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### conversion units

Posted: **Thu May 23, 2019 10:53 am**

by **menica**

Hi,

if I want to convert the current density INA [mA/cm2] in nA I should divide the current in mA/cm2 by the factor (area*(0.01)):

INA/(area*(0.01))

if I want to convert the quantity X [mM/ms] in nA I should multiply by the FARADAY constant expressed in C/mol, by 10(numenbr of electrons in the Na+ ), 0.001 (factor to convert mol in mM), 1000 to convert ms in s, 10^12 to convert from A to nA:

X/((FARADAY)*(10^12))

Is this forumaltion with brackets correct?

Thanks

Best

Menica

### Re: conversion units

Posted: **Mon May 27, 2019 10:38 am**

by **ted**

if I want to convert the current density INA [mA/cm2] in nA I should divide the current in mA/cm2 by the factor (area*(0.01)):

INA/(area*(0.01))

True, assuming that area is in units of um2.

if I want to convert the quantity X [mM/ms] in nA I should multiply by the FARADAY constant expressed in C/mol, by 10(numenbr of electrons in the Na+ ), 0.001 (factor to convert mol in mM), 1000 to convert ms in s, 10^12 to convert from A to nA:

X/((FARADAY)*(10^12))

No. mM/ms is numerically equal to mol/s. Multiplying that by FARADAY converts to coulombs/second == amperes. The complete conversion from mM/ms to nA is simply *FARADAY*1e9.

### Re: conversion units

Posted: **Mon Jun 03, 2019 1:33 am**

by **Darshan**

Hi,

I have a doubt in mA/cm2 to nA conversion:

if I want to convert the current density INA [mA/cm2] in nA I should divide the current in mA/cm2 by the factor (area*(0.01)):

INA/(area*(0.01))

True, assuming that area is in units of um2.

Shouldn't the current in mA/cm2 be multiplied by (area*0.01) to convert it into nA

Explanation:

Code: Select all

```
Multiplying current in mA/cm2 with area in um2
mA/cm2 * area (in um2)
= mA/cm2 * 10^(-8) cm2
= 10^(-8) mA = 10^(-2) * 10^(-9)A
= 0.01 nA
```

**So, mA/cm2 to nA ==> ina (in mA/cm2) *area(in um2) *0.01 **
(Note: Using different Na current notation (ina) than mentioned in question (INA) but similar to NEURON currents)

Similarly, converting nA to mA/cm2 will be done by

**INA (in nA) /(area(in um2) *0.01)**
(Note the different Na current notation)

Also, for this:

if I want to convert the quantity X [mM/ms] in nA I should multiply by the FARADAY constant expressed in C/mol, by 10(numenbr of electrons in the Na+ ), 0.001 (factor to convert mol in mM), 1000 to convert ms in s, 10^12 to convert from A to nA:

X/((FARADAY)*(10^12))

No. mM/ms is numerically equal to mol/s. Multiplying that by FARADAY converts to coulombs/second == amperes. The complete conversion from mM/ms to nA is simply *FARADAY*1e9.

Code: Select all

```
mM/ms *FARADAY (in C/mol)
= M/s *C/mol
= mol/(Litre-s) *C/mol
= C/s *(1/Litre)
= A/Litre
= (10^-9/10^-9) *A/Litre
= 10^9 nA/Litre
```

how to account for factor (1/Litre) or (1/1000 cm3) which is the volume in the conversion?

Thanks,

Darshan

### Re: conversion units

Posted: **Tue Jun 04, 2019 1:46 pm**

by **ted**

Darshan wrote: ↑Mon Jun 03, 2019 1:33 am

Shouldn't the current in mA/cm2 be multiplied by (area*0.01) to convert it into nA

You're absolutely right about this. I was in a hurry to reply and made a couple of very stupid mistakes.

A quick way to derive the conversion formula starts from this fact: for a cell with 100 um2 surface area, a membrane current density in mA/cm2 is numerically identical to net membrane current in nA, i.e. a membrane current density of X mA/cm2 is equivalent to a net membrane current of X nA. For example, if you have a cell with surface area A um2, and you know that sodium current density is Y mA/cm2, the net (i.e. total) sodium current in nA is Y * A/100.

mM/ms is numerically equal to mol/s. Multiplying that by FARADAY converts to coulombs/second == amperes. The complete conversion from mM/ms to nA is simply *FARADAY*1e9.

Don't forget volume and ionic charge--the conversion factor becomes vol*z*FARADAY*1e9 where vol is the volume of accumulation in liters. Surface area of the volume of accumulation plays no role in the conversion.

### Re: conversion units

Posted: **Tue Jun 04, 2019 7:55 pm**

by **Darshan**

Thank you for clearing these doubts!

### Re: conversion units

Posted: **Wed Jun 05, 2019 9:47 am**

by **ted**

We have just seen yet another demonstration of the value of multiple eyes examining the same code.