think of any methods to do this without filtering?