Logarithmic Amplifier

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# Logarithmic Amplifier - PowerPoint PPT Presentation

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

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### 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

I = abs(Uout) + 10^(abs(Uout)*4.121 - 5.98)

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

To calculate the output signal for a given conductance

(sM in nS and U in V):

Uout = (log(sM× U) + 5.83) / 4.05

Observed Current Window: 10 – 100 nA

B1

A1

Observed Current Window: 5 – 25 nA

C1

Observed Current Window: 0 – 10 nA

It is necessary to focus on the relevant current range to resolve the different conductance groups

At high I (1mA) and high U (1 V),

the gap heats up,

measurements are impossible

due to current oscillation.

200 nA

1 mA

Shortcut ~ 100 mA

Conclusion:

For Break-Junction Experiments at high

Voltages, a current limiting resistor may help to facilitate measurements.

Critical Current

Simmons:

f = (b(U)/(a × 10.25))2 + U/2(f in eV; b in nm-1)

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)

Not useful

Sinc I(V) is linear

For small V

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 (result per CH2)

0.153 = dC-C in ODT

f = (b(V) /(a*10.25))2 + U/2(result in eV; b in nm-1)

f = (bN(V) /(a*10.25*0.153))2 + U/2(result in eV; bN per CH2)