Operated in the constant height mode, the STM forms a current image by recording
ID: 2080099 • Letter: O
Question
Operated in the constant height mode, the STM forms a current image by recording the current as the probe is scanned over a surface. A probe, 3A above a gold surface, consists of a double tip, with a second atom 0.5A higher than the lowest atom (which is the one 3A above the surface) gives a 'ghost' image displaced from the main image. What is the relative intensity of this ghost image? (see exercise 2.6 in Chapter 2). The overall gain of an STM controller is 1000x (equivalent to30dB, where dB = 101og_10) at low frequencies. If the piezo resonance occurs at 10kHz, what is the maximum 3dB bandwidth of the amplifier (the frequency at which the gain of the amplifier has fallen by 3 dB from its DC gain)? (The gain falls 2x or -3bB per octave for a simple RC circuit like the integrator in an STM). To do this figure out how many octaves below 10 kHz one must go to get to the full gain, then double the frequency to find the -3dB point (this is how the bandwidth of amplifiers is usually specified.) The input capacitance to an STM amplifier is 2pF. The desired frequency response is 5kHz (i.e., 0.0002 s time constant). Choose the largest current to voltage conversion resistor this constraint allows, and then calculate the smallest measureable current (i.e. just equal to the Johnson noise) at the target bandwidth. The angular frequency, omega, at which the response of an RC integrator falls by 3dB is given by omega RC = 1Explanation / Answer
The scanning tunneling microscope's electronic have been built with purely analog electronic sweep generator and Z feedback loop, and an oscilloscope or chart recorder for taking data. This will not be the approach of choice for any STM project today
The input is a small current Itunnel, and the output is a voltage Uout = Rfeedback * Itunnel. I tried to figure out the bandwidth to be expected from the current amplifier, as well as the way to achieve it. Specifically, I was wondering whether the capacitor in parallel to the feedback resistor Rf, which I found in STM schematics, was really necessary. Besides the "real" components in the circuit, noise sources for amplifier current noise in and voltage noise en. The tunneling junction is represented by a perfect current source it, as well as a resistance Rt and capacitance Ct of the tunneling junction and its connection. In typical STM applications, I estimate
Rt = Ubias/ Itunnel = 0.1V / 1 nA = 100 MOhm,
Ct = 10 pF.
The capacity estimate assumes 3 cm of shielded connection wire (2 pF/cm, conservative estimate), 2 pF input capacitance of the op-amp, and some additional stray capacitance. I hope I'm not totally off base there... In the feedback loop,
Rf= 100 MOhm is chosen for suitable amplification, and
Cf= ?? is to be determined.
To check for stable operation, calculate the noise gain, i.e. the closed-loop amplification for the voltage noise source en
Acl(f) = (Rf+Rt)/Rf* (1 + j f/fz) / (1 + j f/fp) with
fz= 1 / [2 Pi (Rf Rt)/(Rf+Rt) * (Cf+Ct)] zero frequency, and
fp = 1 / (2 Pi Rf Cf) pole frequency.
At high frequencies, the amplification depends on the capacities only, and approaches
Acl(f>>fp) = 1 + Ct/Cf..
Stable operation at optimum bandwidth is achieved when the closed-loop gain at the pole frequency is equal to the open-loop gain of the op-amp, i.e.
Acl(fp) = GBW/fp,
where GBW is the Gain-Bandwidth product. GBW = 5KHz; at GBW = 5KHz. The solution of the resulting quadratic expression, courtesy of Burr-Brown, is
Cf = 1/(4 Pi Rf GBW) [1 + 8 Pi Rf CtGBW]1/2.
With the typical numbers give above, we get Cf < 0.1 pF for both op-amp types. This means that very likely no additional feedback capacitance will be needed - 0.1 pF are easily caused by stray capacitance of the resistor leads, circuit board etc. In fact, the main challenge will probably be to keep that stray capacitance as small as possible: Looking back at the expression for fp, we get
fp = 5kHz for Cf = 0.2 pF, but only
fp = 0.5 kHz for Cf = 2 pF, which might well be more realistic.
A common suggestion is to put the Rf resistor on standoffs to reduce capacitive coupling to the circuit board.
Noise considerations:
Noise on the op-amp output is closely related to this.In fact, the voltage noise en, amplified by the noise gain Acl(f), will turn out to be one major noise source in typical STM configurations
All calculations will assume a resistor value R = 100 MOhm, and a bandwidth B = 5 kHz. The three noise contributions to consider are:
Resistor noise: The feedback resistor itself will produce thermal noise (caused by thermal fluctuations of electrons inside the resistor), no matter how perfect the op-amp. Its contribution is
uresistor, eff = (4 kT R B)1/2,
with Boltzmann's constant k, absolute temperature T, and resistor value R and bandwidth B.
uresistor, eff = 160 µVrms.
Current noise: The shot noise of the amplifier's input bias current. Since op-amps with very small input bias current are used in the tunneling current amplifier (we want to measure small currents!), this error contribution is small, even after being amplified by the feedback resistor R.
ucurrent,eff= in R B1/2.
.ucurrent,eff= in R B1/2.
Voltage noise: the amplifier's input voltage noise will be amplified by the error gain, which depends on the capacitance of tunneling junction and feedback. network.
uvoltage, eff = en(1 +Ct/Cf) .
uvoltage, eff = en(1 +Ct/Cf) .
Current noise is negligible for the low-bias current op-amps typically used in STMs. For the large feedback resistor assumed here, the dominant term may actually be caused by the resistor alone, but op-amp voltage noise may be amplified to comparable values (depending on capacities). The total noise estimate is well below 1mV, corresponding to 10pA of tunneling current, which should be good enough for decent images.
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