{
    "info": {
        "author": "Sebastian Werner",
        "author_email": "wese3112@startmail.com",
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        "classifiers": [
            "License :: OSI Approved :: GNU General Public License v3 (GPLv3)",
            "Operating System :: OS Independent",
            "Programming Language :: Python :: 3"
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        "description": "# eecalpy Python module\n\n[![pypi package](https://badge.fury.io/py/eecalpy.svg)](https://badge.fury.io/py/eecalpy)\n[![Build Status](https://travis-ci.org/wese3112/eecalpy.svg?branch=master)](https://travis-ci.org/wese3112/eecalpy)\n\nThe *Electrical Engineering Calculations for Python* module is a\ncollection of classes for simple to complex electrical calculations, with a\nspecial focus on handling tolerances.\n\n## Installation\n\nThe ``eecalpy`` package is available on the Python Package Index (PyPI). The package needs Python 3+, you can install it with:\n\n    $ pip install eecalpy\n\n\n## Introduction\n\nCheck out the voltage divider below. For both resistors their tolerance and the\ntemperature coefficient \u03b1 are given (\u03b1 in parts per million). \n\n![Simple voltage divider](img/vdiv.png?raw=true \"voltage divider\")\n\nLet's create two variables for them.\n\n    >>> r1 = R(resistance=1000, tolerance=0.05, alpha_ppm=250)\n    >>> r2 = R(2e3, 0.01, 100)\n    >>> r1; r2\n    1.0k\u03a9 \u00b1 5.0% (\u00b1 50.0\u03a9) [0.9500 .. 1.0500]k\u03a9 @ 20\u00b0C \u03b1=250ppm\n    2.0k\u03a9 \u00b1 1.0% (\u00b1 20.0\u03a9) [1.9800 .. 2.0200]k\u03a9 @ 20\u00b0C \u03b1=100ppm\n\nThe formula for the voltage divider factor is `r1 / (r1 + r2)`. To calculate it use `R.voltage_divider(other_resistor)`:\n\n    >>> r1.voltage_divider(r2)\n    0.33 \u00b1 4.0% [0.3199 .. 0.3465]\n\nYou can also use a shorthand notation:\n\n    >>> r1 // r2\n    0.33 \u00b1 4.0% [0.3199 .. 0.3465]\n\nAttention: Do not use the statement `r1 / (r1 + r2)` here, because it would use the tolerance limits\nof `r1` twice (addition and division) and therefore yield a false result.\n\nThe result above is an instance of the `Factor` class. Now only the voltage is missing.\nThese are created using `U(voltage, tolerance=0.0)`.\n\n\nLet's assume the input voltage is 24V with a 1% tolerance the output voltage of the\nvoltage divider then is:\n\n    >>> vin = U(24, 0.01)\n    >>> vout = r1 // r2 * vin\n    >>> vout\n    8.0V \u00b1 5.0% (\u00b1 400.0mV) [7.6000 .. 8.4000]V\n\nNote: 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:\n\n    >>> vin * (r1 // r2)\n    8.0V \u00b1 5.0% (\u00b1 400.0mV) [7.6000 .. 8.4000]V\n\nFor demonstration, let's calculate some of the voltage divider parameters.\n\nCurrent through `R1` and `R2` (to GND):\n\n    >>> vin / (r1 + r2)\n    8.01mA \u00b1 3.33% (\u00b1 266.81\u00b5A) [7.7394 .. 8.2730]mA\n\nPower dissipation of the resistors:\n\n    >>> vout**2 / r1\n    65.46mW \u00b1 21.35% (\u00b1 13.97mW) [51.4842 .. 79.4301]mW\n    >>> (vin - vout)**2 / r2\n    128.26mW \u00b1 12.3% (\u00b1 15.78mW) [112.4776 .. 144.0351]mW\n\nLet's also see how `vout` changes when the ambient temperature is 200\u00b0C:\n\n    >>> r1.at_T(200) // r2.at_T(200) * vin\n    8.14V \u00b1 4.97% (\u00b1 404.16mV) [7.7359 .. 8.5443]V\n\n`R.at_T(temperature)` is the same as `R.at_temperature(temperature)`.\nIt returns a new resistor object at the given temperature (in \u00b0C).\n\nYou can of course also use perfect values, so without the tolerance and\ntemperature coefficient:\n\n    >>> r1 = R(1e3)\n    >>> r2 = R(2e3)\n    >>> vin = U(24)\n    >>> r1; r2; vin\n    1.0k\u03a9 @ 20\u00b0C\n    2.0k\u03a9 @ 20\u00b0C\n    24.0V\n    >>> vout = r1 / (r1 + r2) * vin\n    >>> vout\n    8.0V\n\nBy the way, you can get the series resistance using `+` and the parallel\nresistance using `|`:\n\n    >>> r1 + r2\n    3.0k\u03a9 @ 20\u00b0C\n    >>> r1 | r2\n    666.67\u03a9 @ 20\u00b0C\n    >>> r1 | (R(5e3) + R(3e3)) | r2  # complex statements allowed!\n    615.38\u03a9 @ 20\u00b0C\n\n## Classes\n\nThe available classes are:\n\n* Voltage `U(voltage, tolerance=0.0)`\n* Resistance `R(resistance, tolerance=0.0, alpha_ppm=None)`\n* Current `I(current, tolerance=0.0)`\n* Power `P(power, tolerance=0.0)`\n* Factor `Factor(factor, tolerance)` (unitless factor, example below)\n* squared Voltage (V\u00b2) `Usq(voltage, tolerance=0.0)`\n* squared Current (A\u00b2) `Isq(voltage, tolerance=0.0)`\n\nAll classes do have the following members (example when using a voltage):\n\n    >>> v1 = U(24, 0.04)\n    >>> v1\n    24.0V \u00b1 4.0% (\u00b1 960.0mV) [23.0400 .. 24.9600]V\n    >>> v1.value\n    24\n    >>> v1.min\n    23.04\n    >>> v1.max\n    24.96\n    >>> v1.unit\n    'V'\n\nA unit can also be created using the `.from_min_max(min, max)` classmethod when\nthe lower and upper limit is known (min/max):\n\n    >>> P.from_min_max(3, 4)\n    3.5W \u00b1 14.29% (\u00b1 500.0mW) [3.0000 .. 4.0000]W\n\nAll units feature the add, subtract, multiply and divide operators. The calculation\nonly works if the result's type is one of the classes above:\n\nThis works because the result type is one of the known classes:\n\n    >>> U(10) + U(20)\n    30.0V\n    >>> I(2e-3) - I(10e-3)\n    -8.0mA\n    >>> U(10) * I(2e-3)\n    20.0mW\n    >>> U(10) / I(2e-3)\n    5.0k\u03a9 @ 20\u00b0C\n    >>> U(10) * Factor(2)\n    20.0V\n    >>> I(10e-3) * R(150)\n    1.5V\n    >>> P(200) / U(5)\n    40.0A\n    >>> U(3) * U(3)\n    9.0V\u00b2\n    >>> U(3)**2  # U squared\n    9.0V\u00b2\n    >>> U(3)**2 / R(1e3)\n    9.0mW\n\nThis does not work because voltage divided by power is not a known class:\n\n    >>> U / P\n    Traceback (most recent call last):\n    File \"<stdin>\", line 1, in <module>\n    TypeError: unsupported operand type(s) for /: 'type' and 'type'\n\n\n",
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