Logarithmic Amplifier
Serial number: N9503A
Pin to the right hand side !
General Remarks:
1.) The switch-pin on the amplifier
should always be kept to the right when viewed from front
If the pin is to the left only very small currents up to ~ 10nA
can be measured
2.) The set-point current should be usually kept at 1nA
Higher set-point currents have to be calculated
from the equations given on the following pages
A nominal set point current of 2.5 nA for example corresponds to
A real current of ~ 20 mA !!!!
3.) Break-Junction experiments should not be
performed at voltages larger than 0.4 V
4.) Do not try to resolve the A Group with it
5.) This head is configured for the old (basement) STM
The voltage offset is - 29 mV.
To measure at a real tip-voltage of + 0.1 V you choose
a voltage of + 0.071 V
To measure at a real tip-voltage of – 0.1 V you choose
a voltage of - 0.129 V.
Calibration Curves using 5 different Resistors
15 KOhm
1 MOhm
10 MOhm
100 MOhm
1 GOhm
Tip +
100000
10000
Tip -
I / nA
1000
100
10
1
0.1
All calculated current values were corrected
for the tunnelling Voltage offset of -29 mV
(Old STM)
0.01
1E-3
-3
-2
-1
0
1
2
3
Output Voltage
I = abs(Uout) + 10^(abs(Uout)*4.121 - 5.98)
7
10
M / nS at 100 mV
6
10
5
10
G0
4
10
3
10
2
10
1
10
0
10
-1
10
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0
Uout / V
But only if Uout values are spaced equidistant
The output signal (Uout) which appears as current in the output
file is logarithmic in the range from 10 – 10000 nA where:
For Positive tip Voltages:
I[nA] = - 10(4.05 (Uout) - 5.83) = - invlog(4.05 (Uout) - 5.83)
where Uout is negative (‘nA’);
For example Uout = -2 V → log(I) = 2.27 → I = 186 nA
Real Current / nA
1000
100
10
-2.2
-2.1
-2.0
-1.9
Output singnal / V
-1.8
-1.7
To calculate the output signal for a given conductance
(M in nS and U in V):
Uout = (log(M × U) + 5.83) / 4.05
10000
Tip Negative (10 - 10000 nA):
I[nA] = invlog(-5.90 + 4.08 (Uout))
where Uout is positive
1000
For example Uout = + 2V
I / nA
log(I) = 2.26
I = 182 nA
100
10
1.7
1.8
1.9
2.0
Uout / V
2.1
2.2
2.3
All Current Results BJ -2 nm to 2 nm; 1s; + 0.2 V; 1 nA
100000
20000
1 mA
( = 65 G0 at 0.2 V)
Counts
10000
1000
15000
Counts
100
0
500
1000
1500
2000
2500
3000
3500
Output Signal / 'pA'
10000
5000
0
0
500
1000
1500
2000
2500
Output Signal / 'pA'
3000
3500
All Current Data Histogram - Break Junction - C4DT - + 0.2 V - 1 nA
Focus on C - Group
700
600
C1
C2
Counts
500
B1
400
300
200
G0
B2
A1
A2
100
0
1400
1600
1800
2000
2200
2400
2600
'I' / 'pA'
Resolution is not very good – C and B group cannot be clearly resolved
Linear Amplifier: C4DT
+1000 mV focus on small jumps (I < 15 nA): A1: 3.5 nS
700
600
Counts
500
400
300
200
100
0
1
2
3
log(I/pA)
4
5
SMC of HS-(CH2)4-SH Break Junction Method – Linear Amplifier
-200 mV Large I = 10 - 100 nA ; C1: 199 nS
1600
Observed Current Window: 10 – 100 nA
Counts
1200
A1
800
B1
400
0
1
2
3
4
5
log(I/pA)
-200 mV medium (I = 5 - 20) B1: 50 nS
500
Counts
Observed Current Window: 5 – 25 nA
C1
400
300
200
100
0
1
2
3
4
5
log(I/nA)
150
-200 mV small (I - 0 - 10 nA) A1 4.46 nS
Observed Current Window: 0 – 10 nA
Counts
100
It is necessary to focus on the relevant current
range to resolve the different conductance groups
50
0
1
2
3
log(I/nA)
4
5
Heating and expansion (+ oszillations) at I ~ 1 mA and U = 1 V
8
Cooling and contraction at I ~ 200 nA and U = 1 V
6
z-Voltage
4
2
~ 200 nm
0
-2
-4
At high I (1mA) and high U (1 V),
-6
-100
0
100
200
300
400
500
600
the gap heats up,
700
Current Output / V
Time / s
1
measurements are impossible
0
due to current oscillation.
-1
200 nA
-2
1 mA
-3
Shortcut ~ 100 mA
-4
-100
0
100
200
300
400
Time / s
500
600
700
6
z - Voltage
4
2
At U = 1 V measurements can be
performed up to ~ 10 micro-Ampere
0
-2
-4
Conclusion:
-6
0
200
400
600
800
1000
For Break-Junction Experiments at high
Time / s
Voltages, a current limiting resistor may
help to facilitate measurements.
