Neurons are excited via an external electrode by passing current through it. A neuron at rest is at -70 mV, it needs additional charge amounting to around 15 mV to initiate an action potential. But the external electrode only gives electrons, which are negatively charged, how can they increase the potential inside the neuron? Does the electrode provide charge to the outside of the membrane of the neuron?


2 Answers 2


The electric field of an external electrode provokes the opening of sodium channels, which in turn has the effect that the positive charged sodium ions travel into the cell body and invoke the wanted change to the Membran potential…

reference: Simulation Neuronaler Netze [("Simulation of Neural Networks")] / Andreas Zell. - Bonn; Paris; Reading, Mass [u.a.]: Addison-Wesley, 1994, first edition


The current in the cells is ionic, and the electron motion is negligible. Yes, the increase of membrane potential to the specific threshold causes changes in the channels, which causes a spike. Let's look in more detail how to get to that threshold, before which the channels are closed.

In depolarization, the membrane potential is increased, where membrane potential is a difference between voltage inside the cell and the voltage outside the cell (let's follow this convention). In the simplest case of extracellular DC microstimulation with a monopolar cathode, this would be a result of a decrease of extracellular voltage, and an increase of intracellular voltage.

The passing current on the electrode is negative, this decreases the (extracellular) voltage, but with having less decrease further away from the electrode tip. Thus, it depends on a distance to the cell - if the cell is too far, then the current may identically affect both extracellular and intracellular voltages, without sufficiently affecting their difference.

With the closer distance, however, we should also consider the resistance of the membrane. If it's so large that intracellular voltage is not directly affected by the current, we would still get different extracellular voltages at different sides around the cell (because voltage change is distance dependant as mentioned above, and the cell might be also sufficiently big). At different parts of membrane, the extracellular voltage would be different, but intracellular stay the same. The membrane will attempt to equilibrate, and increase it's intracellular voltage to average out the membrane potential that was made different at different sides by current application.

On the other hand, with little membrane resistance, both the extracellular and intracellular voltages will change almost identically without needed affect on their difference.

Thus for a given current and a given distance(and a given cell size), we need "the right" membrane resistance so that the change provided by the current application is different extracellularly vs intracellularly. There should be a spatial gradient in extracellular voltage, but also a smaller spatial gradient in intracellular voltage. Then one part of membrane would be be depolarized, and another - hyperpolarized.

If there is little gradient in intracellular voltage, even if with significant gradient in extracellular voltage, the resulting membrane potential will not be sufficiently different across membrane. Since in practice we can control the current, and not the membrane size, resistance and distance to it - we want control for not having too much current (otherwise we'll have too little intracellular gradient because of current flow through the cell plasma), and not too little current (because of lack of current flow due to high resistance).


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