# Ion-selective electrode: high input impedance to measure the voltage

I read that an ion-selective electrode has an high impedance ($100 \, M\Omega$) and that it's necessary a high input impedance to measure the electrode voltage.

I'd like to understand the above sentence with a circuit. I tried to draw a circuit with a voltage source and two resistances (input resistance and load resistance):

for the Kirchhoff law:

$$E=(R_i+R_L) \, i$$

At this point it's difficult to me to understand why I need a high Ri. Could you suggest me the right way please?

EDIT

I am studying a biosensor (https://ieeexplore.ieee.org/document/7576627/) and at page 947 there is the sentence: Ion-selective electrodes (ISEs) separate specific ions with a semi-permeable membrane between two electrodes, thereby creating a potential difference (∼0.1–100 mV) proportional to the amount of that ion concentration in the solution. However, due to the nature of these sensors, their impedance is very high, roughly on the order of 100 MΩ, necessitating high resolution sampling of the electrode voltage with a high input impedance.

Thank you in advance.

• You are mixed up a bit. View the electrode as a voltage source with a series resistance. Then, ask your self what happens to the voltage when you put a load on it that is heavier than the series resistor (Rl is a smaller value resistor than Rl.)
– JRE
Commented Jul 8, 2018 at 9:18
• That is, view Ri and V1 as a single element, and consider what happens to the voltage across Rl for various values of Rl.
– JRE
Commented Jul 8, 2018 at 9:19
• Hello @JRE, the voltage across RL is $V_L=R_L \, i$ or $V_L=R_i \, i - E$. If Ri increases, VL will increase. If RL increases, the current will decrease. I am confused. Commented Jul 8, 2018 at 9:32
• An "ion-selective electrode" as used typically in a mass spectrometer usually has zero input impedance. Commented Jul 8, 2018 at 10:56
• Hello @Andyaka, read my edit answer please. Commented Jul 8, 2018 at 12:03

## 1 Answer

In your setup, the assumption is that both E and Ri are unknown. You are trying to measure E with a voltmeter having an input resistance of RL. The voltage seen by the voltmeter, call it Ev, will be the result of the voltage divider formed by Ri and RL, namely Ev = E*RL/(RL+Ri). Since Ri is not known, E cannot be accurately determined unless RL is much greater than Ri. In that case, the equation reduces to Ev = E. That is why your reference calls for a high input impedance voltmeter.

• Hello @Barry, thank you very much for your explanation. Commented Jul 8, 2018 at 15:47
• Glad I was able to help you. Good luck in your studies. Commented Jul 8, 2018 at 16:43