I think the answer is neither here nor on Wikipedia but forgive me if it is, I don't read maths well. This question applies specifically to current density during transcranial stimulation with either DC or AC currents.

I know that current density for DC currents can be calculated as current/area (A/m^2). e.g. if I have 1mA (anodal or cathodal) current running between 2 electrodes with a surface area of 35cm^2 each, I'll get a current density of 1mA/35cm^2=0.0285mA/cm^2 at each electrode.

For alternating currents (with polarity change), however, it is not clear to me whether I should use the peak to peak amplitude (i.e. the difference between max.anodal current and max. cathodal current), or half peak amplitude (the absolute difference between 0 current and max. anodal/cathodal current intensity - which is equivalent if the AC current is symmetrical) in this calculation. Is it the maximum value or the slope (max. value minus min. value) that is important?

e.g. if I apply an AC current, alternating between -1 and 1mA, is my current density the same as for DC stimulation (because of absolute current being maximally 1mA in either cathodal or anodal direction) or do I have to double it (because of a net difference of 2mA between maximal anodal and maximal cathodal current)?

In the academic literature on transcranial AC stimulation either definition has been applied but I suspect only one can be correct.

  • $\begingroup$ I would use the root mean square of the current defined by :$I_{rms}^2 = \dfrac{1}{T} \int_0^T I^2(t) dt$. $\endgroup$ – Joelafrite May 19 '15 at 15:37

Current density (CD) is dependent on voltage (AC or DC), geometry of the conductor, and the material properties. I use printed circuit tracks where the geometry plays the most important role in calculating the CD. For straightforward geometries, I guess hand calculations could be used - but for me, such hand calculations have been incorrect by up to a factor of 10 when verified by the finite element analysis (FEA).

I use FEA and my basic finding about CD is this: "Current density is higher along the shortest path." As an example, the current along a semi-circular track will flow very close to the inner radius of the track and thereby producing the highest CD along the inner radius. You can sketch the geometry and CD description here to visualize what I just explained. For AC in your hand calculations, I believe you may have to use the peak to find the maximum CD, neither the average nor 2x of the peak.

I would suggest to sketch your circuit in enough details, apply the above intuitions to mark the most probable locations for the highest CDs, follow-up with hand calculations and find the maximum CD. I would also suggest to confirm at least one set of your calculations using the FEA. If you have transients, thermal effects, material anisotropy or other complexities, then FEA might make your life easy. Good luck.


On AC a nice measure to take could be the effective current. Its definition is the following: The effective AC current is the current value such that the average power transfer would be the same on an DC current.

So, power is therefore calculated: $$ P(t) = V(t)I(t) $$

For AC circuits where voltage and intensity varies with time, we have a varying power: $$ P(t) = V_0 I_0\cos(\omega t)\cos(\omega t + \phi) $$

The average power is: $$ \langle P(t)\rangle = V_0 I_0 \langle\cos(\omega t)\cos(\omega t + \phi)\rangle = \frac{1}{2}V_0 I_0\cos\phi $$

Where $\phi$ is the phase angle between current and tension. As you can see for maximum real power transfer, this angle must be zero. Assuming that, the effective current $I_{eff}$ is by definition: $$ \langle P\rangle = V_{eff} I_{eff} = \frac{1}{2}V_0 I_0 \quad\Longrightarrow\quad V_{eff} = \frac{V_0}{\sqrt 2}\quad\mbox{ and }\quad I_{eff} = \frac{I_0}{\sqrt 2} $$

So, the power transfer on DC with $I_{eff}$ current and voltage $V_{eff}$ is the same of this AC circuit with this effective current and tension.The effective tension/current is precisily what is given from electrical companies and electrical equipment running at AC power. Some countries, the tension is $110V$ or maybe $220V$. This tension is the effective one. Its not the peak one. For your needs you can define an effective current density.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.