Metadata-Version: 2.1
Name: eecalpy
Version: 0.9.3
Summary: collection of classes for simple to complex electrical calculations
Home-page: http://github.com/wese3112/eecalpy.git
Author: Sebastian Werner
Author-email: wese3112@startmail.com
License: UNKNOWN
Platform: UNKNOWN
Classifier: Programming Language :: Python :: 3
Classifier: License :: OSI Approved :: GNU General Public License v3 (GPLv3)
Classifier: Operating System :: OS Independent
Requires-Dist: lark-parser

# eecalpy Python module

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The *Electrical Engineering Calculations for Python* module is a
collection of classes for simple to complex electrical calculations, with a
special focus on handling tolerances.

**USE AT OWN RISK, I DO NOT GUARANTEE THE CORRECTNESS OF THE CALCULATIONS IN THIS PACKAGE**

## Installation

The ``eecalpy`` package is available on the Python Package Index (PyPI). The package needs Python 3+, you can install it with:

    $ pip install eecalpy


## Introduction

Check out the voltage divider below. For both resistors their tolerance and the
temperature coefficient α are given (α in parts per million). 

![Simple voltage divider](img/vdiv.png?raw=true "voltage divider")

Let's create two variables for them.

    >>> r1 = R(resistance=1000, tolerance=0.05, alpha_ppm=250)
    >>> r2 = R(2e3, 0.01, 100)
    >>> r1; r2
    1.0kΩ ± 5.0% (± 50.0Ω) [0.9500 .. 1.0500]kΩ @ 20°C α=250ppm
    2.0kΩ ± 1.0% (± 20.0Ω) [1.9800 .. 2.0200]kΩ @ 20°C α=100ppm

The formula for the voltage divider factor is `r1 / (r1 + r2)`. To calculate it use `R.voltage_divider(other_resistor)`:

    >>> r1.voltage_divider(r2)
    0.33 ± 4.0% [0.3199 .. 0.3465]

You can also use a shorthand notation:

    >>> r1 // r2
    0.33 ± 4.0% [0.3199 .. 0.3465]

Attention: Do not use the statement `r1 / (r1 + r2)` here, because it would use the tolerance limits
of `r1` twice (addition and division) and therefore yield a false result.

The result above is an instance of the `Factor` class. Now only the voltage is missing.
These are created using `U(voltage, tolerance=0.0)`.


Let's assume the input voltage is 24V with a 1% tolerance the output voltage of the
voltage divider then is:

    >>> vin = U(24, 0.01)
    >>> vout = r1 // r2 * vin
    >>> vout
    8.0V ± 5.0% (± 400.0mV) [7.6000 .. 8.4000]V

Note: the statement `vout = vin * r1 // r2` does not work. It's evaluated from left to right, so python first tries `vin * r1` which is not implemented (voltage times resistance), but you can always use parenthesis:

    >>> vin * (r1 // r2)
    8.0V ± 5.0% (± 400.0mV) [7.6000 .. 8.4000]V

For demonstration, let's calculate some of the voltage divider parameters.

Current through `R1` and `R2` (to GND):

    >>> vin / (r1 + r2)
    8.01mA ± 3.33% (± 266.81µA) [7.7394 .. 8.2730]mA

Power dissipation of the resistors:

    >>> vout**2 / r1
    65.46mW ± 21.35% (± 13.97mW) [51.4842 .. 79.4301]mW
    >>> (vin - vout)**2 / r2
    128.26mW ± 12.3% (± 15.78mW) [112.4776 .. 144.0351]mW

Let's also see how `vout` changes when the ambient temperature is 200°C:

    >>> r1.at_T(200) // r2.at_T(200) * vin
    8.14V ± 4.97% (± 404.16mV) [7.7359 .. 8.5443]V

`R.at_T(temperature)` is the same as `R.at_temperature(temperature)`.
It returns a new resistor object at the given temperature (in °C).

You can of course also use perfect values, so without the tolerance and
temperature coefficient:

    >>> r1 = R(1e3)
    >>> r2 = R(2e3)
    >>> vin = U(24)
    >>> r1; r2; vin
    1.0kΩ @ 20°C
    2.0kΩ @ 20°C
    24.0V
    >>> vout = r1 / (r1 + r2) * vin
    >>> vout
    8.0V

By the way, you can get the series resistance using `+` and the parallel
resistance using `|`:

    >>> r1 + r2
    3.0kΩ @ 20°C
    >>> r1 | r2
    666.67Ω @ 20°C
    >>> r1 | (R(5e3) + R(3e3)) | r2  # complex statements allowed!
    615.38Ω @ 20°C

## Classes

The available classes are:

* Voltage `U(voltage, tolerance=0.0)`
* Resistance `R(resistance, tolerance=0.0, alpha_ppm=None)`
* Current `I(current, tolerance=0.0)`
* Power `P(power, tolerance=0.0)`
* Factor `Factor(factor, tolerance)` (unitless factor, example below)
* squared Voltage (V²) `Usq(voltage, tolerance=0.0)`
* squared Current (A²) `Isq(voltage, tolerance=0.0)`

All classes do have the following members (example when using a voltage):

    >>> v1 = U(24, 0.04)
    >>> v1
    24.0V ± 4.0% (± 960.0mV) [23.0400 .. 24.9600]V
    >>> v1.value
    24
    >>> v1.min
    23.04
    >>> v1.max
    24.96
    >>> v1.unit
    'V'

A unit can also be created using the `.from_min_max(min, max)` classmethod when
the lower and upper limit is known (min/max):

    >>> P.from_min_max(3, 4)
    3.5W ± 14.29% (± 500.0mW) [3.0000 .. 4.0000]W

All units feature the add, subtract, multiply and divide operators. The calculation
only works if the result's type is one of the classes above:

This works because the result type is one of the known classes:

    >>> U(10) + U(20)
    30.0V
    >>> I(2e-3) - I(10e-3)
    -8.0mA
    >>> U(10) * I(2e-3)
    20.0mW
    >>> U(10) / I(2e-3)
    5.0kΩ @ 20°C
    >>> U(10) * Factor(2)
    20.0V
    >>> I(10e-3) * R(150)
    1.5V
    >>> P(200) / U(5)
    40.0A
    >>> U(3) * U(3)
    9.0V²
    >>> U(3)**2  # U squared
    9.0V²
    >>> U(3)**2 / R(1e3)
    9.0mW

This does not work because voltage divided by power is not a known class:

    >>> U / P
    Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
    TypeError: unsupported operand type(s) for /: 'type' and 'type'