0
Current / V
-1
1 nA
10 nA
-2
100 nA
1 mA
10 mA
Critical Current
100 mA
1 mA
-3
-4
0
200
400
600
Time / s
800
1000
Contact Conductance / G0
1000
100
10
-1
0
1
2
3
Penetration Depth / nm
4
5
General Remarks
Tip-Sample Distance (s) under various conditions
1000000
100000
G0
-1:
dlnI/ds = 10 nm
s (10 pA) = 1.6 nm
10000
Aromatic Adsorbates
1000
-1:
dlnI/ds = 5 nm
s (10 pA) = 3.2 nm
I / nA
100
10
1
0.1
0.01
1E-3
1E-4
1E-5
Vacuum
-1:
dlnI/ds = 20 nm
s (10 pA) = 0.8 nm
Air
or Alphatic Adsorbates
1E-6
0
1
2
s / nm
3
4
5
Voltage drop due to Amplifier impedance at close proximity
Simmons:
f = (b(U)/(a × 10.25))2 + U/2
(f in eV; b in nm-1)
100
f / eV
10
1
0.1
a = 0.5
a=1
0.01
1E-3
1E-4
0
2
4
6
8 10 12 14 16 18 20 22 24
-1
b / nm
For a =1 and U = 0 V the above equation is identical with:
f = (dlnI/ds)2 * 9.526 meV where dlnI/ds = -2k = -2(2mf/hbar2)
Different Representations of the Simmons Fit to the ODT I(V) curve
Simmons Data for C8 Single molecule
f = 1eV; a = 0.5
6
4
I / nA
2
0
-2
-4
-6
-2
-1
0
1
2
U/V
0.8
Linearised Representation
Simmons Data for C8 Single molecule
f = 1eV; a = 0.5
4
Simmons Data for C8 Single molecule
f = 1eV; a = 0.5
Vtrans = 1.12 V
0.7
2
ln(I/V )
I/V [nA/V]
3
2
Not useful
Sinc I(V) is linear
For small V
1
0.5
0
0
1
2
2
3
2
U /V
0.6
4
0.4
-2.0
-1.5
-1.0
-0.5
0.0
-1
0.5
-1
U /V
1.0
1.5
2.0
Transition Voltage Spectroscopy on Simmons Curves
0.8
Simmons Data for C8 Single molecule
f = 1eV; a = 0.5
Vtrans = 1.12 V
a = 0.5
2
ln(I/V )
0.7
0.6
0.5
0.4
-2.0
-1.5
-1.0
-0.5
0.0
0.5
-1
1.0
1.5
2.0
-1
U /V
-4.0
Vtrans = 0.53 V
a=1
2
ln(I/U )
-4.5
-5.0
-5.5
Simmons Data for C8 Single molecule
f = 1eV; a = 1
-6.0
-5
-4
-3
-2
-1
0
-1
1
-1
U /V
2
3
4
5
Dependence of I on a
Simmons Fit: f = 1 eV; d = 1 nm; A = 0.0726 nm
a = 0.4
12
1
0.1
9
a = 0.5
Experiment
0.01
6
a = 0.6
1E-3
3
a = 0.7
1E-4
a=1
0
I / nA
I / nA
2
C8: f = 3 eV a = 0.5
C8: f = 3 eV a = 1
1E-5
1E-6
1E-7
-3
1E-8
-6
1E-9
-9
1E-10
-2.0
-12
-1.5
-1.0
-0.5
0.0
U/V
-2
-1
0
U/V
1
2
0.5
1.0
1.5
2.0
No a and hence no m* used:
Area dependent Transmission
b(V) = 10.25 * a(f – (U/2))1/2
(result in nm-1)
bN(V) = 10.25 * a(f – (U/2))1/2 * 0.153
0.153 = dC-C in ODT
f = (b(V) /(a*10.25))2 + U/2
(result per CH2)
(result in eV; b in nm-1)
f = (bN(V) /(a*10.25*0.153))2 + U/2 (result in eV; bN per CH2)
Consistency check for Simmons Model:
2
Current through a Polymer wire vs Length at U = 0.1 V for f = 0.14 eV and A = 0.217 nm
10000
G0
1000
a = 1; f = 0.14; U = 0.1 V  b = 10.25 * a(f-U/2)
a = 0.5; f = 0.14; U = 0.1 V  b = 1.54 nm
100
= 3.1 nm
-1
Slope of Dashed Line ~ 2 nm
Phi-Dashed Line =
2
((dln(I)/ds)/-2)  0.0381 eV
~ 0.04 eV
10
1
I / nA
1/2
0.1
-1
0.01
1E-3
-1
Slope of Dashed Line ~ 3.6 nm
Phi-Dashed Line =
2
((dln(I)/ds)/-2)  0.0381 eV
~ 0.12 eV
1E-4
1E-5
1E-6
-1
Experiment: slope 3.38 nm
1E-7
0
1
2
3
s / nm
4
5
-1
Simmons Model:
2
Current through a Polymer wire l = 3 nm at U = 0.1 V for A = 0.217 nm
f = 0.14 (dlnI/ds - Experiment = 3.38 nm )
-1
I / nA
0.1
0.01
1E-3
Experiment
1E-4
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1
a