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NBE-E4120, Cellular Electrophysiology
Exercise 1
Solving instructions
1. An example solution:
Using trial and error method, we first select a 10M resistor to limit the currents.
Current i divides into two branches, and iin should be ~50 pA. We assume that the
current iin will be small compared to other currents -> i ≈ i2
Let’s try selecting a 100 M resistor to the other branch and then calculating the
resistance for R.
Kirchhoff:
4,5𝑉 − 𝑖 ∙ 10𝑀 − 𝑖𝑅 = 0
→𝑅 ≈ 10𝑘
𝑖𝑅 − 𝑖𝑖𝑛 𝑅𝑖𝑛 − 𝑖𝑖𝑛 ∙ 100𝑀 = 0
i
10M
iin 100M
i2
R
Rin
2. a) See exercise slides.
b) The difference between the recorded voltage
and the membrane voltage less than 1%:
Combine with the voltage division equation:
𝑉
𝑉𝑚
> 0,99
𝑅
𝑖
𝑉 = 𝑅 +𝑅
𝑉𝑚
𝑖
𝑒
→ 𝑅𝑖 ≈ 50 𝐺Ω
c) Now we start considering signals that change over time. Therefore, we must apply
impedances.
Voltage division for the membrane potential and the measured potential (similar
than in purely resistive circuit):
𝑉=
𝑍𝐶
𝑉
𝑍𝑅 +𝑍𝐶 𝑚
1
The impedance of a capacitive component:
𝑍𝐶 =
The impedance of a resistive component:
𝑍𝑅 = 𝑅𝑒
𝑗𝜔𝐶
The equation that describes the dampening of the signal amplitude as a function of
1
the angular frequency ω of the signal:
𝑉
𝐴=| |=
𝑉𝑚
√(𝜔𝐶)2
1
√(𝜔𝐶)2 +𝑅2
𝑒
The dampening is given in dB:
−3 𝑑𝐵 = 20 log10 𝐴
By combining two previous equations, the angular frequency that is dampened by 3
dB can be solved -> 99,8 1/s, which corresponds to 16 Hz. Does this have an
influence on action potential recording?
3. Noninverting amplifier
−𝑉𝑖 = 𝑅𝑖𝑛 𝑖𝑖
𝑅
→ 𝑉𝑜 = (1 + 𝑅 𝑓 ) 𝑉𝑖
𝑖𝑛
𝑉𝑜 − 𝑉𝑖 = 𝑅𝑓 𝑖𝑜
Inverting amplifier
𝑉𝑖 = 𝑅𝑖𝑛 𝑖𝑖
𝑅
→ 𝑉𝑜 = − 𝑅 𝑓 𝑉𝑖
𝑉𝑜 = 𝑅𝑓 𝑖𝑜
𝑖𝑛
4. Familiarize yourself with the extra material of lecture 1 (especially chapter
‘Measuring biological signals’) and exercise task 2, where you can an example
sketches of recording setups. What have you learned (based on task 2 and extra
material) about e.g. input resistance of a measurement device or the effect of a
micropipette on recording high frequency signals?