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Two Simple Formulae: capacitance and inductance

Two relationships can simplify capacitor and inductor voltage and current calculations

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    Issue: Draft 04 of 2016_11_012​

    PREFIX
    Kirchhoff's two laws and Ohms law are the corner-stones of practical electronics, and are easy to grasp. They apply to both direct and alternating voltages and currents.

    You may not even be aware of Kirchhoff's two laws, but if you have done any electronics, you will have been using them all the time. In spite of their impressive and daunting name, Kirchhoff's laws are simplicity itself, even obvious.

    Kirchhoff's first law says that, the sum of all currents flowing into a particular point in an electrical circuit, equal the sum of all the currents flowing out of the same point in the circuit.

    Kirchhoff's second law says that, the sum of the voltage drops in an electrical circuit equal the voltage (power supply) applied to the circuit.

    Ohms law is equally simple and gives the relationship, I = V/R
    Where:
    I = current in Amps
    V = Voltage in Volts
    R = resistance in Ohms

    The formula can be transposed to find any of the three variables, if you know the other two:
    I = V/R
    R = V/I
    V = I * R

    INTRODUCTION
    Kirchhoff's and Ohm's laws are easy to grasp, but the next step, the relationship between the currents and voltages with capacitors and inductors, is much more difficult.

    This article gives two simple formulae to easily derive the absolute and instantaneous rate of change (differential) of voltages and currents in capacitors and inductors without the use of calculus or the exponential function. By correct application, the formulae are also useful for making approximations where the exponential function is involved.

    CAPACITOR
    Q = CV = IT (1)
    Where:
    Q is the charge stored in the capacitor in Coulombs
    C is the capacitance in Farads
    V is the Voltage across the capacitor
    I is the current flowing into the capacitor in Amps
    T is the time, in seconds, that the current flows into the capacitor

    INDUCTOR
    The inverse (wrong word) of capacitance is inductance, so you get:
    Q = IL = VT (2)
    Where :
    Q is the charge stored in the inductor in Coulombs
    L is the inductance in Henrys
    I is the current flowing through the inductor in Amps
    V is the Voltage across the inductor
    T is the time, in seconds, that the voltage is applied to the inductor

    Of course, both (1) and (2) assume perfect components and a constant current and constant voltage respectively. They do not relate directly to the exponential voltage across a capacitor when the charging current is derived from a constant voltage via a resistor. As electronic circuits invariably use constant voltages, it is more common to see an inductor charged by a constant voltage though. In any case, the two simple relationships are very useful, if used correctly, for giving a guide to any of the variables. In other circumstances the relationships are accurate.

    EXAMPLES
    (1) You would like to specify a reservoir capacitor for a simple mains power supply.
    Where:
    Type of rectification = full wave (rectification factor =mains period/2)
    Mains frequency = 50 Hz (period = 0.02s)
    Current drain from reservoir capacitor = 3A
    Maximum acceptable ripple voltage = 1V

    From, Q = CV = IT, by deleting and transposing, you can get, C = IT/V

    Thus, C= (3A * 0.02s)/(0.7V * 2 rf) = 0.04286 F, so the closest standard value, 470mF (470,000uF), will do.
    Obviously, this is a gross simplification about rectifier/capacitor power supplies, but in practice, it surprising how close it is.

    (2) You can do similar calculations with inductors.

    For example, say you need to establish what the maximum current is in an inductor in a basic fly-back switch-mode regulator.
    Where:
    Supply line = 12V
    Switching frequency = 100Hz (period= 0.01s, half period=0.005s)
    Switching frequency mark to space ratio =1:1
    Inductance = 0.01 Henrys

    From Q = IL = VT, by deleting and transposing again, you can get, I = VT/L
    Thus I = (12V*0.005s)/(0.01H * 2) = 3A
    In this case the answer is accurate because the voltage is constant.

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